Introduction to Algebraic Topology [1 + corrections ed.]

Table of contents :
Title
Forewords
Table of Contents
1. Homology Groups of Chain Complexes
2. Affine Spaces
3. Affine Simplices and Boundary Operator
4. The Singular Homology Theory
5. Homotopy Properties of Homology Groups
6. The Excision Theorem
7. Direct Decomposition and Additional Aids to the Computation of Homology Groups
8. The Tensor Product
9. The Functor Hom
10. Categories and Functors
11. Categories, Functors, and the Singular Theory
12. Axioms for Homology and Cohomology
13. Mayer-Vietoris Sequence
14. The Jordan-Brouwer Separation Theorems
15. Finite Cell Complexes
16. Betti Numbers and the Euler Characteristic
17. Complex and Real Projective Spaces
18. Maps of S^n on S^n and Lens Spaces
19. Classification of Surfaces
20. Singular Cup Products
21. The Singular Cap Product
22. The Anticommutativity of the Cup Product
Index
Index of Symbols
Errata

Citation preview

INTRODUCTION TO ALGEBRAIC TOPOLOGY

MERRILL RESEARCH AND LECTURE SERIES Erwin Kleinfeld, Editor

INTRODUCTION TO ALGEBRAIC TOPOLOGY EMIL ARTIN and HEL BRAUN

University of Hamburg

Translated from the notes of Armin Thedy and H61 Braun

by

ERIK HEMMINGSEN Syracuse University

CHARLES E. MERRILL PUBLISHING COMPANY A Bell & Howell Company

Columbus, Ohio

Copyright © 1969 by CHARLES E. MERRILL PUBLISHING Co. ,

Columbus, Ohio. All rights reserved. No part of this book may be reproduced in any form, electronic or mechanical,

including photocopy, recording, or any information storage and retrieval system, without permission in writing from the

publisher.

Library of Congress Catalog Card Number: 69—11166 Standard Book Number: 675—096510—0 AMS Classification Number: 5501

PRINTED IN THE UNITED STATES or AMERICA 12345678910—7372717069

Foreword to the German Edition Professor Emil Artin lectured on algebraic topology during the winter semester of 1959—60. Later, in the summer of 1962 and the winter semester of 1962—63, Professor Hel Braun gave a more ex-

tensive course on the same subject in which she made her preparation in consultation with Professor Artin. The present volume is based on lecture notes of Professor Hel Braun, which she very kindly placed at my disposal. I wish to express my particular gratitude for this. Armin Thedy

Foreword to the English Edition The first nineteen sections of this volume are a revised translation of the German edition. The last three sections are new and are taken from the notes of Hel Braun. Basic changes have been avoided in the hope of preserving, as much as possible, the flavor of Artin’s

original presentation. The reader is expected to have some familiarity with the elements of point set topology and an introduction to the theory of groups, rings, and vector spaces. However, the book is remarkably selfcontained. Proofs are given of many of the algebraic and topological theorems that are central to the subjects discussed.

The early sections introduce exact sequences of modules, the algebraic structure of homology groups of chain complexes, and the geometry of affine spaces and simplices. The authors then proceed directly to the discussion of singular homology, for which the theorems corresponding to the Eilenberg-Steenrod axioms are

proved. The middle third of the book is devoted to tensor products, categories, functors, axiomatic homology theory, and the MayerVietoris Theorem. The last third contains a number of geometrical V

vi

Forward

applications (the Jordan-Brouwer separation theorem, finite cell complexes, betti numbers and euler characteristics, projective spaces, degree of mappings of spheres, lens spaces, and the classification of compact separable two—manifolds) together with an introduction to cup and cap products. However, not all applications

occur at the end of the book. From the very early pages, there are extensive geometrical applications which serve to give the reader a good intuitive grasp of the meaning and effectiveness of the tools that have been developed. Erik Hemmingsen

Table of Contents Chapter 1

Homology Groups of Chain Complexes Chapter 2

Afline Spaces

10

Chapter 3

Affine Simplices and Boundary Operator

17

Chapter 4

The Singular Homology Theory

27

Chapter 5

38

Homotopy Properties of Homology Groups Geometrical Consequences of the Homotopy Theorem

Application to Graphs

50 53

Chapter 6 The Excision Theorem

57

Chapter 7

Direct Decomposition and Additional Aids to the Computation of Homology Groups Computation of H0(X, A) Homology Groups of a Point P The Sphere vii

70 72 73 75

Table of Contents

viii Homology Groups of Graphs

80

Degree of a Function f:S’I —> S’'

81

Chapter 8

85

The Tensor Product 91

Tensor Products of Functions

Chapter 9

98

The Functor Hom 101 103

R, S—Modules Quotient Modules

Chapter 10

107

Categories and Functors Chapter 11

115

Categories, Functors, and the Singular Theory The Functors ® M and Horn (

,M) in the

Singular Theory Homology Cohomology Application of Homology to Function Theory

118 119 120 122

Chapter 12

Axioms for Homology and Cohomology

129

Chapter 13

Mayer-Vietoris Sequence

137

Preliminary Considerations with Respect to Modules 137 Chapter 14

The Jordan-Brouwer Separation Theorems

149

Table of Contents

ix

Chapter 15

Finite Cell Complexes Spherical Complexes Products of Spherical Complexes

158 164 166

Chapter 16

169

Betti Numbers and the Euler Characteristics Chapter 17

175

Complex and Real Projective Spaces Groups of the Complex Projective Space CP" Groups of the Real Projective Space P"

177 178

Chapter 18

182

Maps of S" on S" and Lens Spaces Application to Lens Spaces

187

Chapter 19

193

Classification of Surfaces Simplicial Complexes

194

Chapter 20

Singular Cup Products

201

Chapter 21

The Singular Cap Product

209

Chapter 22

The Anticommutativity of the Cup Product

216

Index

225

Homology Groups of Chain Complexes

Let R be a commutative and associative ring with unit element

1. By a (unitary) R—module is meant an additively written, abelian group A, for which a product R X A ——> A is defined such that r(a + a’) = ra + ra’,

r(r'a) = (rr')a,

(r + r’)a = ra + r’a

r, r’ E R

1a = a

a, a' E A.

Let A, B be R-modules. A function f:A ——> B of A into B is called

an R-homomorphism (or merely a homomorphism, if it is clear which R is under discussion) if and only iff(a + a’) = f(a) + f(a’) and f(ra) = rf(a) hold for a, a’ e A, and r e R. The set of all a E A with f(a) = 0 is called the kernel off and designated by

KerU) = f“(0).

Introduction to Algebraic Topology

2

The set f(A) is designated by Im(f). The homomorphism f:A —~> B is called an epimorphism if and only iff(A) = B. It is called

a monomorphism if and only if Ker f = 0, and it is called an isomorphism if and only if it is both an epimorphism and a monomorphism. The kernel of f:A —> B is a submodule of A. If C is a submodule of A, then A/C is made into an R-module, the factor module of A mod C, by defining the sum and product operations in terms of representatives :

(a+C)+(a'+C)=(a+a’)C,

r(a+C)=ra+C.

To each f:A ——» B corresponds the canonical decomposition f = fififi, where I. fozA —> A/f"(0) is defined by f.,(a) = a +f“(0) and is an epimorphism,

II. f,:A/f ”(0) ——>f(A) is defined by f,(a + f“(0)) =f(a) and is an isomorphism, III. f2:f(A) ——> B is defined by f2(f(a)) =f(a) and is a monomerphism.

It is easy to prove:

1.1. Lemma: Let f:A ———> B be a homomorphism of R-modules, C a submodule ofA, and D a submodule ofB. Then f induces a homomorphism, f:A/C-—>B/D with f(a + C) =f(a) + D if and only if f(C) C D. The set of all homomorphisms of A in B is denoted by HomR (A, B) or by Hom(A, B). Hom(A, B) becomes an R-module when

the definitions

(f +f')(a) =f(a) +f’(a). (W = rf0!) are made for allf,f’ e Hom(A, B), a e A, and r e R. Let Z be the set of integers. A collection of R-modules Aa and R-homomorphisms jg, q e Z such that fur fl ...—)Aq+l—>Aq—>Aq_l—>...

is called a sequence. Here, (1 can run through all the integers, or else the sequence can terminate on the right, on the left, or on both sides.

It is also possible for the arrows all to be directed from right to left.

Homology Groups of Chain Complexes

3

A sequence is called exact at the location q if and only if Im(f,,+1) =fi,+1(AM) =f;‘(0) = Ker(fi,). A sequence is called exact if and only if it is exact at all locations having one arrow leading in and one arrow leading out. Exactness at the location AG is illustrated schematically by Figure 1. A,

Aq+1

AP.

Figure 1

To prove exactness at the location Aq, the following must be established: (1) fq+,(AqH) c Ker(f,,); that is,f.,fi,+1 = 0; and (2) f“,(Aq+,) c: Ker(13,); that is, to each an e A“, for which fa(aq) = 0, there is an a“, e A"+1 for whichfa+1(aa+,) = ad.

The module consisting of the zero element only is designated by 0. By 0 ——> A is to be understood that injection of the zero module into A for which the image is the zero element of A. Furthermore, B ——> 0 designates the function under which all elements of B

have as image the element 0. The following special cases of exactness are important: 1

(l)

0 —> A ——> B exact means exactness at A; that is, that Ker(i) = 0 and that iis a monomorphism.

(2)

A —> B ——> 0 exact means exactness at B; that is, that

J

j(A) = B and that j is an epimorphism. (3) 0 ——> A —-> 0 exact means A = 0. t

(4)

0 ——> A ——> B —> 0 exact means that i is an isomorphism.

(5)

0 —> A —t—+ B —j—> C ——> 0 exact means that i is a monomorphism, j is an epimorphism, and that Im(i) = Ker j. Thus,

C is isomorphic to B/Im(i).

Introduction to Algebraic Topology

4

A sequence ~~ —> A,“ L A, 1» A,,_l ——> ~- in which 3G3“; = 0 for all q is called a chain complex. The elements of a;1(0) = Ker(3q) are called cycles, the elements of 84,104,“) = Iméiq+1 are called boundaries, and 8,1,1“,+ l) c Ker(aq). In most

cases the subscript q will be omitted from an, and the symbol a will be called a boundary operator. If A is a chain complex, then the departure from exactness at the qth location is measured by the factor module Hq(A) = Ker aq/6q+1(Aq+1) = Keraq/Iméiq+l of cycles modulo boundaries. The R-module H.1(A) is called the qth homology group of A. By a homomorphism g:A —> B of a chain complex A into a chain complex B is to be understood a sequence of R—homomorphisms gq:Aq —> Be for which g3 = 8g; that is, for which 6q = g,,_,a,, for all q e Z. For g:A —>B to be a homomorphism of chain complexes thus means the existence of a diagram ---—>AaH—>Aa-—>Aq_l——>...

1a..

in

la.-

-——>B

—->B,,—>Bq_1—>--- s Q+l

in which the condition, 3n = gq_,3,, for all q, is described in words by saying that the diagram is commutative at each of its rectangles. Each homomorphism g:A —-> B of chain complexes induces homomorphisms g“: Hq(A) ——> Hq(B) which are defined as

follows: For aq + 304“.) e Hq(A),

gum + 3(Aq+1)) = gq(aq) + 3(Bq+l)' The right-hand side is an element of H«(3), since

33001.1) = sweat = 3a-:(0) = 0By Lemma 1.1, the function g“ is well defined since

gq*3(Aq+l) = aq+1(Aa+l) C 33m. The index q is usually omitted from g" to yield the notation

g* :Hq(A) —> Haw).

Consider the diagram 0 —> A —‘—> B —’—> C —> 0 of chain

Homology Groups of Chain Complexes

5

complexes and of chain homomorphisms i, j. Written in detail, this becomes

where the diagram commutes at each square and a is a boundary operator; that is, 33 = 0.

t j The sequence 0 ——> A —> B —> C —> 0 of chain complexes is called exact if and only if each of the sequences 0 ——> A“ ‘—'> Ba 1* C7 —> 0 is exact. The homomorphisms i*:Ha(A) —’ Ha(B) and j* :Ha(B) ——> Hq(C) that correspond to iand j have already been defined. Now let

0 —> A —> B —> C —> 0 be an exact sequence. Then a set of homeomorphisms 60*:H0(C) —> Hq_,(A) is defined as follows: The elements of Hq(C) have the form ca + ace“, where 8cq = 0. Since jg is an epimorphism, there is a ba 6 B‘, for which jq(bq) = ca. From now on, i will be written instead of I}, and j instead of jg. On account of j(8b,,) = 3(jbq) = Bea = 0 and the exactness at the qth location, there is an aq_, in A".l for which 1210.] = 3b,,. I

ba_—)Cq aq-i ‘_’ aba

The element 3”,.‘(cq + a) is defined to be (1.1-1 + 3A“. This is in Hq_,(A) because ii)a,,_l = 31%-, = Bab” = 0 and thus, since i is

Introduction to Algebraic Topology

6

a monomorphism, 3a,.l = 0. The symbol 3,, will be written in place of a,,.. Now it must be proved that 3* is well defined: Let c; = c, + 8c,” Then there is a b,“ e B,+1 for which q,+1 = jq and C"; = Cu + 31'q =j(bq + 3174+!)-

Since Kerja = i(A.,), the most general bf, e B, for which jb; = c; has the form b; = b, + 3b,,+1 + ia, for a suitable a, 6 Ag. Then

3b; = 3b,, + aid, = ia.H + iaaq = i(aq_l + 3a,).

Since i is a monomorphism, it follows that 8*(c; + 3cm) = a‘,_l + 3a., + 8A, = a“.1 + 8A“. Therefore, 8* is well defined. It is trivial that 3*:HQ(C) —+ Hq_,(A) is a homomorphism, since all calculations were linear.

1.2. Lemma: Let A, B, C be chain complexes. Let f:A —+ B and g:B —> C be chain homomorphisms. Then gf is also a chain homomorphism, and (gf)* = g*f*. Let f,g: A ——> B be chain homomorphisms. Thenf + g is a chain homomorphism, and (f + g)* = f* + g," Furthermore, R1,, = Id and 0* = 0.

Proof:

Lemma 2 follows immediately from the definition of *.

1.3. THEOREM: If 0 —+ A —‘+ B —’—> C ——> o is an exact sequence of chain complexes, then i.

J.

' - ——> How) —> Ha(B) —* Ha(C) a. (HHa—(A) —-> H 4(3) —-> - -is an exact sequence of R-modules. Proof: The two conditions for exactness required in the definition will now be established. (1) From Lemma 2, j*i* = (fl), = 0,. = 0. (2) Let b, + 33w 6 Hq(B). Then

ab, = 0 and j*(b, + 33,“) = j(b,) + ac“. In accordance with the definition of 3*, the element b, is chosen

Homology Groups of Chain Complexes

7

as an inverse image of j(bq), and an element add E A”.l is found for which i(aq-1) = 3b,. Hence, i(aa_l) = 0. Since i is a monomorphism, it follows that aw = 0. Therefore, 3*j*(ba + 830+!) = a*(j(ba) + ace“) = “(1—1 + aAq = 0 + aAq-

(3)

This means that 3*}; = 0. Let ca = EC“. 6 Hq(C). If Ca = j(bq) and 6b. = i(aq_,), then 3*(cq + ace“) = aw + 3A.. The image of this under 1} is consequently

i(a,,_.) + 38,, = 3b. + as, = 33,. This is the zero element of Hq_1(B), and therefore, i*3* = 0. Thus, it has just been proved that the sequence is a chain

(4)

complex. Let (1" + 3A.,“ 6 Ker(i*); that is, 3(aa) = 0. Then i(a¢) = 3b“, for some be+1 6 Ba“. Let ca+1 =j(bq+1). Then, aka“) = aj(b¢:+1) =ja(bq+l) =ji(aa) = 0 since ji = 0. Therefore, ca+1 + 6G,,+2 e Hq+1(C) and 3:4c + vH) = aq + aAaH-

(5)

Let bq + 83¢+1 e Ker(j*); that is, j(bq) + 8C0“ = 0 and 8b = 0. Then j(bq) = 36.,“ for some ca“ 6 Ca“. Since j is an epimorphism, there is a bu.” for which j(b,,+1) = can. Let b; = b,, —— 3b,“. Then, Kb» =j(ba) "ja(bq+l) =j(ba) _ 36.,“ =j(ba) _j(bq) = 0'

Consequently, there exists an ad for which i(aq) = b;. From

i301a) = 3i(aa) = 3011.) = 3(bq) = 0 it follows that 3(aa) = 0, since i is a monomorphism. Hence, a, + 3Aq+1 E Hq(A). Then, i*(aq + 3/141“) = i(aq) + 33“, = bi + aBa“ = be + aBa+l;

(6)

in other words, Ker(j*) c Im(i*). Let ca + 6C“, 6 Ker(a*); that is, 3*(c. + ace“) = 0 and 8cq = 0. From the choice of bu and ab, such that ca = j(ba) and 3(bq) = i(aq_1) there fOIIOWs

Introduction to Algebraic Topology

8

8*(cq + BCqH) = a“; + 3A. = 0. Hence, there is an aa 6 Ag for which a._1 = aaa. For bf, = be — i(aq) it follows, therefore, that

3073) = a(bu) — 311%) = 30%) - Kilo—x) = 0Hence, b; + 334+; e Hq(B). Furthermore,

J'*(b.', ‘1' 334m) =J'(b£) + 3C“, =j(ba) _ji(a41) + v+1 = on + v-H-

This completes the proof of Theorem 1.3. The following theorem is used in connection with Theorem 1.3. 1.4. THEOREM:

In the diagram of chain complexes 1

0——>A —:—>B —>C —>0

1, . is , l»

0——>A'——>B’——>C’—>0 let the two rows be exact, and let each square be commutative; then in the diagram of R—modules ' _’

1. 1. a. l. 4(A)—’HH(B) _’Hq(C) _’Ha-1(A)“’ Ha—1(B) —’ ' ' ' I.

E.

’l’.

' —’

h.

f.

1'.

a.

30

i;

«(AI) _’ Hq(Bl) —’ H0(CI) _' Ha—I(AI) —’ Hq—I(BI) —" ' ' '

each square commutes. Proof: Theorem 1.3 is not used here. (1) From i’f= gi and Lemma 1.2 follows (Ff). = (gi)* = 11f, = g*i*. The commutativity of the second square follows in the same way. (2) In the completion of the proof, there remains the commutativity of 3.

Ha(C ) —* Ha-1(A ) n.

I.

How) 1» H -. 0. By “the center of gravity of the points a, with mass distribution ,u‘” is meant the point S = SLOW/Jim. Here [Lg/[It 2 0 and 2 (la/p) = 1. Clearly, the convex set of the example is just the set of the centers of gravity obtained from all possible mass distributions. 2.1 Lemma: Let Ybe a convex set and a" a2, - ‘ -, a, G Y. Then each center ofgravity ofa1, a2, - - - , a, is a point of Y.

The proof is carried out by induction on r. The case r = 1 is trivial. Suppose now that the lemma has been proved for r — 1. Let pl, pt}, - - -, p, be a mass distribution with at least one non-zero ’14, say, ,u, > 0. Let s’ be the center of gravity of a,, a2, - - -, a,_l with the mass distribution ,uq, pa, - - -, [1.7-1, for which FI’=/‘1+/‘z+'”+/llr—1-

Then s = (#’/#)S’ + (er/M17, or s = tS’ + (1 - Oar, where t= ,u.’/;L and 0 g t g 1. By inductive hypothesis, s’ e Y; and hence, by the definition of convexity, s e Y. By the convex hull of a set X c E is meant the smallest convex subset of E that contains X. Its existence is trivial since the intersection of convex sets is convex and E is convex. 2.2. THEOREM:

The convex hull of r points al, a2, - - -, a, of E

consists precisely of all the points of the form x = 25m, where 2E4=1and§,20.

Introduction to Algebraic Topology

14

This set is convex; and, according to Lemma 3, is conProof: tained in every convex set that contains the points a,, a,, - - -, a,.

The convex hull of r + 1 independent points is called a geometrical simplex of dimension r. The convex hull of r + 1 dependent points is called a singular geometrical simplex. The r + 1 points are called vertices of the simplex. Let E, E’ be afline spaces and let V, V' be the corresponding vector spaces. The vectors (P, Q) and 4>(P', Q’) will be written (P, Q) and (P', Q’) from now on.

Definition: A function f:E ——>E’ is called afi‘ine (or an affine map) if and only if it has the two following properties:

(1) The (2)

equation (P, Q) = (P1, Q1) implies (f(P). f(Q)) =

(f(Pl), f(Q.» and

~

~

The induced mapping f :V—-> V’ defined by f(P, Q) =

(f(P), f(Q)) is a vector space homomorphism.

By (1), the function f is well defined. Actually, it is possible to formulate the definition of an affine map somewhat more weakly. For instance, it is sufficient, in place of (2), to require only that f~(.f- x) = Effie) for x e V and real 5. For

f(x + y) = f(x) + f(y) is obtained immediately from (1) as follows: There exists R E E with x = (P, Q), y = (Q, R), yielding x + y = (P, R). Hence,

706 + y) = f(P, R) = (f(P),f(R)) ~ = (f(P),f(Q)) + (f(Q),f(R)) = f(X) + f(y)Obviously, iff:E -—> E’ and n’ —> E” are afline maps, then gf:E ——> E” is an affine map. Let P0, P1, - - -, P,, P, 0 he points of E and let 0P=2I=o 54(01") 6

V, where E E, = 1. Since f is a homomorphism, it follows that

«(0), 1‘0» = mm = x more = g. gamma». For an arbitrary 0’ from E’,

(03f(1’)) = (0mm) + «(0), f(P))

= £30540” f(0» + gamma»,

and thus,

(021m) = g 540mm).

Afline Spaces

15

Let x0, x1, . - -, x,, x be the coordinate vectors of P0, P1, - - -, P" P

with respect to 0 e E and let f(xo), f(xl), --~, (x,), f(x) be the coordinate vectors of f(Po), f(P,), - - -, f(P,) and f(P) with respect to 0’ e E’. This proves

2.3.

Let x = 2L, Em, where 2.5. = 1. Then this relation holds

with the same E. for each choice of O and implies the relation f(x)

= 2 51f(x.) for any choice of 0'. From this follows immediately

2.4. THEOREM: The afi‘ine image of an r-dimensional geometrical simplex is a (possibly singular) geometrical simplex. In the case r = 2, statement 2.3 becomes

2.5.

f(tx + (1 — t)y) = tf(x) + (1 — t)f(y) for z real.

Thus, 2.5 is a consequence of (l) and (2) of the definition of afiine. To prove the converse, let f:E —> E’ be a function which satisfies 2.5 for some choice of coordinates in E and E’. Here, x and y denote the coordinate vectors of points P and Q of E, and f(x) and

f(y) are the coordinate vectors of their images. Then (P, Q) = y — x, and similarly (P,, Q,) = y1 — x1. Proof of (1): From (P, Q) = (P1, Q1); that is, y — x = y1 — x1, it follows that Jix + iy, =

%x1 + 1}y. From 2.5, it follows that

1‘rf(x) + %f(.v1) =f(%x + H1) = %f(M) + %f(y). and hence, that f(y) —— f(x) =f(y.) —f(x,). Thus, (1) is proved. Furthermore, the function f” :V——> V’, given by

f(P, Q) = (f(P),f(Q)) =f(y) —f(X) is well defined. For (2), it suffices to prove that f(éx) = Eflx). From

f(EX) =f(Ex) —f(0) = Ef(x) + (l - «El/"(0) —f(0) = Ef(X)

-

2.5, for y = 0 e V, it follows that

£f(0) = 57(36)-

Condition 2.5 implies that the afline image of a convex subset of an afline space is convex, for the line segment between x and y has as affine image the line segment between f(x) and f(y).

2.6. THEOREM:

Let E and E' be afi‘ine spaces, with dim E = q.

Introduction to Algebraic Topology

16

Let a- be a q—dimensional simplex having vertices P, with coordinate vectors at, i= 0,], - - -, q. Let P2, 1': 0,1, - - -, q, be points ofE'

whose corresponding coordinate vectors are b,. Then there is precisely one afi‘ine map f:E —> E’ for which f(P,) = P,’. Proof:

Assume that f is affine, and that f(P,) = P1, i = 0, 1, - - ~,

q. Each coordinate vector x has a unique representation x = 2 an; with real coefficients 5; for which 2 E: = 1. By 2.3, it follows that f(x) = E af(at). Thus, f(x) is uniquely determined by the vectors f(a‘). To prove the existence of f, it is therefore suflicient to let f(x) = 2 ab. and to prove 2.5. From ”5 +(1— 0y = ’2 £1“: +(1— t): 77:“; = 2 (”:4 + (1 —‘ 077031: it follows that

N): + (1 — tly) = 2 (IE. + (l — t)7/:)b. = IEEtbt +(1— 0271:”:

= f(x) + (l — t)f(y)The affine map f just defined sends P; on P; for each i, as required. Since the Si are preserved, the image of a under f is the

possibly singular simplex whose vertices are Pi.

Affine Simplices and Boundary Operator

A few preparatory remarks are needed so that a suitable formalism can be set up to handle affine mappings of geometrical

simplices. For this purpose, it should be remembered that R is a fixed ring with unit element, and that all R-modules are unitary. Let I be an index set and Mt, i e I, be R-modules. The set of all functions on I with values f(i) E M‘, where i E I, is called the direct product of the modules Mi; it is denoted by M = H Mi. The elements of M are written x = (~ - ~, xi, - - .), M becomes an R-module upon the introduction of the operations x + y = (- - -,x, +y¢, - - -)and rx=(- - -,rx‘, -- -), where r e R. The set of all (- - -, xi, - - -) in which xi=0 for all but a finite set of indices 1' is a submodule S of M called the direct sum of the modules M1, and is 17

Introduction to Algebraic Topology

18

written S = 2 M1 or S = 2‘9 ML. The elements x, and (0, - - -, 0, x., 0, - - -, 0) are identified. Consequently, x = E x‘ for x E S. When I is finite, S = M. Otherwise, M has a higher cardinality than S.

Definition:

By a generating system of an R-module A is meant

a subset A0 of A such that each x e A can be written in the form

x = Z rad, with r“ e R and all but a finite set of r“ zero. acA.

If each x e A has only one such representation, A0 is called a basis of A. A module which has a basis is called free. That not all modules are free can be seen from the example of an arbitrary finite

group taken as a Z-module. If A is a free R-module with basis A0 and a 6 A0, then Rd and

R are isomorphic as modules because they are free. Here, A is isomorphic to the direct sum of a set of submodules, each isomorphic to R, which are in number the cardinality of A0. A can be identified

with this sum and written

Conversely, from each set A0, a free module A can be constructed with A0 as basis: To each a E Ao associate a module R. isomorphic to R, and define A as the direct sum of all R“, (x 6 A0. Designate by

rd the image of r e R under the isomorphism of R on R“. Then not = “a if and only if r1 = r2. Furthermore, (rl + r,)a = net + rad; r1(2a) = (raga; and A consists of all finite sums Enhaa.

A is called the R-module generated by the basis A. It is trivial to prove 3.1a. THEOREM: Let A = 2am. Ra be a free R-module with basis A0. Let {b,} be a subset of an R-module B indexed by A0. Then there is precisely one R-homomorphism f:A —> B for which f(a) = bu for a 6 A0. Let A, B, and C be arbitrary R-modules. A function s X B

—> C is called bilinear if and only if

f(a + a" b) =f(a, 17) +f(a', b) f(a. b + b') =f(a. b) +f(a, 12')

Afline Simplices and Boundary Operator

19

and

fifth b) = rf(a, b) =f(a, rb) for a e A, b e B, r e R. If f is interpreted as a composition and f(a, b) is written as the product ab, then bilinearity implies distributivity from both sides and the homogeneity condition

(ra)b = r(ab) = a(rb). It is trivial to prove 3.1b. THEOREM:

Let A, B be free R—modules with bases A0 and

Bo, and let C be an arbitrary R-module. Let {cw} be a subset of C indexed by A0 x 3,. Then there is a unique bilinear function f:A x B——> C such thatf(a,,3) = ”fora 6 A0, ,8 6 B.,. Let A, B, and C be free R-modules for whose elements there are defined distributive and homogeneous products ab, bc, (ab)c, a(bc) [with values in certain modules, not of any interest at the moment]. Then it is easy to see that the associative law a(bc) = (ab)c holds precisely when it is satisfied by the basis elements. In this volume, the composition will always be of the following sort: Let X, Y, and Z be sets. Let the free module A have as basis certain functions sending Y in Z, and let the free module B have as basis certain functions sending X in Y. The successive ap-

plication of these functions defines a composition ab Whose values lie in a module generated by certain functions sending X in Z. The associative law holds for functions and, therefore, for this composition.

Special symbols are used for particular choices of the spaces and functions: When X and Y are topological spaces, then the free module generated by all continuous maps of X in Yis denoted by

C(X, Y). If X or Yis empty, C(X, Y) is defined to be 0. Let X and Y be convex subsets of affine spaces. A map of X in

Y is called afi‘ine if and only if it is obtained from an affine map of the affine spaces by restricting the domain to X and the image to Y. A map f:X —> Y is thus afline precisely when it satisfies 2.5 of Chapter 2. The set of afl‘ine maps of X in Y will be denoted by A°(X, Y). The free R-module generated by all afiine maps of X in Y

will be denoted by A(X, Y); it has An(X, Y) as a basis. If X or Yis empty, A(X, Y) = 0 by definition, as before. At the end of this

Introduction to Algebraic Topology

20

chapter the affine spaces will be topologized. Thereby, A(X, Y) will become a submodule of C(X, Y).

Consider an affine countably infinite dimensional space whose corresponding vector space is generated by the vectors do, d1, - - -. , dq (more precisely, the points having coThe points do, d1, ordinate vectors d1, d2,

, d, with respect to the arbitrary fixed

point 0) span a q-dimensional subspace whose points have the coordinate vectors x = 29:, ad” where 2 E, = l. The convex hull of the points do, all, - . - , d, is a q-dimensional geometrical simplex;

it will be denoted by A“. It has been proved that there is a one-to-one correspondence between the points of A, and their barycentric coordinates (En, .51, ~ . -, Ea), where E, 20 and 25, =1. This correspondence was

independent of the choice of 0. Therefore, A, can also be defined as follows: Consider an arbitrary set of q + 1 objects {(10, d1, - - - , dc}

and all the maps of this set in the real numbers such that, for the images 5, of L1,, both Er 2 0 and E E, = 1. The individual maps are called points of Aq, and A, is the collection of all its points. Let E be a convex subset of an afline space and let (1,, a,, - ' . , a, be points of E. The afline mapsa —> E for which f(d,) = 11,, for i = O, 1, - - - , q, is unique by Theorem 2.6 and will be denoted by (a0, a1, - - - ,

an). The map (a0, a,, - - - , aq) will be called the afiine simplex of E with the (ordered) vertices a0, (1,, ~ . ~ , (1,. By Theorem 2.6, the image of A. under (a0, a,, - - - , ac) is the (possibly singular) geometrical simplex with the vertices an, a“ - - -, a,. This proves 3.2. THEOREM:

The afline simplices of E form a basis for the

free R-module A(Aa, E). So far, A(Aq, E) has been defined as a free module for q = 0, l, 2, - - -. Purely formally, the symbol A(A," E) is defined, for q = —-l, as the free module with a basis consisting of precisely one element, which is to be denoted by (.). Thus, A(A_l, E) = R(.). The element (.) is called the empty simplex. Now, two homomorphisms will be introduced that are used in all auxiliary constructions but not in the principal results. Unnecessary parentheses will be omitted when they are clear from the context.

Let E and F be convex subsets of affine spaces and p e A0(E, F); that is, p is an affine mapping of E in F. The image of x e E under

Affine Simplices and Boundary Operator

21

p will be denoted by x" = p(x). For q2 0 and (ao,a,, ---,a,,) e A0(Aq, E), this means that P(aoa at: ' ‘ ’, aq) = (03: 0?, ' ' '2 gap) 6 A0(AQ) F)‘

For q = —1, p (.) is defined to be (.). Thus, there is a composition between the basis elements of A(E, F) and those of A(Aq, E), with values in A(Aq, F). This composition is unique by Theorem 3.1b. The map p can also be interpreted as a module homomorphism p:A(Aq, E) ——> A(Aq, F). By Theorem 3.1a, each x e X can be interpreted as a homomorphism of A(Aq, E) in A(AQH, E) by means of the definition x(a0’ah"'saq)=(xaa0aah"’)aq)

forqzo

for q= —1.

x(.)=(x)

Simple propertics of these two homomorphisms will now be found:

(1)

For p e A0(E, F), px = xpp. Here,

A(Aq, E) —‘> Am“, E) i» A(AM E)

A(Aq, E) —’> A(A.,+1,E) —’> A(Aq+1,E). It is sufficient to show that the image of a basis element is the same in both cases. Clearly, Px(a0) an ' ' '3 aa) 2 (xpa all, ' ‘ ‘a a5) = xPP(aoa al: ' ' '9 ac)-

(2)

Let d) be a homomorphism of A(Aq, E) into an arbitrary module B. Then x is a homomorphism of A(Aq_1, E) into B by means of z

4)

A(All-l: E) ——> A(Aq’ E) '—') B-

This leads to

3.3. Lemma: Proof:

The relation ¢x = 0, for all x, implies that 4): 0.

By Theorems 3.1a and 3.2, it suffices to prove that (Mao,

a“ - - -, aq) = 0 for all affine simplices of E. This follows from

Introduction to Algebraic Topology

22

(#010, at, ' ’ ‘t ac) = (#00011, ' ' ': at!) = 09

which, in turn, holds since it is true for all a0 6 E.

3.4. Lemma: Let E and F be convex subsets of afi‘ine spaces. Let g,:A(Aq, E) -—> A(A,,, E), where p 2 0, and h:A(A¢, E) —> A(A,,H,f), where p 2 ——1, be homomorphisms, where the homo-

morphism g3 can vary with x. Furthermore, let A be an afi‘ine map of E into F. Then there is exactly one homomorphism f:A(Aq+n E) —> A(A¢+,, F) such thatfx = h + x" :for aIIx e E. Proof:

By Theorem 3.1a, it suffices to define f on a basis of

A(Aa+1, 13) Let f(ao, an ‘ ' 's (1“,) = Mar, ‘12, ' ' 'a aq+1)+ aigaxau an: ' ‘ '9 ace-1)-

Then, for each basis element (a0, a1, - ' -, a,,) of A(Aq, E) and each

x e E, the definition implies that fz(a0’ at, ' ‘ ‘9 “(1) =f(x) an, ' ‘ ‘s an) = h(a0’ ' ' 'i an) + xlg1(a0’ ' ' '1 “0)

Since this holds for each affine simplex, it follows that fx =

h + x‘gz. If there were an fl with fix = h + x‘g," then (f—fi) = 0, for all x e E. By Lemma 3.3, this would imply that fl = f, so that

f is uniquely determined. The Boundary Operator 3,: A homomorphism 3,:A(Aq, E) ——> A(A.,_1, E) will be defined recursively, for q 2 0, as follows:

30x = Identity on A(A_,, E), and

3.5. aq + x34 = Identity on A(A.,, E) for q 2 0. This definition is consistent since 6'

1

A(Aq, E) —> A(A.,_,, E) -—+ A(A¢, E) and

A(A., E) i» Am“. E) 1‘» «Ac, :3).

Affine Simplices and Boundary Operator

23

Suppose that 6, has already been found and is unique. From Lemma 3.4 it follows, for q = p + l, x = Identity, h = Identity, and g2 = —8,,, that am: Ame“, E) —> A(A,, E) is a uniquely determined homomorphism. It is worth knowing a, explicitly, particularly for the basis elements of A(Aq, E):

For q = 0, 3.,(ao) = 301M.) = Id(.) = (.).

= (a1)

-

For 4 = 1’ a1(“0: at) = ala0(al) = (Id "‘ ao30)(a1)

ao(-) = (a,) * (do).

It should be noticed that the portion of the definition of 3a which reads: 60x = Identity on A(A_,, E) could be replaced by 30x = 0 on A(A_,, E) to yield a new boundary operator. This would change

the explicit calculation on basis elements for the case q = 0 to 30(ao) = 0, but would not affect calculations for q > 0. The relation 60(110) = 0 can be interpreted intuitively as saying that the boundary of a point is zero. If the change in definition is maintained, it becomes necessary for the preservation formula 3.6, which follows in the next paragraph, to make the convention that (.) = 0. The reader can, if he cares to, follow the changes in the behavior of the case

q = O that result from the change in definition. They will be ignored here until the introduction of singular homology in Chapter 4. When q 2 0, the symbol (a0, an - ~ -, d“ - ~ ~, (1,) will denote the (q — l)-dimensional afline simplex (a0, (1,, - ~ ~,a,_,, am, - - -, a”), obtained from the affine simplex (amal, ---,a.,) by deleting the vertex point a,. Then it follows, for q _>_ 0, that 3.6.

Proof:

8"(ao, . . ., aq) = $0... 1)i(ao, . . ., fit, . . ., ad).

This has already been checked for q = 0, 1. The proof now

proceeds by complete induction: Suppose that 3.6 has already been proved for q —— 1. Then, aa(a0: ' ' " ac) = 300(01, ' ‘ '9 aq) = (Id _ aoaq-lxal: ' ‘ '9 aa) =(a19 ' ' '9 an) _ ao§(_l)‘_l(ah ' ' 'sdb ' ' 'aaq)

= g(—1)‘(ao, ”35‘, ~~~,a.,). Formula 3.6 implies that, apart from signs, the boundary of an affine simplex is the sum of its boundary simplioes.

Introduction to Algebraic Topology

24

For x e E andq 2 0, 3,, x 8,, = 3a, and 8,8,,“ = 0.

3.7.

By definition, 30(ao) = (.), and

Proof:

aoxaowo) = 30x(.) = 3°06) = (-)

Suppose that the formula 3.1x?" = a, has been proved for some 4 2 0. Multiplication of 6q+,x + x3. = Id on the left by 8,, yields 3,134“): + aqxa, = a." and thus aaamx = 0 for all x. Lemma 3.3 implies that 813“, = 0. Now multiply 8“,): + x3, = Id on the right by a“, to obtain equate“, + xaqaq“ = 3“,; that is, eggs,“

= 3“,. The composition between A(E, F) and A(A,,, E) with values in A(A,,, F) was taken (Theorem 3.1b) to be the one defined through linear combinations of the “products” obtained by writing in succes-

sion a basis element of A(E, F) and a basis element of A(A.,, E). These basis elements are mappings Aq—> E and E —>F, respective-

ly. In the special case where E: A, and q 2 1, let (do, ---,¢i,, ---,da) denote a particular basis element of A(Aq_l, A“). Then, forao,a,, - - -,a,, e Eandqz l, 3'8'

Proof:

(a0! a1) ""aa)(d0’ " 'a‘ih "',dq) = (a0: "'96” " Hag)-

Both sides of the equation are affine maps A44 —-> E. By

Theorem 2.6, these are equal if they yield the same images of a set of q independent points, say do, all, . - -, dq. The map (do, (A, ~ - - , (3}, ~ ~ ., dq) sends the point d, on d, for v < i and on d“, for u 2 i, v = 0,1, - - -,q — l. The point d, is thus sent by (do, an "',au)(d09 d1) “3J1: "':dq)

on a, when u < iand on a,“ when u 2 i. The map (a0, ---, d,, ~--

(1,) has the same efi‘ect. Notation: In the module A(Aa, A), the map (do, db - - -, d) is the identity. It will be denoted by A3. Forq 21 and c e A(A.,, E), 3.9.

C(3q) = ace.

Notice that on the left the module element c e A(Aq, E) and 3.,A; e A(A,,_,, A.) are composed. Property 3.9 is important; for it

Afiine Simplices and Boundary Operator

25

says that the identity of A(Aq, A4) is the only element for which a, needs to be computed.

Proof of 3.9:

Multiply 3.8 by (— l)‘ and sum over i to obtain

A (a.,, a“ . . ., ac); (—1)‘(do, ..., d,, ..., dc)

= g(_1)£(ao, ”.43“ NM") or (a0, a1, ---, aq)3qAI, = (Mao, a,, . ~ -, a“). Now 3.9 follows for each c e A(A., E) by the taking of linear combinations. For q 2 2, a twofold application of 3.9 yields 0 = 30—13(10) = 30-,(ca,,(A;)) = €3a(A3)3«—1(Afz—l)-

The last expression arises in the composition of three modules; and parentheses are omitted since mappings, and thus module com-

positions, are associative. Now specialize E to A0 and c to c = A}, e A(Aq, Ac). Then,

304A}.l e A(A.,_2,Aq_1) and ag e A(Aq_,, A“).

Since the factor A; = Id is superfluous, it follows that

3.10.

a.(A:,)aq_,(A;_,) = o, for q 2 2.

The Topologization of Afline Spaces:

An affine space E of finite

dimension n is metrized as follows: Let eh e2, - - -, en be a basis for

the corresponding vector space, and define lal, for a = 2 me, e V,

by |a| = M273. Then V becomes a normed vector space. If P and Q are points of E with coordinate vectors x and y, then d(P, Q) is defined by d(P, Q) = I y — xl. It is easy to check that this makes E

a metric space. The metric depends upon the choice of the basis

26

Introduction to Algebraic Topology

e1, e,, - - -, em but the topology does not because each spherical ball

defined in terms of one basis contains one defined by each arbitrarily chosen basis. (The reader should consider this statement as an

exercise to be checked.) Subspaces, such as A, always have the induced topology. The metric is introduced in infinite dimensional affine spaces by analogous devices, but here the topology is a function of the metric. If A, is embedded in the (q + l)-dimensional affine space that was defined by the barycentric coordinates £0, 15,, - - -, Ea, then the complement of Ag consists of the points for which one of the conditions E. 20, 25; = 1 fails. The complement is therefore open; hence, A" is closed. Since E is homeomorphic to R4“, and since Ag is bounded, it follows that A”, and thus its continuous images, are compact. Each affine map is linear and therefore continuous; hence,

each geometrical simplex is compact.

The Singular Homology Theory

Let X be a topological space. Since A, is also a topological

space, it is possible to define the free R-module C(Aq, X), q 2 0, generated by the set of all continuous maps a: A” —> X. Each such map a is called a singular q-simplex of X. To each integer q, a free R-module 5,,(X), whose elements are called chains (or q-chains), is defined by C(Aq, X) 4.1.

Sq(X)=

foqO

R(.)

forq=—1,ifX¢ Qf

0

forq Swan —> $.00 3'» s -.(X) —> --For q = 0, S0(X) is the R-module whose basis elements are the continuous maps of d0 in X. These maps can be identified with the points of X. Thus, So(X) can also be conceived of as the free R-

module generated by the points of X. The boundary of a O-simplex; i.e., of a point in X, is then (.) by definition. Each singular l-simplex is a continuous map of the interval Al into X; and thus a continuous curve in X together with its parametrization. Therefore, S.(X) is the free R-module whose basis is the set of continuous curves with given parametrization. Since 3,(A}) = (d1) — (do), a composition with azA, ——> X yields 0((d1) — (do)) = 0(d.) — «7(do); that is, 3,0 = a(d,) — «(110). This can

The Singular Homology Theory

29

be stated as: The boundary of a curve a is the difi”erence obtained by subtracting the endpoint from the beginning point. This difference is meaningful in the free module generated by the points of X. By 3.6 it follows that, when c is a singular q-simplex and q 2 0,

an = §(—1)*a(dm ..-,at., “ad-2" This formula shows that when q 2 1, then lam c la], and more generally, that |a,c1 S( Y, B) induces a homomorphism f* 2H,,(X, A) ——> Hq(Y, B) for all q. It follows immediately from the definition of f# that fif is the identity whenfis. Forf:(X, A) ——> (Y, B) and g:(Y, B) —-> (Z, C), the definition yields (gf)# = g#f#. Lemma 1.2 now yields 4.7.

(gf)* = g*f*, and f = Id impliesf. = Id.

Let (X, A, B) be a space triple with B c A c X. The identity map Id :X —> X induces the continuous maps

i:(A, B) ——> (X, B) and j:(X, B) —’ (X, A). 4.8. THEOREM:

The sequence ‘41:

#

o —> 5.,(A, B) —> S.,(X, B) —’—> S,(X, A) —> o is exact.

Proof: Four cases will be considered: (1) X = z: Then Sa( , ) = 0, for all q and all three entries in the parentheses. Thus, Theorem 4.8 is correct in this case. (2) X¢ QandA=B= Q}: Then Sa(A,B)=0and i*=0forall 4. Furthermore, Sq(X) = S),(X, A) = Sa(X, B) and j# = Id. Hence, the sequence is exact. (3)

Xi Q,A ¢ Q,andB= Q:Whenq20,gc=ldc= cistobe

introduced in the definition off#. Then if," is the injection and

The Singular Homology Theory

35

jff is the canonical map on the factor module, and the sequence of Theorem 4.8 is exact. The exactness is trivial here for q < —l and, in the non-augmented case, for q s ——1, since all

So are zero then. In the augmented case if,:R(.) ——> R(.) is the identity and jf,:R(.) ——> 0, so that exactness also holds here. The case B = o is precisely the previously treated case of a

space pair where 0 ——> S,(A) ——> Sa(X) —> SAX, A) —-> 0 is exact.

(4)

B :2 Q : When q < 0, the sequence is exact since all the modules

are zero. There remains the principal case, in which B at z and q 2 0. Here, the exactness of

o —> s..(A)/s.(B) 3» Sq(X)/S.(b) i S.(X)/S.(A) —+ o is under question. In this sequence, i#(c + Sq(B)) = c + Sa(B) e Sa(X, B) for c e Sa(A),

and j#(d + Sa(B)) = d + Sq(A) for d e Sq(X). Thus, i# is an injection andj# is an epimorphism. For c e Sq(A),

j#i#(c + 54(3)) = c + S.(A) = S.(A) and thus, j#i# = 0. If d e Sq(X) and the element j#(d + Sa(B)) = d + Sa(B) is the zero of S,(X, A), then d e Sa(A), and hence, d + Sq(B) e Sq(A, B). This implies i#(d + 511(3)) = d + 50(3); in other words, Ker(j#) c Im(i*).

4.9. Corollary:

The sequence: 3

in

- —+ H...(X, A) 1+ H.(A, B) ——> H.(X, B)

a 3. —j—>H,(X, A)—*H._1(A, B) __> . ..

is exact by Theorem 1.3. 4.10. THEOREM:

Homeomorphic pairs ofspaces have isomorphic

homology groups. More precisely: Iff:(X, A) ——> ( Y, B) is a homeomorphism; that is, iff induces a homeomorphism f of X on Y such thatf(A) = B, thenfinduces an isomorphism ofHq(X, A) on Hq( Y, B).

Introduction to Algebraic Topology

36

Proof: The map g =f ‘1 is a homeomorphism g:( Y, B) ——> (X, A) such that gf = Id and fg = Id. Then, by 4.7, g*f* = Id and fig. = Id. Hence, f* and g* are isomorphisms.

4.11. THEOREM: Let B c A c X, B’ c A’ c X’ and let f: (X, A, B)——>(X’,A,'B’) be continuous. Then f induces maps g:

(A, B) —* (A', 3'). ’1: (X, B) —> (X', B'), and k-‘ (1". A) —"

(X', A') such that the following diagram is commutative: #

#

0 —’ Sq(A, B) —‘-> S,(X, B) —’—> 5.,(X, A) ——> o 1%

1/#

l h#

j’#

1k#

0 —> $4.43 3') —> Sq(X’, B’) —> SU(X’, A') ——+ 0. The theorem is trivial for the several cases: X = a; Proof: q < —1; non-augmented complexes when q = —l; q = —l and B ¢ $; and q = —l and B’ = Q , since in these cases the upper or

the lower row of modules are all zero. In the augmented case when q: —1, Xi Q, B=B’ = Q, and A at g, it follows that A’ at Q , and that the same sequence

0—>R(.)—->R(.)—->0-—->O

occupies both the upper and lower rows. Furthermore, since g#, h*, and k# reduce to the identity, the commutativity is trivial. For q = —1, there remains the case for augmented complexes where q Q and A = B = B’ = Q. When A’ = Q, the same sequence 0 —-> 0 ——> R(.) —> R(.) —-> 0 occupies both rows and com-

mutativity is trivial. When A’ ¢ g, the terms S_1(A, B) and S_,(X’, A’) are both zero and the diagram commutes. There remains

the principal case: q 2 0 and X eh a . Here, Y :é Q5. (a)

Let c e Sq(X). Then

h“1"‘(0 + Sq(B)) = h"(0 + $43)) =fc + 543'). i'*g*(c + Sa(B)) = i’#(f€ + Sq(B')) =fc + Sa(B'). where, to avoid the introduction of further notation, the letter

f is also used to denote the map of X in X’. (b) Let c e Sq(X), then

k*j#(c + 34(3)) = “(c + Sa(A)) =fc + Sa(A'),

j'#h#(c + 5.02» =j'#(/c + S.(B'» =fc + S..(A’).

The Singular Homology Theory

4.12. Corollary:

37

Under the hypotheses of Theorem 4.11, each

square in 5t

‘4

i.

0(A, B)

_’

g.

a.

—>

j.

at

0(X, B) _’

«(X: A) _’ Ha—1(A’ B) _’ ' ' '

h.

k.

g.

r. 5a. q(A’, B’) —> Hq( Y, B’) —> Ha( Y, A’) —+ Hq_1(A', B') ——> - - -

is commutative.

This follows immediately from Theorem 1.4. Later, in axiomatic study of homology, Theorem 4.11 will be

taken as an axiom.

flomotopy Properties of Homology Groups

The affine case will be considered first; and, as in Chapter 3, the

proofs will be carried out for reduced homology. Let E and F be convex subsets of affine spaces. The set of all afiine maps of E in F was denoted by A°(E, F), and A(E, F) was the free module generated by Ao(E,F). For q < —-l, A(A.,, E) = 0 by definition. In

Chapter 3, the notation x“ = p(x) was introduced for the image of x e E under the affine map p:E —> F. A homomorphism p:A(A., E) —> A(A.,, F) was induced by p:E —* F. If q 2 0 and a = (00, - - -, a“) 6 AAA”, E) is an affine simplex, then, by definition, P(ao: ‘ ' " ac) = (as, ' ' 'y as) E A0(Ad9 F)‘

For the elements 0 = Ema of A(A,,, E) it follows that p(c)= 38

Homotopy Properties of Homology Groups

39

2 r,p(a). For q 3 —1, p:A(Aq, E) —-> A(Aq, F) is defined as the identity. In addition, it was proved in Chapter 3 that, for q 2 —1, each x e E could be interpreted as a homomorphism x:A(Aa, E) —> A(Aq+1, E). It was for this purpose that the definition x(ao, - - -, ac) =

(x, a0, ---, ac) was made for q 2 0 and for each affine simplex ((10, -~, aq) e A0(A,,, E). For q = —1, x(.) = (x), by definition. It was then easy to see that px = xPp. Let x, [1. be two afline maps of E in F. With the resulting homo-

morphisms 7t, ,u:A(Aq, E) —> A(Aq, F), a recursive homomorphism

«1».mm... E) —> Am... E) is defined by means of 5.0.

"Pa = 0

for q s —1

5.1.

«Ix-“1x + x‘qpq = x‘x“,u. = x‘px

for q 2 —l.

The formulas are meaningful since the homomorphisms of 5.1 are defined on A(A,,, E) and have images in A(Aq+,, F). Indeed, the diagram

A

«A. E) —’+ AA... E) —”> Am... F) 1» Am... F) We I

describes the situation. The map «pa can be found explicitly as follows: For q g -1,

«p4 = 0 by 5.0. Forq = —1, formula 5.1 yields «pox + x‘O = x‘x‘m. By means of A(A_,, E) = R(.), it follows that «pox(.) = x‘x“,u,(.) or «1).,(x) = x‘x“(.) = (x", x"). Hence, «p. is given by 1Po(ao) = (a3, a3). Now suppose that \pq_1 has already been defined for q 2 1. Then, for the affine q-simplex (a0. - ~ -, a”), formula 5.1 yields “#0010, ' ' 'a ac) = “#1100011, ' ' 's ac) = (130575011, ' ' ’9 ac) _ “twig-10h: ' ' ‘9 ac)

= (at. at. 111‘. at. - . -, at) — awnmah a2. - - ma.)

Introduction to Algebraic Topology

40

It will now be proved that 5'2'

“Paolo: a1! ' ' Wad) = g(_1)t(a011 ’ ' Wail: a4" ' ' Hag)

for q 2 0.

Proof: (By induction.) For q = 0, formula 5.2 reduces to 4,0010) = (a3, as‘). This has already been proved. Suppose that q 2 1 and that the formula has already been proved for q — 1. Then it follows that «#4010! an ' ' '9 a4) = (a3; (15‘, ail: ' ' 'a (1:) _ ag‘l’q—l(als a2, ‘ ' '9 ac) Q

= (03,05, ' ' 'rag) _ aa‘E(—1)l_l(allr ' ' Hall: “is ' ' 'ra‘a‘) =1

= i: (—1)1(aoly ' ' '9 all) ai‘: ' ' '3 a60{=0

Geometrical Interpretation for Small q: For q = 0, 1P0(ao) = (a3, a3) is the line segment connecting a3 with as‘. For q = l, "I’l(a09 a1) = (“011 all)" ai‘) _ (“3’ all, ail)-

Thus, the diagram is

Figure 4

For q = 2,

«Mam an a2) = (00", at, ai‘, at) — (at, at, at, at) + (a&, at, at at). which can be illustrated by

Homotopy Properties of Homology Groups

41

Figure 5 This geometrical interpretation assumes that the points (12,“, at a}, at‘,

af, a; are distinct. It is given only to strengthen the intuition and will not be used explicitly. Even when q 2 2, the construction yields a triangulation of the prism with signs on the simplices whenever the vertices a3}, . - - , a}, and a5, - - ~ , a: are located in a “reasonable” manner. 5.3. Lemma:

For the boundary of a subdivided prism, 3“l + "l’a-iaa = ,u. — 7L for aIIq.

Remark on 5.3: Let 0' be a q-simplex. Then a,“ 1]»,(0) is the boundary of the simplicially subdivided prism on a. According to formula 5.3, this boundary is p.(¢r) — Ma) — «p,_,a,(a). Here (with signs disregarded) [1.(0') is the top, M0) the base, and «pq_16,(a) the

simplicially subdivided lateral faces. Thus, 5.3 says that inner boundaries vanish.

For q g —1, both p. and 7t are defined to be the Proof of 5.3: identity, and therefore, ,1. —- 7t = 0. Since «I», = 1,4,4 in this case, 5.3 follows for q s —1. If «In, is applied to the simplex (x), where x e E, then «p,(x) =

(x‘. X“) and 3mm“) = (X“) — (X‘) = MX) — MX) Since «Ir—x = 0, 5.3 holds for q = 0. Suppose that 5.3 holds for some q 2 0; then 3.5 and 5.1 yield

Introduction to Algebraic Topology

42

Bau'gbqflx = ecu-7‘10“ " 1Pa) = (Id — x13¢+.)(,ux _ ’4’4)

= I” — "I’a _ xiiaanx + xxaan'q’aIn the last equation, use was made of ya”. = aqflp, which is an immediate consequence of 3.6. From 5.3 for q, from 5.1, with q — 1 used for q, and from 3.5, it follows that

ana‘P‘q — [‘3‘ = —‘Pa _ xA/‘aanx + x*(/b — h — hire—lac)

= w. — mam + xm — xx + («p.x — mm.) = —7\.x — «mad — xaq) = —)\.x — «pqaaflx.

An application of 3.3 yields 5.3 for q + l. The computations will now be interrupted until after Theorem 5.5 so that an indication of their purpose may be given.

Let I denote the interval 0 g t g l, and let A c X and B c Y be topological spaces. Definition:

Two continuous maps f, g:(X, A) —> (Y, B)

are

called homotopic if and only if there is a continuous map F(x, 1): (X X I, A X I) —> (Y, B) such that F(x, 0) =f(x) and F(x, 1) = g(x) for all x e X. This is written f ~ g. Intuitively, it means that the map f can be deformed continuously into the map g. The

maps 7t, ,u:(X, A) ——> (X X I, A X I) which are defined by Mx) = (x, 0) and g(x) = (x, l) are homotopic since the identity G(x, t): (X X LA X I) ——>(X X I, A X I) is continuous and satisfies both G(x, 0) = Mx) and G(x, 1) = ,u.(x). Some steps will now be taken toward proving that if f, g:(X, A) —> (Y, B) are homotopic, then f... = g*. This will be completed later, in 5.14. If F defines a

homotopy betweenfand g, thenf = FA and g = Fp. From A... = y,” it follows from 4.7 that fat = (F70: = F*>"* = Fella: = 8:»

Thus, it is sufficient to prove that A... = M. A step in that direction is

5.4. THEOREM: Given two chain complexes A, B and a chain homomorphism f:A —-> B. If there are homomorphism: qAq —* Ba for which 5.4a.

then f... = 0.

aq+lDa + Da-lad =fi’

Homotopy Properties of Homology Groups

43

Proof: The situation discussed in this theorem is described in the diagram at“

A+1‘—’Aq ——>A_ —>

1“ ,. 1 »~-’1'

-—> Bail -—> B“—> B- —> an!

Let c e Hq(A); that is, c = a + a,+,(A,+,), where a is a cycle in

A. By 8a = 0 and 5.4a, it follows that aq+,Dq(a) = fl,(a), and hence, that fl,(a) e 33“.. Therefore, f*(c) =fi1(a) + aa+1(Bq+x) = aq+lBa+h

where 3,,“ Ba“ is the zero element of Hq(B). Thus,f* = 0. Comparison of 5.3 and 5.4a shows that preparations are under way for the construction of a D0 which will suffice to prove (/‘_7\')*=I‘*—7V*=0-

Definition:

Two chain homomorphisms a, B:A —> B are called

chain homotopic (written (1 ~ ,8) if and only if there are homo-

morphisms D, for which 8”q + Dwa, = a, — [3,.

From

it ~ B, it follows by Theorem 5.4 that 61* = 13*.

5.5. THEOREM:

The relation ~ isan equivalence relation. When,

for a, a’1A —> B and ,8, [STA ———> B, the relations oz ~ 01’ and

[3 ~ 3’ hold, then rd + 5,8 ~ rd’ + s,8’ for r, s e R. When, for a, a’zA -—> B and 13, [9’18 ——-> C, the relations a ~ (1’ and ,3 ~ 8' hold, then ,Boz ~ fi’a'. Proof:

The relation at ~ a is obtainable by the choice D, = 0,

for all q. If a ~ (2’ and D, is replaced by —D.,, then (1’ ~ (1. If 6D,, + 0,43 = a, — a; and

aDp+Dza=aL—dfl then use of D, + DI, shows that oz ~ 01’. Thus, ~ is an equivalence relation. If oz, (1’, B, ,8':A ——> B, and if

am+mfi=m—¢mwm+mfi=m—%

Introduction to Algebraic Topology

44

then q + SD; yields the desired result. For the final conclusion of 5.5, suppose that

an. + D443 = a. — a", and 30; + 03-19 = 13 — ’3’. Let 5., = 33+q + Dgaq. Then, 35,, + Dana = 3,934). + apgaa + 1830...?) + D;_1aq_la = IBKaDa + Da—la) + (3D; + D’_13)da

= 19m — as) + (Bo — Blade = Bode — 18:1“;

because 3/3",+1 = 3.33 and baa = dq_13 since a and B’ are chain homomorphisms. In the following computation, an abbreviated notation will be used: Let X and Y be topological spaces and let ¢:X ——> Y be a

continuous map. Then $1X X I —> Y X I shall denote the map which is defined by $((x, t)) = (¢(x), t) or by (x, t)‘ = (x’, t). The

continuity of 4) implies continuity of $. When X 1» Y —0-> Z, it is easy to verify that (F0313) = 5% The association 4) —> $ and a linear extension yields a monomorphism of C(X, Y) into C(X x I, Y X I). If X and Yare affine spaces and s,(X x I, A x I) where ##(c + S,,(A)) = ,uc + S,(A X I) for q 2 0 and t {3. Then 5.12 implies that

5.13.

amp, + owe, = #1: — x:

for q 2 0.

Equation 5.13 also holds when q < 0, since the left side is 0 and on the right either p, = 7»: = 0, or else (in the augmented case when q = —1 and A = a) ,q = Mi = Id. Now Theorem 5.4 can be applied, with 1]», in place of D, and p." —— V in place of f, to yield IL:- = M. This proves the Homotopy Theorem (5.14):

5.14. THEOREM:

Let 7», p:(X, A) -—+ (X x I, A X 1) be given

by 7t(x) = (x, 0) and p(x) = (x, 1). Then the induced maps A,” p,:Hq(X, A) ——> Hq(X x I, A X I) are equal; i.e. [5* = x...

5.15. Corollary: f1: = g*-

If f,g:(X, A) ——> (Y, B) are homotopic, then

Introduction to Algebraic Topology

48

Proof:

The proof was carried out just prior to 5.4.

5.16. Lemma: Let X and Ybe topological spaces and let X,, X2 be closed subspaces of X. Let f,:X, —+ Y be continuous for i = 1, 2 and let f,(x) =fi(x) for x e X, n X,. Then the function f:X, U X; —> Y, defined byf(x) = f,(x)forx e X,, is continuous. Proof:

Let A be a closed set in Y. Then f:‘(A) is closed in

X,. Since X, is closed, f;‘(A) is closed in X. Hence, f“'(A) =

f{1(A) U f{‘(A) is closed in X1 U X2, andfis continuous. The lemma and proof remain correct if the word “closed” is replaced by “open” throughout. The hypotheses can be weakened,

although this is not of interest here. The next theorem is an application of the lemma.

5.17. THEOREM: Iff; g, h:(X, A) ——-> (Y, B) are continuous and iff~ gandg ~ h, thenf~ h. Proof:

There exist continuous functions F, G:(X X I, A X I) —>

(Y, B)

with

F(x, 0) =f(x), F(x, 1) = 80‘),

G(x, 0) = 80‘),

G(x, l) = h(x). Let I’ be the interval 0 gigs and I” be the interval % g tg 1. Then X X I’ and X X I” are closed subspaces

of X X I whose intersection is the set {(x, {7) | x e X}. The function defined by F(x, 2t) is continuous on X X I’ and that defined by G(x, 2t — 1) is continuous on X x I”. By 5.16, the function H defined by

H(x, t) = {

F(x, 2t)

for

t e I’

G(x, 2t — l)

for

t e I”

is continuous. Furthermore H(x, 0) =f(x), H(x, l) = h(x). Thus,

f ~ g5.18. Corollary:

The relation ~ is an equivalence relation.

Proof: The transitivity has just been proved. For reflexivity, define F(x, t) = f(x). For symmetry, use F(x, 1 — t) in place of F(x, t).

5.19. THEOREM:

Let

f.,j;:(X, A) —> (Y, B)

and

g,, 3,:

(Y, B) —-> (Z, C) be continuous. Letfl ~ f, and gl ~ g,. Then garl ~ 82/;-

Proof: Let F:(X X I, A X I) —> (Y, B) be a continuous function for which F(x, 0) =f,(x) and F(x, 1) =fi(x). If g:(U, 0) —> (X, A)

Homotopy Properties of Homology Groups

49

and h:( Y, B) —> Im(h) are continuous, then the maps defined by F(g(u), t) and h(F(x, t)) are continuous irrespective of Im(h). Because

F(g(u). 0) =f1g(u),

F(g(u), 1) =fzg(u), h(F(x. 0)) = hf:(X), and

h(F(x, 1)) = hf,(x), it follows that fig ~ fag and hfi ~ hfi. For the maps of the hypothesis it therefore follows that g,fl =g2fi, and g,fl ~ g,f,. Transitivity then yields g,f1 ~ ggfi.

Definition: If, to a continuous function f:(X, A) ——> (Y, B), there is a continuous function g:( Y, B) ——> (X, A), such that gf:(X, A) —-> (X, A) and fg(Y, B) —> (Y, B) are both homotopic to the identity, then g is called a homotopy inverse off.

5.20.

The following holdfor homotopy inverses:

(1) If f has a homotopy inverse g and iff’ ~ f, then g is also a homotopy inverse for f’; because f’g ~ fg ~ Id and gf’ ~ gf ~ Id. (2)

If g is a homotopy inverse off and g’ ~ g, then g’ is also a

homotopy inverse for f. The proof is as in (l). (3)

Ifg and g’ are homotopy inverses off, then g ~ g’. This follows fromg ~ gId ~ gfg’ ~ Idg’ ~g.

Definition: A functionf:(X, A) —> (Y, B) which has a homotopy inverse is called a homotopy equivalence. Two pairs of spaces (X, A) and (Y, B) for which there is a homotopy equivalence f:(X, A) -—> (Y, B) are called homotopic. Let fi(X, A) —> (Y, B) and f,:(Y, B) —>

5.21. THEOREM:

(Z, C) be homotopy equivalences. Then fm:(X, A) —>(Z, C) is a

homotopy equivalence. Let g1 and g2 be the respective homotopy inverses of f, Proof: and f2. Then fifiglg, ~ f2 Id g2 ~ fag, ~ Id, and similarly, glg,f,fl ~ Id. Thus, glg2 is a homotopy inverse off,fl. 5.22. Corollary: relation.

Homotopy of space pairs is an equivalence

The transitivity was proved in 5.21. Symmetry follows Proof: from an interchange of f and g. Reflexivity follows from letting f=g=Id.

Homeomorphic space pairs are homotopic since 5.23. Remark: f '1 is a homotopy inverse off.

Introduction to Algebraic Topology

50

By the homotopy theorem (5.14) it follows for homotopic 5.24. maps f, g: (X, A) ——*(Y, B) that f* = g*. If f has a homotopy inverse g, then it follows that (fg)* = Id and (g)... = Id. Hence, f,:Hq(X, A) ——> H,( Y, B) is an isomorphism. Thus, homotopic space pairs have isomorphic homology groups. Example of a Homotopy Equivalence: Let 7t:(X, A) —->(X X I, A X I) be defined by Mx) = (x, 0) and pr:(X X I, A X I) ——> (X, A) by pr(x, t) = x. That )t is a homotopy equivalence with pr as homotopy inverse can be seen as follows:

Clearly, pr 7x, = Id. The function F:(XXIXI,AXIXI)—>(XXI,AXI) which is defined by F(x, 1, s) = (x, ts) shows that x pr ~ Id since

A,pr(x, t) = (x, 0) = F(x, t, 0)

and Id(x, t) = (x, t) = F(x, t, l). 5.25. Corollary:

by

Mx) = (x, 0)

The function 7t:(X, A) ——-> (X X I, A X 1) given

defines

an

isomorphism

7t.:Ha(X, A) —*

H,(X X I, A X I).

Geometrical Consequences of the Homotopy Theorem Definition: Let Y be a subspace of X and r a continuous function r:X ——> Ywhich is the identity on Y. Then r is a retraction of X on Y and Yis a retract of X. Not every subspace is a retract. For example, let X be a closed interval with interior points and let Y be the pair of endpoints of X. Since continuous images of connected sets are connected, a retract of X would need to be connected. Thus, Y cannot be a retract of X,

for Y is not connected. Definition:

Let (Y, B) and (X, A) be space pairs with (Y, B)

c (X, A); that is to say, Y c Xand B c A. Letr:(X, A) —->(Y, B)

Homotopy Properties of Homology Groups

51

be a continuous function for which r(y) = y for all y. Then r is called a retraction of (X, A) on (Y, B) and (Y, B) is called a retract of (X, A). Let 1‘:( Y, B) -—> (X, A) denote the injection. Then, for the retraction r, ri:( Y, B) -—> (Y, B) is the identity. Thus, r is the homotopy inverse of i if and only if ir:(X, A) —-—> (X, A) is homotopic to the identity. Definition: A retraction r of (X, A) onto (Y, B) is called a deformation retraction, and (Y, B) is called a deformation retract of (X, A) if and only if iris homotopic to the identity. From the Homotopy Theorem (5.14), it follows for a deformation retraction r:(X, A)——>(Y, B) that r;‘ = i* and that

i*:Hq(Y, B) —> Hq(X, A) is an isomorphism. (The reader will remember that isomorphisms are epimorphisms.) A retraction r:(X, A) —> (Y, B) is a deformation retraction if and only if there is a continuous function F:(X X I, A x I) —->

(X, A) with F(x, 0) = x and F(x, 1) = ir(x) = r(x). For y e Y,

F(y. 1) = r(y) = yDefinition:

Let r:(X, A) —> (Y, B) be a deformation retraction.

Then (Y, B) is a strong deformation retract of (X, A) if and only

ifF(y, t) =yfor ally e Yandt e I. 5.26. Examples: (1)

LetP¢ Q be two points andlet X=PU Q, Y=P, A = Q. Then Y is a retract of X but not a deformation retract since F

maps the connected set Q X I on a connected set (and hence, a point) and it therefore is impossible that both F(Q, 0) = Q and F(Q, l) = r(Q) = P. (2) Let X consist of two circles which have a single common point. Let Y be one of the circles and let A =z. Folding shows that Y is a retract of X. It will be proved later that Yis not a deformation retract of X. (3) Exercise: Let X consist of the points (x, y) of R” for which either ngl and y=0, or x=0 and Ogygl, or Let Ybe the x= l/n, Ogygl, where n= 1, 2, 3,

point (0, 1) and A = Q. Then Y is a deformation retract, but not a strong deformation retract of X.

Introduction to Algebraic Topology

52

Y

||lll

|Illl Illll

I||||

0m”?

, (l, 0)

(l/n, 0) Figure 6

(4) Let X be the ball |x| 3 r, that is, the set of all points of R" with position vector x and |x| g r. Let A = {25 and let Ybe the origin. Then Y is a strong deformation retract of X, as can be seen from F(x, 1) = tx.

(5) Exercise: Let X be the solid torus. The central circle is a strong deformation retract of X.

(6) Exercise: Let X be a torus (surface) in which a small circular (7)

hole has been cut. Describe a pair of circles which have just one common point that form a deformation retract of X. The contraction of a spike is often employed: Let X = A U S be a subset of R" where A n S = origin and S = {x | x e R", nlgl, x.=0 for i=2, ---, n}. Let F(x,t)=tx for

x e S and F(x,t)=x for x e A. ThenF:(XX I)-——>Xis continuous and satisfies F(x, 1) = x for all x e X and F(x, 0) = r(x) = x for xe A and F(x, 0) = r(x) = 0 for x e S. Thus, A is a deformation retract of X = A U S.

(8)

Let H be a hypersurface in R" with the following property: To each vector x at 0 there is precisely one intersection point

h(x) of H with the half line )ix, 7t > 0, and h(x) is a continuous function of x. Then H is a deformation retract of each of the three sets:

X0: {x l X¢0}, X1 = {x I IXI 2|h(x)l}, and X, = {x I xqéO, |x| 3 |h(x)|}. Proof:

The function

FM—WW) _

l+t h( )

is continuous for x ¢ 0, 0 g t S 1. Furthermore,

F(x, 0) = Ilia?) = x

Homotopy Properties of Homology Groups

53

F(x, 1) = h(x), and F(h(x), z) = h(x). Thus, H is a strong deformation retract of X0. Since the in-

equality |x| g |h(x)| implies that

M + tU106)! Silxl +|h(X)|, which implies that |F(x, t)| g |h(x) I, the set X, x I is mapped onto AG by F(x,t) for i=1, 2. Hence, H is also a strong

(9)

deformation retract of both X1 and X2. As an application of (8), consider R3 with (—1,0,0) and

(1, 0, 0) removed. By (8), the sphere lxl = 2 is a strong deformation retract of {x | |x| 22}. Let the point (—1, 0,0) be surrounded by a surface Hl which consists of the hemisphere l = 2, x1 < 0 and the disc |x| 32, x, = 0. By (8), H1 is a strong deformation retract of the punctured interior of H1. If H, is defined analogously so that it surrounds (1, 0, 0), then H1 U H2 is a strong deformation retract of the twice-punctured R3.

Application to Graphs Definition: A graph is a connected union of a finite collection of topological line segments which have at most end points in common. A topological circle is therefore one example of a graph. Graphs can be embedded in R3, but in many essentially different ways; for instance, the circle and the trefoil knot are homeomorphic. The nature of the embedding of a graph is a difficult question. However, the problem of the homotopy and homology of a graph is easily

settled.

(1)

(2)

A segment with a free end point, a spike in other words, can be

eliminated by a deformation retraction upon the other end point as described in 5.26 (7). Let P, Q with P :1: Q be end points of a segment of the graph

T. Let P be an end point of at least three segments. On one of these which is not PQ choose a point R so that R, P, Q become vertices of a triangle which has only the sides RP and PQ in common with the graph. Let T’ be the graph that results when one connects R with Q instead of with P. The diagrams

show that both T and T’ are deformation retracts of the union

Introduction to Algebraic Topology

54

-

‘—

Q

—‘

Q

Q

P

P

P

P

P

T

—’

Q

Figure 7

of T with the triangle RPQ. Hence, T is homotopic to T’. In the formation of T’, the number of segments leaving P has

been reduced by one. By a repetition of this procedure, the number of segments leaving P can be reduced to two. Such a result is equivalent to the contraction of P to Q, and can be

repeated until all segments have the same point Q as their beginning and end points.

(3) It was just proved that each graph is homotopic to a union of finitely many topological circles that have exactly one point Q in common.

Figure 8

(4) This procedure also shows that a line segment with the same number of attached circles is homotopic to the figure arrived at in (3). The number of circles will be proved to be a homotopy invariant. It is also a homology invariant as will be shown by a computation of the homology groups.

T

Q

Homotopy Properties of Homology Groups

55

000000 Figure 9

5.27. The “Five” Lemma:

In thefollowing diagram let the rows be

exact, and let the rectangles be commutative. Furthermore, let a, [3,

8, e be isomorphisms. Then y is also an isomorphism. in

O!

is

M

Ai‘—)A2'_)A3—_’A4’_>As

1“,, 1”,, 17,, 18,. l“

B, —>B,——>B,—>B,——>B,. Proof: (1) Let -y(a,) = 0. Then «[r,vy(a,) = 8¢,(a3) = 0, and hence, Am“, E),

is defined by Sda = Id for q s 0 and Sd,,(a) = b,q_18q(a) for q > 0 and a" e A0(Au, E). Thus for a 0-simplex an, the definition states that Sdo(ao) = a... For a 1-simplex a = (a0, a,), the definition

yields Sdl(a0, 4|) = ba((a1) — (00» = (be, a1) '— (but, do);

pictorially, 00

b,

al

Figure 10 For a 2-simplex a' = (a0, a1, a,), Sd,(a) has the appearance

0:

Figure 11

The homomorphism R¢:A(Aq, E) -—> A(A¢+., E) is defined recursively by Ra = 0 for q s 0, and Rq(a-) = b,(Id — Sda — R¢_,a¢)(¢r)

for q > 0 and 0' e Ao(Au, E). 6.1. THEOREM:

(1)

Sd is a chain homomorphism; that is,

Sd,,_,aq = 3.,q,

The Excision Theorem

59

and is chain homotopic to the identity because

(2)

aaflRa + qaq = Id — q.

Proof: Because of 84 = 0, both formulas hold for q g 0 in the non-augmented case. Suppose that (1) has been proved for some q 2 0. Then, aqaaqH = Sdmaqa‘,+1 = 0, and hence,

aWSdMor) = mumsauamoy) = q3a+,(o) since 3.5 can be written amb, = Id — L3,. Now, suppose that

(2) has been proved for a q 2 0. Then, aqam = (Id — Sd¢)3q+, = 3q+l(Id — qH), and, since amb, = Id — bag“, it follows that a«+2Ra+l(a') = a+2bvad _ q+1 _ Raaqnxa') = (Id — 8d,,+1 — R,,a,,+l)(a). 6.2. Proof:

For p e A0(E, F), both da = qp and q = Rap hold. The assertion is correct for q s 0 because q = Id and

Ra = 0. If da = Sdap has already been proved for a q 2 0, then it follows for a' e A0(Aa+,, E) that daHo = pb,qaq+la' = bZSdaaaHpo' = bMqaqupo = qHpo.

Thus, dQH = dp. If q = Rap has already been proved for q 2 0, then pRaHa' = pb,(Id — qH — Raamyr

= ”(Id — Sda+1 — Realm)!” = Ra+IP°'Thus, pRaH = Rq.

E will now be specialized to A" and o' to the identity, denoted by Anq ——> A” for q 2 0. Then p e Ao(Aq, E) is a q-simplex of F. Since pA; = p, the formulas (l) and (2) of 6.1 yield

mm.» = qo) and Mama) = no).

Introduction to Algebraic Topology

60

For arbitrary c e A(A., F) it then follows that

ca A!) = Sda(c) and c

6.3.

z, = Rq(c).

For c = 34A}, and q 2 I, it follows that

3u(A£)Rq—1(A3-1) = mam). According to 3.9,

Rq(A$)3q+1 i = 3q+q(A3)Addition and (2) of 6.1 yield 6'4'

ac(A&)Ra—1(A4'1—l) + Ra(ADaa+i(A'q+1) = A7: _ Sda(AD'

Definition:

Let (X, A) be a pair of topological spaces. In the

non-augmented case, the functions Sda2Sq(X, A) —-> Sq(X, A) and qSq(X, A) —-—> SW(X, A) are defined by q = Id and R, = 0, for q < 0;

Sda(c + 52(4)) = c.(A:.) + S404). for q 2 0; and Ra(c + S.(A)) = CRa(AD + Sun“). for q 2 0. From

Sdo(A3) = Id(A5) = Id = A3 and Ro(A3) = 0 it follows that Sdo = Id and R0 = 0. By Lemma 1.1, q and Ra are

well-defined homomorphisms. Formula (2) of 6.1 will now be established in the general case. It is trivial for q g 0 since, then, 3., = 0, R0 = 0 and Sdo = Id. For q 2 l, formula 6.5 yields

(aaHRa + Ra—laaxc + Sa(A)) = C(Ra(A¢i)aa+1(A¢;+l) + 30(A3)Ru—1(Az—|) + S000

= C(Ai — Sda(A.',)) + SKA) = (Id — qXc + SKA».

The Excision Theorem

61

Hence, for topological space pairs, 6.1(2’).

3¢+1Rq + Roda, = Id — Sda.

6.5. Lemma: Let A and B be chain complexes. For each q, let a homomorphism Dq :Aq —> B“. be given. Iffl, is defined by

6.6.

f. = awn. + Dada”

then fqa —> 3., is a chain homomorphismfor each q.

Proof:

Clearly, fa is well defined. From 6.6, it follows that aafa = aaDq—lau =fh—laa-

6.7. Corollary: division qcq.

If c,, is a cycle, then ca is homologous to its sub-

Proof: Application of Lemma 6.5 to f; = Id — Sda = aaHRq + qaa shows that am, = q_18q. Hence, Sd:S(X, A) —+ S(X, A) is a chain homomorphism that is chain homotopic to the identity

by 6.1(2’). By Theorem 5.4, (q)* = Id. 6.8. Corollary: Let q 2 0 and let a E C0(Aq, X) be a singular q-simplex. Then the carriers of3.0-, Sdacr and Rqo lie in l 0' |. Proof:

For a e C(Y, A0, ea is in C(Y, X). It is immediate that

the carrier laal of ca (a pointset in X) is a subset of | a- |. Now choose a = 80(A3), a = q(A;) or a = Rq(A;), and the corollary follows.

At this point, there is a digression into properties of certain metric spaces.

Definition: By the diameter 8(M) of a non-empty set M in a space with metric d(x, y) is meant

sup d(x. y). 3(M) = 1.14:1 The metric of a euclidean vector space E is written Ix — y |. Let a' = (a0, a1, ---, a.) e A0(A.,, E). The carrier Io'| is the geometrical simplex with the vertices a0, a1, ' - ., a.. Let x, y e lo-|;

that is, x = 2m, y = 2 ma. where s. 2 o, m 2 0. 2 s. = 277,=1,andi= 0,1, ---,q.Then

ly-xl = l2£g(y—a.)I£MaXIy—a.l,

Introduction to Algebraic Topology

62 and hence,

Iy—xISMfin—atISI‘EflIaJ—at-

(a)

Since a, e lo I, it follows that

3(Icrl) = N‘Iglx Ia: — aJI'

(b) More precisely,

IJ’ _ asI = “$01)“: _ “DI SO — ’71) Max I“: _ atI

= (l - m)3(|0|)Then for

y = b, = Eta/(q + D)“:

it follows that I b, — a. I g (q/q + ])8(I 0' |), and hence, according to (a),

(c)

lbw—xISq—_i1_—13(|_ 0,

Sd(c + Sa(A)) = cSd(A&) + SM)Then, by 6.10,

Sd"(c + 5.104)) = 48mm" + SKA) = c"(A£) + SM)The standard simplex A, is a compact, metric space because, for instance, it is closed and bounded. The following theorem will have application to simplices:

64

Introduction to Algebraic Topology

6.11. LEBESGUE COVERING THEOREM:

Let M be a com-

pact, metric space. To each covering of M by open sets B, there is an

e > 0 such that each subset of M whose diameter is less than 6 lies entirely in one of the sets B. Each x e M lies in a B. Since B is open, there is an open Proof: ball around x of radius r, which is contained in B. Let K, be the

open ball around x of radius {yrI > 0. M is covered by the set of all KI. Since M is compact, M is covered by a finite collection {Km}, i=1,2, ---,n.Lete=Minr,,‘,A X is a continuous function (here for the first time the continuity of the map a is used), each of the sets o"(B) is open; and

the collection of these sets is a covering of A,. Since A, is a compact, metric space, there is an e > 0 (called the lebesgue number) such

that each subset of A, with diameter less than e is contained in one of the sets o‘1(B). For this 6 choose an n such that 5 I ~r| < e for each 7 that occurs in Sd"(Aa) with coefficient different from zero. Then

each |r| is in o“‘(B) for some B, and hence, |zrr| c B c: B. When c e C(Aq, X), the symbol c e Q will be used to indicate that to each simplex o occurring in c there is a B e a such that la-l c B. It has just been proved that to each singular simplex a' there is an n with Sd"(o) e w. Here, as before, Sd°(o) = 0'. With each singular q-simplex 0' of X there will be associated a natural number n(a') which satisfies the following conditions: (a) For a e w, n(a) = 0. (b) For each simplex a, which appears in 30(0) with non-vanishing coefficient, n(o,) _(X, A)

induces a chain homomorphism i*:S(X — U, A — U) —> S(X, A) in accordance with Theorem 4.6. Since

i#(c + Sq(A —- U)) = to + Sq(A) = c + Sq(A),

it follows that jk = 1'". Let p = k"‘7\,. Then p:S(X, A) —> S(X — U, A — U),

pi“ = k‘mjk = k“Id k = Id, and

1",) =jkk"7\. =17». Consequently,

Id—jh=3D+Da.

The Excision Theorem

69

Theorem 5.4 yields (“PM = i*P* = Id-

From pi# = Id, it follows that ml" = Id; therefore, z} is an isomorphism. Notice that U did not need to be an open set here. Remark on the augmented case: By definition, Hq(X, A) = Ker aq/Im am where aa:Sq(X, A) —> S _,(X, A). For q > 0, the augmented groups Sq(X, A) and the homomorphisms 6‘, are identical

with those that occur in the non-augmented case. Hence, the groups H,,(X, A) are the same in both cases, provided that q > 0 or q = 0

and A at Q. For q < 0, Ha(X, A) = 0 by Theorem 4.4. There is then a difference between the augmented and the non-augmented homology groups only in the case X 72 a, A = :2), q = 0; that is, in the computation of H0(X, (a) = H0(X). However, the Excision Theorem is meaningful only when A ¢ g . Therefore, the Excision Theorem can also be employed in the augmented case except when q = 0 and A — U = Q. This exceptional case seldom arises in applications and, because of U c A, can only occur when A is both open

and closed.

Direct Decomposition and Additional Aids to the Computation of Homology Groups 7.1.

Let G be the direct sum of R-modules Gk, k 6 J. For each k,

let Hk be a submodule of Gk. Then the sum H of the Hk is direct. The direct sum 26’0d is isomorphic to G/H. Proof: Let

i(2(g;c + H») = 23;; + Hi and 112g); + H) = 2(gk + Hi) define homomorphisms i :ZQGk/Hk —-> G/H and s/H ——> Zea/km. Since the sums are direct, i and j are well defined. Since ij = Id and 70

Direct Decomposition and Additional Aids

71

ji = Id, 1' is an isomorphism. Let isk/H,‘ —> G/H be defined by ik(gk + H.) = gk + H. The situation will be indicated by writing

1' = 21). directly. Now let (X, A) be a topological space pair. Let X = U X,‘ be a disjoint decomposition of X into subsets X,” and for each k and each x e X. let X,‘ contain the path-component of x; that is, the

set of points connectible to x by a path. Here it should be recollected that a path is a singular l-simplex. For A. = A n X,” the union A = U A. is disjoint. The injections i,‘(X,., Ak) —> (X, A) induce

homomorphisms 200*: 2911,,(Xk, A) —> Hq(X, A).

7.2. THEOREM: phism.

In the non-augmented case, 2(5),. is an isomor-

Proof: Let 0q —> X. The set [a'[ is the continuous image of Ag and is therefore pathwise connected. Hence it is contained in precisely one Xk. Consequently,

5,,(X) = $5414.) and saw) = $54.4), for q 2 o. By definition, Sq(X, A) = Sq(X)/Sq(A) and SAX,” A.) = Sa(Xk)/Sq(Ak). For if:S,,(X,c, A.) ——> Sq(X, A), 7.1 shows that git: ZGSAX,” A.) ——> Sq(X, A) is an isomorphism. The sum 29s,,(xk, Ak) forms a chain complex with a defined for each component. Since the if commute with a, the sum 2:17.t does also.

Direct computation of the image of an element of Ha(2@Sq(Xk, Ak)) under

(EitHAZeSAXm AD) —> H0(X: A) in analogy to the proof of 7.1 shows that (Sit). is an isomorphism. By 7.1, there is also an isomorphism of EQHJX," Ak) on Hq(2©Sq(Xk, Ak)). It is easy to check that the composition is an isomorphism

203*: 2919416.. Ax) —> H406 A)For q < 0, the theorem is trivial in the non-augmented case. In the augmented case, the theorem is false since S_1(X, A) can be different from 2933403., At); for instance, when A = Q and X has

several path components.

Introduction to Algebraic Topology

72

If {Xk} is a decomposition of X into its path components, then Theorem 7.2 permits the computation of Ha(X, A) to be reduced to computation of the groups H,(X,,, Ah).

Computation of Ho(X,A) Let r be the cardinality of the set of path components X,‘ of X for which X,, n A = Q.

7.3. THEOREM: Ho(X, A) is a free r-module. It has r basis elements when A 75 Q or when the group is non-augmented andA = Q . It has r — 1 basis elements when the group is augmented and A 7’: Q . This theorem yields the following corollary. 7.4. Corollary: By choice of a suitable ring R, for instance R = Z or R = Q, the number ofpath-components of X can be determined from H0(X). In any case, for an arbitrary ring R, X is pathwise

connected only if in the augmented case Ho(X) = 0. Proof of 7.3:

(1) The non-augmented case. By 7.2, X may be taken to be pathwise connected. From 30 = 0, it follows that each element of

S0(X, A) is a cycle. The bounding cycles will now be determined: (a) Let A at Q and let P e A. Each Q e X can be connected to P by a path 0. Then

30 = 03(A!) = 0(d1) — 0(do) = (Q) - (P). where (Q) denotes the 0-simplex d.J ——> Q. Furthermore,

3(«I + S1(A)) = 30 + So(A) = (Q) — (P) + So(A) = (Q) + So(A)The cosets (Q) + So(A) generate S0(X, A). Therefore, each element of S0(X, A) is a boundary, and H0(X, A) = 0. (b) Let A = Q. Let c = 2am be a l-chain. Denote the beginning and end points of a, by P. and Q” respectively. Then 3,c = Za,((Q.) — (P,)). A O-chain d = 2b,(R,) e So(X) is therefore a boundary only if 2b, = 0. Conversely, let d = 2b.(R,) e So(X) and let 2b, = 0. Choose P e X. Then d = 2bs((Rt) — (P)) = 3(21’101),

Direct Decomposition and Additional Aids

73

where a" is a path from R,i to P. The association :1 ——> 2b; defines a homomorphism of So(X) on R whose kernel consists precisely of the boundaries. Hence, H0(X) g R. (2) The augmented case. (a) Let A 9/: z . Then by 6.18, the Hq(X, A) are the same in both the augmented and non-augmented cases. They therefore have the same number of basis elements. (b)

Let A = (23. The proof can be carried out either by a

direct computation of cycles and boundaries or else as follows: The group S0(P) of a point P in the augmented case is generated by (P). From 6a(P) = a(.), a e R, it

follows that only the zero element is a cycle. Therefore, H0(P) = 0. Now let X at Q be an arbitrary space and let P e X. The exact sequence

0 = Ho(P) ——> Ho(X) ——> H0(X: P) -—> H—1(P) = 0 shows that H0(X) is isomorphic to H0(X, P). However, (X, P) was a case considered in (2a), and P lies in exactly

one path-component of X. Therefore the number of basis elements is r — l.

Homology Groups of a Point P Theorem 7.3 shows that H0(P) = 0 in the augmented case and that H0(P) ; R in the non-augmented case. The next theorem

handles all other dimensions.

7.5. THEOREM:

For q :2 o, Hq(P) = 0.

The result is already known for q s 0. For q > 0, the Proof: only singular simplex in P is aa:Aq ——> P and Sa(P) is generated by it. The boundary 3q is an alternating sum of q + 1 terms, each of which is a (q — l)-simplex; that is, each term is 04-1. Consequently, 840,, = 0 for 4 odd and 3q = “42—: for q even. When q is odd, a'q = 30‘,“ and each element of Sq(P) is a boundary. Hence, H,,(P) = 0. When q is even, 3a(aaq) = aa and am, is a cycle only if a = 0. Hence, Hq(P) = 0. 7.6. Corollary: If X is homotopic to a point, then in the augmented case Ha(X) = 0 for all q.

Introduction to Algebraic Topology

74

Examples: (1) Let X be the set of rational numbers in the usual topology. Then the individual points form the path-components. Hence, Hq(X) = 0 for q ¢ 0. H0(X) is a free module with countably many generators.

(2)

Let X be the set of points (x, y) e R2 with xy at 0; in other words, the plane without the axes. Each of the four quadrants is homotopic to a point by means of a defamation retraction. Hence, Ha(X) = 0 for q :x: 0 and Ho(X) ”=2: R + R + R in the augmented case.

Definition:

Let (X, A) be a topological space pair and let U c: A.

Then the injection i:(X — U, A — U) —>(X, A) is called an excision if and only if, in the non-augmented case,

i,.:Hq(X — U, A — U) —> Ha(X, A) is an isomorphism for each q. Notice that no assumption is made on the openness of U. For instance, i is an excision when U c A (Excision Theorem).

7.7. Lemma:

Let Vc U c A, and suppose that V defines an

excision by means of the injection (X — V, A — V) —> (X, A). Furthermore, let (X — U,A — U) be a deformation retract of

(X — V, A — V). Then U defines an excision.

Proof: Consider the injections il:(X — U, A — U) —> (X— V,A — V) and i,:(X— V,A — V)—>(X, A). By the comments preceding 5.26, (i,)* is an isomorphism. Since 1', is an

excision, (i,),. is an isomorphism. Hence, for i = i,i,, the induced homomorphism i,.. = (i2)*(i,)* is an isomorphism.

An immediate consequence of the properties of exact sequences is the following lemma: 7.8. Lemma:

In the exact sequence of the triple (X, A, B)

t.

1.

.

- - —>Ha(A,B>—>Hq(X.B>—>Hq(2cA)i>Hq-.(A,B)—>- -let Hq(. ,~) = 0 and Hq+l(-,-) = 0 for one of the space pairs. Then the two intermediate homology groups are isomorphic.

7.9. Corollaries: (1) If A is homotopic to a point and thus Ha(A) = 0 for all q in the augmented case, then H,,(X) is isomorphic to Ha(X, A) for all q.

Direct Decomposition and Additional Aids

75

(2) If X is homotopic to a point, then 3*: Ha(X, A) ——> Hq_1(A) is an isomorphism in the augmented case. 7.10. Lemma: Let M and N be R-modules and let j:M -—> N be an isomorphism. Let g:M ——> M be defined by g(x) = mxfor a fixed m in R. Let the diagram .1

M——>N

,1

l»

M—,—>N

be commutative. Then h(y) = myfor y e N.

Pr00f-'

Clearly, h(y) =J'gj“(y) =J'(mj"(y)) = mji"0') = my-

The Sphere In euclidean space R"+1 the n-dimensional sphere S" is defined by

s"={x I x e Rn+l,x:+

+xz=1}.

The northern hemisphere E2 is defined by E2={x | x e S:,x,,g0},

the southern hemisphere E! by

E2 ={x | x e S",x,.$0}, the equator by E20E3={x | x e S",x,.=0},

and the equatorial n-cell E" by E" = {x I x e R"+1,|x|gl, x” = 0}. Clearly, S" = E3: U E2, and the equator can be identified with S"".

Furthermore, S" = (a; S° = {+1, —l}, E11={+1}; E2 = 0, S' is a circle, and E1 = {xlxl = 0, |xo| g 1}. E" is also referred to as an n-ball.

Introduction to Algebraic Topology

76

Figure 12 7.11. Lemma: sion.

The injection e:(E", S"“) ——> (S", E3) is an exci-

Proof: Remove from the n-sphere S" a neighborhood, V = [xl x e S", x,, < —--.1r}, of the south pole. Since it is the intersection of S” with a half-space, V is open. By the Excision Theorem,

(5" — V, EI' — V) —> (S", E3) is an excision since V e 51‘. Let U = E2. It is sufficient to prove that (S " — U, 151' —— U) is a deformation retract of (5" — V, E! — V). For this purpose, define the projection function by p(xo, ~--, x") = (x0, ---, x,._,, 0), and for 0 g t g 1

and x e S" - Vdefine F(x, t) by x, for x e E1;

F(x, 0 =

(1 ._ t)x + tp(X) for x e E’.‘ — V. |(1 — 0): + tp(x)|’

For x e E? n (E:' — V) = S", both p(x) = x and [x] = 1. Hence,

F(x, 1) is well defined. The denominator is not zero since the vectors

Direct Decomposition and Additional Aids

77

x and p(x) have different directions for x 9E 5"”. Hence, F is continuous. For x E El‘ — V, the numerator is a point of the line

segment joining x with p(x). Division by the absolute value induces a radial projection of this point on S". Hence, F is a mapping of (S" - V) x I in S" — V. Clearly, F(x, 0) = x/lxl = x. Furthermore, F(x, 1) = x for x e E2, and F(x, 1) =p(x)/|p(x)| 6 S"" for x e E1‘ — V. Thus (E2, 5"”) is a deformation retract of (S" — V, EL' — V). Therefore e:(E2, S"") ———> (5", E2) is an excision, and e*:Hq(E1, 5"") —> HAS", E2) is an isomorphism (even in the augmented case, since E’1 — V 7": Q). The projection p of R“1 upon the hyperplane defined by x" = 0 yields a homeomorphism of both E1; and E? with E". Since the origin is a deformation retract of E", each E", E2, and EL‘ is

homotopic to a point. By Lemma 7.11, j*:Ha(S") -——> Hq(S", E2) and 8*:H,,(EI,‘, S"") ———>Hq_l(S"") are both isomorphisms in the augmented case. Hence, 3*e;‘j* is an isomorphism of HAS") on

Hq_1(S""), and therefore HAS") ; H‘,_,,(S°). The set So is the point pair 31:1 and therefore has two path-components. In the augmented theory, this means that Ho(S°) ;R and H(S°) = 0 for q 0. Consequently, HAS") ; R and HAS") = 0 for q 9b 71. The only difference between the augmented and non-augmented theories occurs when q = O. For n > 0, the number of path-components is always one. Hence, H0(S") ; R for n > 0 in the non-augmented case. Later, the Mayer-Vietoris Sequence will furnish a simpler way to compute H,,(S"). The results of these computations can be summarized in

7.12. (1)

The homology groups of S“ are as follows: Nomaugmented

Ho(S") ’=" R and HAS") ; R for n > 0.

Ho(S°) ; R (B R, HAS") for q .7: n, 0. (2) Augmented

HAS") ; R, HAS") = 0 for q 7: n. 7.13. Corollary:

Spheres of different dimensions are not homotopic

to each other, to E", or to a point.

Introduction to Algebraic Topology

78

Proof: This follows from 7.12 and from the fact that H,(E") = 0 for all q. Let f:R"+1 —-> RMl be the reflection of R"+1 in the hyperplane

xo = 0; that is, f(xo, x1, ..., x.) = (—160, x., w. x.)- For n 2 0,

f induces a function f:S" —> S".

7.14. THEOREM: If f is the reflection in x0 = 0, then in the augmented casef* :Ha(S") ——> HAS") is a multiplication by — 1 . Proof: Induction is employed here. For n = 0, S° consists of the two points P = (l) and Q = (—1). For q 7': 0, H,(S°) = 0. Hence, the assertion need only be proved for f*:H.,(S°) ——> H,(S°). In this case the cycles have the form r(P) — r(Q) with r e R. Sincef(P) = Q

and f(Q) = P, the sign is changed by f#, and hence, by f*. The diagram «EU.

HASrfil) a_., H _1(S1t)

if.

f.

Hm“) ~> H _.(s") is commutative. Then Lemma 7.10 yields the theorem for q = n + 1 whenever it is true for q = n 2 0.

7.15. Lemma: Let g be an orthogonal transformation of R"+1 and let Detg = +1. Then the induced function g:S"——>S’I is homotopic to the identity, and hence, g* = Id. Proof:

For n = 0, the only orthogonal transformations are

g = Id and g = ~Id; from Detg = +1 it follows that g = Id. Now suppose that the lemma has already been proved for n — 1 _>_ 0. Let E be the plane determined by the north pole a = (0, - - -, 0, l),

g(a) = b, and the origin, in case those are not colinear. Otherwise let E be an arbitrary plane through the three points. Let E0 be the orthogonal complement of E in 11”“. Let h be the rotation of R’l+l which leaves E point-wise fixed and for which h(b) = a. It is easy to see that h ~ Id; that is, h is homotopic to the identity. Since a is

fixed under hg, the subspace E’ perpendicular to a is mapped onto itself by hg, and so is S"", where S"" = S" n E’. From Deth = Detg = 1, it follows that Det hg = l. The determinant of the orthogonal transformation f induced in 5"" by hg is therefore also +1. By inductive hypothesis, f ~ Id, from which it follows that hg ~ Id, and then that g ~ Id. 7.16. THEOREM:

Let g:S"—>S" be an orthogonal trans-

Direct Decomposition and Additional Aids

79

formation. Then g* is obtained from multiplication by Det g. (The determinant is the only homotopy invariant of the orthogonal transformations.) Proof: By Lemma 7.15, it follows from Det g = +1 that g, ~ Id, and then that g,“ = Id. Therefore, let Det g = -—l. The function f2S’I —> Sn defined by f(xo, x,, ' - -, x") =f(—xo, x1, -- -, x") is

an orthogonal transformation having determinant —1. Hence, Det fg = +1. Hence, fg ~ Id and fig, = Id. Since f... is the multiplication by —1 (see Theorem 7.13), g. is also the multiplication by — 1. As an example, consider the antipodal function a(x) = —x. It is orthogonal; and, since Det at = (—l)"“, 05* is the multiplication by (_1)n+1_

A vector field is said to be defined on S" if and only if to each point x e S" a vector a(x) at 0 is associated such that the function

a(x) is continuous and a(x) is orthogonal to x. [If a(x) is localized at x, then a(x) is tangential to S".] In this definition, S" may not be replaced by an arbitrary homeomorphic image; only diffeomorphic

images may be employed. 7.17. THEOREM: (1) 8"” has a vector field. (2) S2" has no vectorfield (The cowlick theorem). (3) Let a(x) be a vector field on 52"“. Then the function x —> a(x)/l a(x)| defines a continuous function 32"“ ——> S2"—l which is homotopic to the identity. Proof:

(1) F0' x = (x0, xx,

xzm) e 5"“

let

a(x) = (_xh x0: -‘X3, x2, ' ' 'a x2n—l)'

From x #5 0, it follows that a(x) 7b 0.It is trivial that a(x) is continuous and orthogonal to x.

(2)

Let a(x) be a vector field on S". Let

F(x, 1) = x cos(t7r) + (a(x)/la(x)|) sin(t7r) for t e I. Since [F(x, t)| = 1, F is a mapping of S" x 1 into S". Clearly, F is continuous, F(x, 0) = x = Id x, F(x, {7) = a(x)/l a(x) I, and F(x, 1) = —x = a(x). Therefore a ~ Id, and n + l is even. Hence, n is odd.

(3)

The function F yields the required homotopy when 0 g t g gr.

Introduction to Algebraic Topology

80

Homology (iroups of Graphs By the Direct Sum Theorem (7.2), it is sufficient to consider connected graphs X. It has already been proved that each such graph is homotopic to a line segment with a finite set of attached circles Sf", i = 1, 2, - - -, r. Denote this line segment by g.

Q:

P,

Pl-H

(3600 Figure 13

The augmented theory will be used here. Since g is homotopic to a point, H,(g) = 0; and hence, H,(X) = H,(X, g) by Lemma 7.8. Between P, and PM, choose points Q, and Q. The union of the open segments Q.Q1 will be denoted by V. The Excision Theorem

shows that V defines an excision (X — V, g — V) ——> (X, g). Now denote the union of the open segments RPM by U. Since (X — U,g — U) is a deformation retract of (X — V, g —— V), U

also defines an excision by Lemma 7.7. Therefore, Hq(X, g) ; H,(X — U, g — U). For q 9t 0, the Direct Sum Theorem (7.2) can now be used to yield Hq(X -— U, g — U) = ZQHJSI“), P,). Lemma 7.8 shows that R for q = l

HASP, Pt) E HASP) = {

0 for t] at 0, 1.

Since X is pathwise-connected, it can now be seen that

Hm ;

RC—BRC—D .. - ®R(rsummands)forq=l 0

7.18. 0 in the augmented theory

H0(X) ; R in the non-augmented theory.

for q 9’: 0, q 9’: 1.

Direct Decomposition and Addition] Aids

81

If R is chosen to be a field, then H1(X) is a vector space of

dimension r. Thus, the number of circles is a homology invariant and indeed, it is the only one. In the theory of function of a complex variable, there often occur plane domains which are bounded by finitely many simply connected (usually piecewise smooth) curves as in the following figure:

Figure 14 If the boundary curves are joined as in Figure 14, then it can easily be seen that the domain is homotopic to a graph that consists of the inner boundary curves and the connecting paths. The homology

group of such a domain is therefore known by 7.17. A domain of this type is called n-tupIy connected if and only if H1(X) has n — 1 direct summands R. It should be noted that the term n-connected is reserved for another concept.

Degree of a Function f:S"——>S" Let f:S" ——> S" be continuous. Let R be the ring of integers. Then HAS") = Z, and HAS") is a free Z-module with one generator

c. For the homomorphism f*:H,,(S") ——>H,.(S"), it follows that f,(c) = mc for some m e Z. Then f*(rc) = rf*(c) = rmc. In other

words, f* is a multiplication by the fixed integer m. This integer m is called the degree off. Obviously, degfg = degf- deg g. Theorem 7.16 can therefore be written: The degree of an orthogonal transformation f:S" ——-> S“ is Det f. For a homeomorphism f:S" —> S",

Introduction to Algebraic Topology

82

degf-deg)“I = deg Id =1. Hence, deg f = i1. Obviously, homotopic functions have the same degree. By a theorem of H. Hopf, which will not be proved here, it follows from deg h = degf that h ~ f.

7.18. Lemma:

Let P be a point of S". Then S" — P is homotopic to

a point.

Proof:

Take P to be (0, 0, -~, -1). For n = 0, 5° -— P is {+1}.

Now let n 2 1. Example 8 following 5.25 showed that 5"“ is a

strong deformation retract of E2 — P since E2 -— P is homeomorphic to E" — 0. Consequently, E3: is a strong defamation retract of S" — P. In turn, the north pole is a strong deformation retract of E: and therefore of S" — P.

7.19. THEOREM: then deg f = 0. Proof:

If f:S" ———> S" is continuous andf(S”) :5 S“,

Let P e S" and P at f(S"). Then f induces a continuous

function g:S" ——> (5" — P). Since S" — P is homotopic to a point, HAS: — P) = 0, and the homomorphism g* :Hn(S") ——> Hn(S"

—P) is the zero homomorphism. Let i:(S" — P)——> S'I be the injection. Then f = ig and f, = i,..g..g = 0. Thus, deg f = 0.

7.20. Lemma: Let n 2 1, let E" be the closed ball |x| 3 1 in R", and let S"" be its boundary. Let f:E’I —> 5"" be continuous and let g be the restriction off to S"“. Then g has degree zero. Proof: Define F(x, t):S’"l X I —> S"‘1 by F(x, t) = f(tx). Then F(x, 1) =f(x) = g(x) for x e 5"", and F(x, 0) =f(0). Hence, g is homotopic to the function x —>f(0), which has degree 0 by Theorem 7.19. Therefore, deg g = 0. 7.21. Corollary:

5"“ is not a retract of E".

Proof: If f:E" —-> 5”" is a retraction, then f(x) = x when x e S"". Hence, g = Id and g... = 0 = Id. This contradicts HAS") .7: 0. There are other ways of proving 7.21; for instance, 7.22. THEOREM: Let Y e X be a retract of X and let i : Y—> X be the injection. Then i. :Hu( Y) —> Hq(X) is a monomorphism.

Direct Decomposition and Additional Aids

83

Proof: For a retraction f:Y—-> X, it is true that fl' = Id and then f*i* = Id. Hence. i* is a monomorphism since it follows from i,(c) = 0 that f,,i,.(c) = 0 and c = 0.

7.23. Corollary: If there is a q for which Hq( Y) ¢ 0 but Ha(X) = 0, Y is not a retract of X. Special choices of X and Yyield the following results: (a)

S"‘1 is not a retract of En since Hn_,(S"") #- 0 and Hfl-1(En) = 0

(b)

If S: is a subspace of S" homeomorphic to S’ for r < n, S: is not a retract of S" since H,(SI) i 0 and H,(S") = 0.

7.24. THE BROUWER FIXED-POINT THEOREM: f:E" —> E" is continuous, then f has a fixed point.

If

Proof: There is nothing to be proved for n = 0 since E° is a point. Let n 2 1. Suppose thatf(x) at x for all x e E". For x e E",

define g(x) as the intersection of S"" with the open half-line which starts at f(x) and contains x. An equation for this half-line is w =f(x) + t(x —f(x)), t> 0. The continuity of g will now be proved. Let f(x) = b and x —f(x) = a. Then, if w e S"“, w2 = a’tfi + 2to(ab) + b’ = l, where ab denotes the scalar product. Since to > 0, the only solution under consideration is

t, = %(—ab + W). This root is real, because b2 g 1. For b’ < 1, it follows that to > 0. For b’ = 1, it follows from x — b = a that ab=xb—b’=xb-—l,

and then from xb g |x|-|b| g1 that xb -1g0. From xb =1, it follows that x2 = b” = l, and therefore that

(x—b)’=x’—2xb+b’=0, but then x = b =f(x) contrary to assumption. Therefore, to > 0 even when b2 = 1. This has proved the existence of a unique intersection point of the half-line with S"". The formula shows that the intersection is a continuous function of a and b, and hence, of x.

For x e 5"", the intersection point is g(x) = x. Hence, g is a re-

Introduction to Algebraic Topology

84

traction of E" on S"". This contradicts 7.21. Therefore, f has a

fixed point. Remark: Proofs of the Brouwer Fixed-Point Theorem in analysis often involve stronger assumptions aboutf.

7.25. THEOREM: Iff and g are continuous functions on S“ into S” and f(x) ¢ g(x) for all x e S", then degf = (—1)“1 deg 3.

Proof:

Let

_ tf(X)-(l —t)g(X) F(x’t)_|t(x)—(l—t)g(x)|

.. forxeS,0gtgl.

The denominator is non-zero for all values of x and t, for it follows from |f(x)| = [g(x)| = 1 and tf(x) = (1 — t)g(x) that t= l —t and f(x) = g(x). Hence, F(x, t):S" X I ——> S" is continuous. Since F(x, 0) = —g(x) = a-g(x) and F(x, 1) = f(x), the functions f and org are homotopic. Therefore, deg f = deg atg = dega-deg g =

(—1)"+1 deg g7.26. Corollary:

If f:S" —-> S" has no fixed points, then degf =

(_1)n+1.

Proof:

Let g = Id in 7.25.

An example of a fixed-point free function is the antipodal map or. 7.27. Corollary: Iff:S" ——> S" is continuous andf(S") :1.- S", then f has a fixed point, and there is a point which is mapped by f on its antipodal point. Proof: By Theorem 7.19, degf = 0. If ,8 is an orthogonal transformation, then deg B = i1 at (—1)"“ degf. By Theorem 7.25, there is an x e S" for which f(x) = ,8(x). The choices ,8 = Id and B = a yield an x e S" for which f(x) = x and a y 6 Sn for which f(y) = —y, respectively.

The Tensor Product

Let R be a commutative ring with identity. Let A“ A,, - - -, A, and C be unitary R-modules. Let ILA, be the cartesian product of the At.

Definition:

A function f31194; —> C is multilinear if and only if

for eachj,1gjg n, and each choice ofah a2, - - -, a,_,,a,+1, ---, a",

where a, e A‘, the function f is an R-linear function sending A , into C; that is,

f(---,a,+b,, --~)=f(---,a,, -~)+f(~-,b,, ...) and f(...,ra’, ...)='f(...’aj,...)

85

Introduction to Algebraic Topology

86

for a,, b, e A, and r e R. In the case n = 2, a multilinear function f:Al >< A2 ——> C is also termed bilinear. If f is interpreted as a product, then bilinearity means that the product is distributive. The image of a multilinear function does not need to be an Rmodule. The multilinear function f is called surjective if and only if Im fgenerates C. In this case, the relation ”((11, (1,, ‘ ' “a an) =f(ral’ (1,, ’ ' ‘9 an)

shows that each c e C is the finite sum of images f(a,, a,, - - -, an). Definition:

By a tensor product of the modules A4, 1' = l, 2, - - -, n

is meant an R-module M together with a multilinear function F: ILA. —> M with the following properties: (1) (2)

F is surjective. To each multilinear mapping fILA, —> C, there is a homo-

morphism f :M ——> C with f = fF. In other words, the diagram

Huh—1M \ if C

is commutative. Obviously, f is only surjective if the homo-

morphism ] is surjective in the usual sense; that is, if f is an epimorphism. If each A, is replaced with an isomorphic B,, then M is also the tensor product of the B,. 8.1. THEOREM: Let M1, F, and M,, F, be tzvo tensor products of the modules A,, i=1, 2, ~--, n. Then F,:M,—>Ml and

F,: M1 —> MI are mutually reciprocal isomorphisms.

Proof: Fl = 1511'} and F2 = F317,, and therefore Fl = FIRE. For each c e F1(a,, ~-, an), it therefore follows that F1132c = 0. Since M is generated by such elements c, it follows that 1711::2 = Id. Similarly, if, = Id. Theorem 8.1 shows that a tensor product of the A,, i = 1, - - -, n

is unique up to an isomorphism. For the sake of convenience, the proof of the existence of such a product will be postponed. Usually one writes Al 69 ~ - - ®A,| = ®l=1 A, instead of M and a1® -- -

® a, instead of F(a,, ~-,a,,). The formula f = fF then becomes

The Tensor Product

87

f(a1, ---, an) =f(a, ® --- ® an). Since f:M——> C is a homomorphism, it follows that

1720; ®

69%) = 2f(a1, ---,an).

8.2. THEOREM (The Associative Law): If B = @311 3,, and A ® B exists, then the function

A = ®r=1Ah

FzfltA, >< H,B,——>A®B which is defined by Fall: ' ' Ham bl" ' 'abm) = (al®' ‘ '®an)®(bl®' ' '®bm)

yields a tensor product ofAl, . ~ ~, A," B1, - - ~ , BM. Proof:

(1)

The elements of A (8 B are finite sums E (1 ® b, where a e A and b e B. Here,

a=2a.®

®anandb=2bl®

@bm.

Multilinearity then yields 2a®b :2(al®"'®an)®(bl®"'®bm) =2F(ab "'sambl, "'9bm)-

Therefore, F is surjective.

(2)

Now let f:H,A, >< 11,8, —-> C be an arbitrary multilinear function. The problem is to find a suitable homomorphism of A®B on C. For [71, ~ - -,b,,, fixed, f(al, - - -,a,,,b,, - - -,b,,) yields a multilinear function of ILA, into C. Therefore, there

is a homomorphism f,,...,,_ :A -—> C with fDi--~b..(al ® ' ‘ ‘ ® an) =f(ah ' ' '9 am bl, ‘ ' ': bm)‘

For a = 2 a1 (9 - - - ® a,,, it follows that fp,...b.(a) = 2f(an ' ' ') am bl, ' ' 'a bra)-

For a fixed a, the right-hand side indicates that fb,...,_(a) is a multilinear function sending 11,3, into C. Therefore, there is a homomorphism fazB ——> C such that

Introduction to Algebraic Topology

88

hot 8» - -® b...) = f.....,_@ = Ef(a19 ' ' Ham b1, ‘ ‘ 'rbm)

When b = 2 b1 69 ~ . ~ ® b,,, e B, it follows that fa(b) = 2f(al, "'aam bl, ‘ ' 'sbm)‘

The function A x B—->C for which (a,b)——> a(b) is trivially bilinear. Hence, there is a homomorphism

s ® B ——> C for which f(al®"'®an®bl®"'®bm) =f(al, ...,ambl ... bm)

Therefore, A®B= (A,®

®An)®(31®

($3...) is

atensor product ofAl, ---,A,.,B,, ---, Bm.

8.3. Corollary:

If the tensor product of any two factors exists,

then the tensor product of n factors exists. Furthermore, the as-

sociative law holds up to an isomorphism. 8.4. THEOREM (The Commutative Law):

If A ® B exists, then

it can be employed as the tensor product B®A by means of the function F(b, a) = (1 ® b. Proof: (1) (2)

F is surjective, since A ® B is generated by the terms a 8) b. Suppose that f:B >< A—-> C is bilinear. The same f can

be interpreted as a bilinear map of A x B into C. There-

fore, there is a homomorphism f :A®B——>C such that

f(a ® b)f(b. 11)Warning! can

be

It has only been proved that, with the proper F, A ® B interpreted

as

B ® A,

or

in

other

words,

that

A 8) B _=_ B ® A. Even in the case B = A, it usually happens that a, ® a, 7'.- a2 ® a1 for al and a, in A. Otherwise,f(a., a2) =f(a,, a.) for each bilinear mappingf, and this is false.

8.5. THEOREM:

Let A = 2 $9 A. be a direct sum of an arbitrary

collection of R-modules A,. If the tensor product A. ® B exists for the R—module B and each A‘, then M = 2?“; ®B) is a tensor

product A ® B provided that a ® b is defined to be 24a. ® b) for a=2,a, e Aandb e B.

The Tensor Product

89

Proof:

(1)

Since the sum is direct, the term a® b is well defined. The function is obviously multilinear. The elements a; GE b generate A‘ ()9 B and consequently, also M.

(2) Let f:A x B ——> C be multilinear. A function f must be found. Let )2 be the restriction off to A. x B. To the bilinear

fl there is a homomorphismf¢:A‘ ® B —> C such that

74a. 03 b) = Man I)) = f(at, b)For x e M, where x = 2 x; and x, e A) 69 B, define f(x) by Suffix.) This defines a homomorphism s —> C. When a = 2a, 6 A, then a 69 b = 21at ® b) and therefore,

.701 ® 17) = 1724“; ® 11) = Edd“; ® 17) = 24 (at, 17) =f(a, 17)8.6. Corollary:

By commutativity, Theorem 8.5 also holds for

direct decompositions of the second factor. By associativity, it holds for a finite number offactors. 8.7. THEOREM:

Each R-module B is also a tensor product

R ® B ifthe definition r ® b = rb is madefor r e R andb e B. Proof: (1)

The function ®:R x B——>B is well defined, bilinear and,

as r = 1 shows, surjective.

(2) Let f:R X B ——> C be bilinear. Define f :B——> C by f(b) = f(1, b). Then f is a homomorphism, and

flu b) = rf(1. 17) = 70b) = 1706917)8.8. Corollary:

By commutativity, each R-module B is a tensor

product B ® R. By Theorem 8.5, the product A ® B exists when one of the factors is a free module (the direct sum of modules isomorphic to R). If A is free, then A 69 B is the direct sum of just as many modules isomorphic to B as there are basis elements in A; this number need not befinite. 8.9. THEOREM:

Suppose that B® M exists. Let A be a sub-

module of B and let H be the submodule of B ® M generated by all

Introduction to Algebraic Topology

90

a ® m, where a e A and m e M. Then thefactor module (B ® M)/H becomes a tensor product (B/A) 6) M by means of the definition

(b+A)®m=(b®m)+H.

8.10. Remark:

In Example (4) following Corollary 8.17, it will be proved

that in general H 7'.- A ® M. The restriction of (9:8 X M —>

B X M to A x M yields only an epimorphism of A (>3) M on H. In case A is a direct summand of B, then H = A ® M.

Proof of Theorem 8.9: The homomorphism of (B/A) X M into (B ()9 M)/H given by 8.10 is well defined and trivially bilinear. It is surjective because (B ® M)/H is generated by the images (b ® m) + H. Now let f:(B/A) x M —> C be bilinear. Define

F:B X M -—> C by F(b, m) = f(b + A, m). To the bilinear function F there corresponds a homomorphism [7 :B ()9 M —> C such that

F~(b 69 m) = F(b, m) =f(b + A, m). For a e A, this means that

F(a®m) =f(a + A,m) = 0. Hence F vanishes on H, and induces, according to Lemma 1.1, a

homomorphism f(B ()9 M)/H —-> C by means of f‘(x + H) = 17‘(x) for x e B ® M. Clearly,

f(b ®m + H) = F(b @m) =f(b + A, m)Then 8.10 yields

f((b + A) 69 m) =f(b + A, m). 8.11. Lemma: free module. Proof:

Each module is isomorphic to a factor module of a

Let x,, i e I be a generating system of the module N.

Then N = 2 Rx” where the sum does not need to be direct. Form

the free module L = 29R)!“ Define an epimorphism of L on N by y, ——> x.. Then N = L/A where A is the kernel. 8.12. THEOREM (Existence of the Tensor Product):

For each

two R-modules N and M, there exists a tensor product N 69 M.

The Tensor Product

Proof:

91

N is isomorphic to L/A, where L is a free module. By

Theorems 8.6 and 8.7, the product L ® M exists. By Theorem

8.9, (L/A) ()9 M exists and is a tensor product N ® M. This completes the proof of the existence of the tensor product of a finite set of R—modules (see Corollary 8.3).

Tensor Products of Functions

8.13. THEOREM:

Let f:A —> A’ and g:B ——> B’ be homo-

morphisms. Then there is a homomorphism f ® g:A ® B ——> A’ ® B’

which is defined by (f 69 g)(a 69 b) = f((1) 6)f(b) and which is called the tensor product offand g.

Proof:

Define h:A X B ——> A’ CA) B’ by h(a, b) = f(a) 69 g(b).

Since ()9 is bilinear, h is also bilinear. Consequently, there is a

homomorphism

h:A 69 B —> A’ ® 3’

such that h(a ()9 b) =

f(a) ® g(b)- Letf® g = 7:. The following computational rules follow easily by applying the relevant functions to a CA) b:

(a)

Forf,f,:A —+ A’ and g, glzB ——> 8’, both

(f+fl)®g=f®g+fi®g and

f®(g+g1)=f®g +f®g1-

(b) For A i» A’ J; A” and B i> B’ —"> E”, (f' (>9 g’)(f® g) = (f’1’) ® (g’g)(c)

For r e R,

r(f®g) = (00693 =f® rgRemark:

The function (f, g) ——> f 8) g is a bilinear function

Hom(A, A’) X Hom(B, B’) —> Hom(A ® B, A’ Q) B’). However, the induced homomorphism

Hom(A, A’) ® Hom(B, B’) ——> Hom(A ® B, A’ (8 B’)

is not necessarily either a monomorphism or an epimorphism.

Introduction to Algebraic Topology

92

8.14. THEOREM:

Let

A —‘—> B —’—> c —+ o be exact and let M be an R-module. Then {81

J®l

A®M——>B®M—>C®M—>0 is exact.

Proof: Since j is an epimorphism, (j ® l)(b (8) m) = jb ® m shows thatj (8 1 is an epimorphism. Furthermore,

(J'®1)(i®1)=(fi)®1=0®1=0Hence, it remains to be proved that to h e Ker(j ® 1) there is an x e A ® M such that (i ® 1)): = h. Since Kerj = M, j induces an

isomorphism s/iA ——> C by means of f(b + M) = jb. Let H be the submodule of B ()9 M generated by the set of all 1'(a) ()9 m,

a e A. By Theorem 8.9, (B® M)/H is a tensor product of B/iA

and M under the definition (b + M) ® m = b ()9 m + H. Since f“ exists,

the

isomorphisms f 8) 1:(B ()3 M) / H —> C ® M

and

f'1 ® 1: C69 M ———> (B® M)/H are mutually reciprocal. For f ()9 1, the definitions show that

(i®1)(b®m+H)=(i®l)((b+ILA)®M) =j(b+iA)®m=jb®m. For j ® 123 ® M ——> C ® M, the definition yields

(J'®1)(b®m)=jb®m. Since f® 1 is an isomorphism, it follows that Ker(j® l) = H. If h e Ker(j ® 1), it can be written as

h = 2110) 09 m = 20‘ ® l)(a ® m)Choose x = 2a 69 m to find the desired element.

8.15.i, THEOREM: y Let A L» B J» c —+ o and A’ ———> B’ —+ C’ ——> 0 be exact. Then the “nine-diagram”

The Tensor Product

93

ten m1 A®A'—>B®A'—+C®A'——>0

11,111 1,1,1 A®C’—®3>B®C’£> ®C’-—>0 10 1 1 0 0

A®B’—g> ®B’£> say—>0

is exact and each rectangle commutes. Furthermore,

8.16. (A ® B’) ea (B®A')MB® B’ flC® C’ —+o is also exact.

Proof:

Theorem 8.14 shows the exactness of the “nine-diagram.”

The commutativity follows from

(f®1)(1®g)=f®g = (1 ®g)(f® 1)The exactness of 8.16 remains to be proved. For this purpose, consider the diagram, 4 A,—+A,——>0

1w 1,,

B,—x—>Ba——>Ba—>0

1,

15

c,—'—>C, 0

whose rows and columns are exact and whose two rectangles commute. It needs to be checked that

B19142 £3,135» C,—>o is exact: Since «p and 8 are epimorphisms, 8"]; is also. Furthermore,

SMxC-Ba): WCBW =0®pya =0®0 =0. Finally, let b2 6 B2 and Sal/(b2) = 0. Because 8(1p(b,)) = 0, there

Introduction to Algebraic Topology

94

is an aa 6 A, for which Baa = 1p(b,). To as there is an a, for which qS(a,) = (1;. Hence, 111%,) = ,8¢(a2) = WG’J— Since «Mb, — a(a,)) = 0, there is a bl GB, for which x071) = b, — a(a2). This shows that

b2 is the image of (b1, a,) under X e a. 8.17. Corollary: Let A be a submodule of B and A’ a submodule of B’. Let izA —* B and i’:A’ —-> B’ be the injections, and let H be the submodule

H = (0' ® 1) 69 (1 ® i’))((A ® 3’) 69 (B 69 A')) of B ® 8’. Then,

(BIA) ® (B’/A’) ; (B ® B’)/H. Proof:

For the exact sequences

LLB—wm—w and

wiwzam'—w the exactness of sequence 8.16 provides the exactness of

(A 09 B’) G (B 69 A') ——> B ® 3’ —> (B/A) ® (B’/A’) —> 0. This means that

(B/A) ® (B’/A’) E (B (9 B’)/HHence H is the submodule of B®B’ generated by all a®b’, a e A, b’ 6 Band all b®a’, b e B, a’ e A’. In the case A’ =0 and B’ = M, the lemma reduces to Theorem 8.9. It should be

noticed again that H = (1‘ ® l)(A ® M) is not in general isomorphic to A ® M. Examples:

(1)

Let A be an R-module and .52! be an ideal, hence a submodule of R. By Theorem 8.7, R 09 A g A. By Theorem 8.9,

NW®figm®mmn

The Tensor Product

95

in which H is the submodule generated by the set of all elements 01 ® a = aa where a e .52! and a e A. Therefore, R/(Ja! Q) A) ; A/a/A.

(2)

Let .9! and Q be ideals in R. By Corollary 8.17,

RM ® RM ; R/H where

H=((i®l)®(l®i’))(.szl®R)@(R®b) =JR+wR=M+fl It follows that

R/d ® R/g = R/(Jz! + 3)-

For relatively prime ideals 4a! and .9, it is true that .52! + Q = R. Hence, R/M ® R/fl = 0.

(3)

Consider, for instance, the

case R = Z,.2/ = (2) andfl = (3). Then (Z/(2)) ® (Z/(3)) = 0. The vanishing of A ® B means that there are no non-trivial bilinear functions on A X B. Let R be a principal ideal ring. By the main theorem for abelian groups, each finitely generated R-module has the form of a finite direct sum 2,9R/(do. Theorem 8.5 implies that

(SW/(a0) ® (SEER/(181)) = 2?:R/(3u) where (8”) = (on) + (,8,) in accordance with Example 2. This

(4)

yields a survey of all tensor products of finitely generated modules over a principal ideal ring R. No such survey is known unless R is a principal ideal ring. Consider the special case where R = Z, B = Z/(4), A = 22/(4) c B, and M = Z/(2) are substituted in Example (3). Then A = Z/(Z); and, consequently, A ® M = (Z/(2)) ® (Z/(2)) = (Z/(2)). Let H be the submodule of B 69 M generated by all a 69 m, where a e A and m e M. Since a = 2b,, whereb e Z, it follows thata®m = 2b ® m = b 69 2m = 0 and therefore, that H = 0. Thus, A ® M ;/_- H.

Definition:

Let R be an integral domain. An R-module M is

called divisible if and only if to each m e M and r e R, r i 0, there

Introduction to Algebraic Topology

96

exists an m’ e M for which m = rm’. Here, m' is not necessarily uniquely defined by r and m. An R-module A is called a torsion module if and only if to each a e A there is an r e R, r .7: 0, for

which ra = 0. Let M be divisible and let A be a torsion 8.18. THEOREM: module. Then, A ® M = 0.

Let a e A and m e M. Choose r e R such that r ¢ 0 Proof: and ra = 0. Find m’ e M for which m = rm'. Then it follows that

a®m=a®(rm’)=(ra)®m=0®m=0. As an example, let Q be the Z-module of rational numbers. Since, for 0 g p/q < 1 and any integer n _—,¢ 0, it is clear that p/q = n(p/nq), it follows that Q/Z is a divisible Z-module. Furthermore, since q(P/q) E 0, Q/Z is a torsion module. Therefore, (Q/Z) ®Z (Q/Z) = 0. Here, the notation (>9,2 indicates that the tensor product depends on the ring. That this is so in the case at hand can be seen from

(Q/Z) ®o/z(Q/Z) = Q/Z q: 0. Exercise:

Show that Q®z Q = Q under the definition a®b=ab

8.19. THEOREM:

fora,beQ.

Let F be a free R-module. From the exactness

0f

o ——> A —‘—> B —’> c —-> 0 follows the exactness of £81

)8]

0—’A®F——->B®F——>C®F-—>0. If i(A) is a direct summand of B, then (i ® l)(A ® F) is also a direct summand of B ()9 F. (Hence, in this special case, H = A ()9 F.) Proof: By Theorem 8.14, it is sufficient to prove exactness at A (8 F. Let {xk} be a basis of F; then F = 29R“. By Theorem 8.5, A ® F = 29A 69 k. Hence, a e A 69 F has a representation

a = 2 a,‘ (9 xk. From (i 69 l)a = 0 it follows that 2 i(a,,) ® xk = 0.

The Tensor Product

97

By Theorem 8.7, the isomorphism M ()9 k = M is given by

i(ak) ® rxk = ri(a). Hence, for all k, i(ak) = O; and then (1,; = 0 since i is a monomorphism. It follows that a = 0. Hence, i® l is a monomorphism. By Theorem 8.5, B®F= 29(B®k)

and, by the definition of the tensor product of functions,

(1' Q) 1)(A ® F) = 290A 69 m); from which it follows by use of Theorem 8.7 that B 8) k = B and M 69 Rx,‘ = M. Hence, if i(A) is a direct summand of B, then

(i 69 1)(A ()9 F) is a direct summand of B 69 F.

The Functor Hom

It was proved in Chapter 1 that the R-homomorphisrns of an R-module A into an R-module B form an R-module Hom(A, B).

Let 4;:A’ -—> A and «s ——> B’ be R-homomorphisms. The function which associates with each f e Hom(A, B) the homomorphism «4»q e Hom(A', B’) is to be designated by Hom(¢, '4»). Since

W+ m = «m» + «m and «Hr/34> = rw¢ for r e R, the function Hom(qs, «[1‘) is a homomorphism of Hom(A, B) into Hom(A’, B’). Hom is called contravariant in the first argument and covariant in the second because Hom(4’h 1P1) Hom(‘bzs 1P2) = Hom(¢a¢l: "l’i‘l’fl98

The Functor Horn

99

Further rules for computation are Hom(¢l + ¢29 1,1“) = Hom(¢1, ‘1’) + H0m(¢3, ‘1’), Hom(¢, "P‘l + W2) = H0131“): 1P1) + Hom(¢t, 1P2):

and

Hom(r¢, 1]») = r Hom(¢, m1») = Hom(tf), mp) for r e R. Particularly important are the functions J: = Hom(4>, Id) in case B = B’, and 17» = Hom(Id, 1p) in case A = A’. For these, 4';f = f(1: and W) = 1,!»f. The rules for computation yield directly $1 + $2 = $1 + Hom(A, B”) is exact.

Proof: (1)

Iffe Ker r: then i~(f) = if: 0, and hence,f= Osinceiis a monomorphism. Therefore, INis a monomorphism.

Introduction to Algebraic Topology

100 ~ ~

N

~

(2) ji=ji=0=0. (3)

Let f6 Ker 7"; that is, j~(f) = jf = 0. The problem is to find g:A ——> B for which i~(g) = f, and therefore, ig = f. To a e A there is, since jf(a) = 0 and the given sequence is exact at B’, a unique b e B with ib =f(a). Define g by g(a) = b. Then g e Hom(A, B) and ig(a) = ib = f(a) for all a e A. Therefore,

is = f9.3. THEOREM:

Proof:

IfA = 25911., then Hom(A,B) gfl, Hom(AhB).

Denote the restriction off e Hom(A, B) to A. by f,. To f

associate the function (~ - ~ , fl - - -)6 IL Hom(Ai, B). This is an isomorphism of Hom(A, B) on H. Hom(A,, B). For to each set of functions {fi:A. —>B}, i e I, there is an f:A ——>B which is

defined by f(a) = Zf,(a‘) for a = Eai e A. Since the sum is direct and only finitely many of the a, are non—zero, f is well defined. 9.4. THEOREM:

IfB = HkBten Hom(A, B)= II,c Hom(A, 3,).

Proof: Let ps——>B,¢ be the projection on the kth component. If f e Hom(A, B), let fl. 2 pks —*B,c. To f associate (- ~ -,fk, -~ ') e Ht Hom(A,B,¢). This is an isomorphism of Hom(A,B) onto Hk Hom(A,B,‘); for, to each set of functions

{sAk-—->B}, k6 I, there is an f which is given by f(a)= (””134“), .. .)_ 1 1

9.5. Let 0 ——->A ——> A’ —> A” ——> 0 be exact and let M be a direct summand of A’. Then

0 N is called an R, S-homomorphism if and only if f is both an

R-homomorphism and an S-homomorphism. Until now the ring R has usually been fixed in considerations of A ® B and Hom(A, B), and consequently it has usually been suppressed in the notation. Where the ring matters, the notation will indicate the ring involved, as in A 69,; B and HomR(A, B).

9.7. THEOREM:

Let A be an R-module and B an R,S-modu1e.

Then A 6),; B, HomR(A, B) and HomR(B, A) can be interpreted as R,S-modules by defining for s e S:

(a) 3(2 a ®R 5) = 2&1 6912 sb), (b) (Sf)(a) = 3001)). (C) (sg)(b) = 3(sb), wherea e A, b e B,fe HomR(A, B) andg e HomR(B, A).

Introduction to Algebraic Topology

102

Proof:

Only in the case of (a) is there a problem about whether

the symbols are well defined: The function h :A x B —> A 6933

given by h(~a, b) = a ®R sb 1s R—bilinear. Hence, there 1s a homomorphism h.A 8),, B -—> A 6),, B for which h(a ®R b) = a ®R sb. Thus the symbols of (a) are well defined. That the relevant modules are S-modules is clear. That they are R-S-modulcs follows from

r(S(a @112 b)) = 70! ® sb) = m ® 317 = S(r(a 09 b)). r(bfl)(a) = r(Sf((1)) = S(rf(a))) = (3(rf))(a). and (r(Sg))(b) = (Sg)(rb) = 80019)) = 80012)) = ('g)(sb) = (5(rg))(b)9.8. THEOREM: Lets —> A’ be an R-homomorphism of the R-modules A and A’ and let n —+ B’ be an R,S-homomorphism of the R,S-modules B and B’. Then f(9,; g:A 69 B ——> A’ ()9 B', HomRU, g):HomR(A’, B) ——> HomR(A, B') and HomR(g, f): Horn3 (B’, A) —> HomR(B, A') are R,S—homomorphisms.

Proof:

The functions in question are R-homomorphisms. There-

fore, it needs only to be proved that they are S-homomorphisms: Let s e S. Then,

(f(9 3X50! 69 b)) = (f® 3%: ® sb) = f(a) 69 g(sb) = f(a) 69 sg(b) = S((f® 901 69 b)). If h’:A’ —-> B and a e A, then

(HomzzU. g)(sh'))(a) = (8(Sh’)f)(a) = 3((Sh’)(f(a))) = g(S(h’f(a))) = 5(g(h’f(a))) = S(gh’f(a)) = (3 HomRCf, 8))(h’)(a)Finally, if h:B’ ——> A and b e B, then

(Homx(g,f)(Sh))(b) = (f(sh)g)(b) = fh(5307)) = fh3(51’) = (5(fhg))(b) = (8 Homn(g.f))(h)(b)9.9. Generalized Associative Law: Let A be an R-module, B an R,S-module, and C an S-module. Then there is an R,S-isomorphism of (A ®R B) ®s C on A ®R (B ®S C) under which (a ®R b) ®s c

goes into a @912 (b (9.; c). Proof:

For a fixed c e C the function fizB ——> B ®s C, defined

by fi(b) = b ®S c, is an R, S—homomorphism. It was proved in 9.8

The Functor Hom

103

that (1 ®R fi):A ®R B ——> A ®R (B ®s c) is an R,S-homomor-

phism. By definition (1 @1212)“ 691! b) = a ®R (b ®s c).

This defines a function F:(A ®R B) 69 C —> A ®R (B ®s C). ~ Since

F is

S-bilinear,

there

is

an

S-homomorphism

F :(A (>91: B) 693 C —> A (>91, (B (>9s C), for which F((a ®3 b) ®s 0) = a ®R (b 8).? c)-

For r e R,

from: on b) ®s c) = F HomS(B @R A, C)

for which F(f)(b (911 a) = f(a)b for f: A ——> Homs(B, C), a e A andb e B. Quotient Modules The torsion submodule T(A) of the R-module A is Definition: the set of all a e A for which there is an r e R such that r at 0 and ra = 0. Clearly, T(A) is a submodule of A. In case T(A) = ()5, A is called torsion free. The factor module A/T(A) is torsion free but is not in general free. A is called a torsion module when T(A) = A.

Introduction to Algebraic Topology

104

Let R be an integral domain and K its quotient field. Since K is an R,K-module, A ®R K is an R,K-module and therefore a vector

space over K. The dimension of the K-vector space A 6),, K is called the rank of A. Let A be embedded in A ®R K under the R-homomorphism qS:A ——> A 6),, K defined by B is called an equivalence if and only if there is a «I»: 3—» A such that «M: = e4 and ab = es. If «p’qS = 5.14 and

W’ = e, hold for «p’: B ——> A then

«.0 = w = w = «We» = «VHence 1]» is uniquely determined by 4), if it exists. Then morphism «Ir, if if exists, is called the inverse of = 9.4A category I is called linear if and only if each .51 (A, B) is an additive group and the distributive laws («In + *1/94) = "P‘id’ + 111%!)

and

111(4’1 + 4):) = W1 + W2

hold for C. In a linear category,

the sets .51 (A, B) are not empty since 0 e .9!(A, B). The space pairs and their continuous functions form a nonlinear category. Examples of linear categories are:

(1) The class of objects consists of the R-modules; the morphisms are their homomorphisms. (2) The class of objects consists of the R~chain complexes of R-modules Au together with those of their R-homomorphisms 3‘, that satisfy amen = 0, where 3n:

at

.—->Aq+l——>Aa—>Aq_,——>

Let A, B be chain complexes. Then .5104, B) consists of all chain homomorphisms f# :A ——> B. Such a chain homomorphism is not a function, but a set of homomorphisms

Categories and Functors

109 3m

8!

——>Aq+1—>Aq—->Aq_1-—> if?“

if: 3m

aBiz“

lit, 8.

93¢

aBa—l

"3

where the fif are R-homomorphisms and the rectangles commute.

(3)

Each R-chain complex A has a corresponding set H(A) of homology groups Hq(A), q 6 Z. These sets H(A) are the objects. Each chain homomorphism f#:A —> B corresponds to a set f* of module homomorphisms f*a:Ha(A)—>HQ(B). The set e52((H(A), H(B)) consists of all f*.

Definition: Let .1” and 3!” be two categories. A covariant functor T (of one variable) of f in .1” is a function which associates with each object A e 1’ an object B e .1” and with each morphism ¢:A —> B a morphism T(¢):T(A) —> T(B) such that T(eA) = em)

and Tow» = T(¢)T(¢).

A contravariant functor T of .1” in .1” is a function which

associates with each object A e .9? an object T(A) e .1" and with each morphism ¢:A ——> B a morphism T(¢):T(B) —> T(A) such that T(eA) = 9N4) and TOP?» = T(¢)T(‘l’) If f and .9f’ are linear categories, then a functor T of .91” in

1" ’ is called additive if and only if

Tu» + w) = T(4>) + T01») for 4;, «1r 6 MM, B).

Application of a functor T to a commutative diagram

A —‘—> B

l~ is

at C ——> D

yields another commutative diagram. In case T is covariant, the image diagram is

N) T(A) —‘* T(B) T(nt)

C

T(B)

T (1’)

——> D.

Introduction to Algebraic Topology

110

Since

T(/3)T(¢) = T(.8¢) = Third) = T(1P)T(d), the diagram commutes. In case T is contravariant, the image diagram is T“)

T(A) (— T(B) TM)

7(5)

T(C) 353 T(D). Since

T(497'(I3) = T(B¢) = TOP“) = T(d)T(‘P), this diagram also commutes. Example:

Let .9!” be the linear category of Example (2) and let

X’ be that of Example (3). The mapping T which associates with each chain complex the set of its homology groups, and with each

f# its image T(f#) = f,' is a covariant, additive functor of 1’ in X’: For by Lemma 1.2,

T(Id) = Id, T(g*f*) = (3‘7“). = gift = T(3*)T(f*), and

T(f# +13") = T(f*) + T(s*)Definition:

A sequence - ~ - Ag.l is associated a T(aq). When T is covariant, T(a,,):T(A¢) —> T(Aq_l); when T is contravariant, T(aa):T(Aq_,)—>T(Aq). Since T is additive, T(a,,_,aa) = 0. When Tis covariant,

T(3q_,)T(aq) = T(3a-1 a) = 0. Hence, T0041)

1'0.)

' —" 1"(14q4-l)——> T(Aa) ”—> T(A'4a—1)'—> ‘ ' ‘ is an R-chain complex T’(A). When T is contravariant, T(aq)T(aa—l) = T(aa—laa) = 0

and, therefore, T(3lol)

1'(a

- B, is exact. (2)

Let Al AA AB, be exact. Let a e A, where p1(a) = 0

and p,(a) = 0. There is an al 6 Al such that a = i1(al). From 0 = pli,(al) = k1(a1) it follows that (11 = 0. Therefore, a = 0. For (x = 1, 2 it is clear that

had — ilkr‘p. — izkr‘paxx) = pa(x) — p..(x) = 0, where x e A. Therefore, Id = ilkf‘pl + izkg‘lpz. 10.5. Lemma: The sequence 0 ——> A 4 B -p—> C ——> 0 is exact, and i(A) is a direct summand of B if and only if there exist homomorphisms i’ and p’ such that the diagram A

y’

p

C

rd l\ B /l Id

A '

:’\C

commutes and the equations p’i’ = 0, pi = 0, and ip’ + i’p = Id hold. Proof: ‘ P (1) If such i' and p’ exist, then A ——> B ——> C is exact by Lemma 10.4. By Lemma 10.3, i is a monomorphism, i(A) is a direct summand of B, and p is an epimorphism.

Introduction to Algebraic Topology

114

(2)

Let 0——>A—‘>B—p+C——>0 be exact, and let B=

i(A) 6-) B’. Since i(A) = Ker p, B’ is mapped isomorphically on C by p. Now define p’(i(A) + b’) = a and i’(c) = b' when p(b’) = c. Then it is easy to see that the triangles are commutative and the equations are fulfilled. 10.6. THEOREM:

Let A, B, and C be S-modules. Let 0 ——> A

—‘—> B —n—> C ——> 0 be exact and i(A) be a direct summand of B. Let T be an additivefunctor of the S-modules into the R-modules. ()

If T is covariant, then 0 —> T(A) 2» T(B) 1")» T(C) —+ o is

exact and T(i)T(A) is a direct summand of T(B). If Tis contravariant,

then 0 T(B) ——> T(C) ——>0 is exact, and that T(i)T(A) is a direct summand of T(B). If T is contravariant, it follows that

T(i)T(P) = 0, T(i')T(P') = 0. T(i)T(p’) = Id, T(i')T(p) = Id, T(p')T(i) + T(p)T(l") = IdLemma 10.5 then shows that

N) m) 0 (Y, B) the set {T’(f*)},, which is abbreviated f*. The elements off. are the R-homornorphisms

fan = (T'(f*))*a=Ha(X, A) —> HAY, 3)For covariant functors T, the induced functor of .1”, in .1”. is covariant.

Cohomology:

Let T be contravariant. Then T’ is contravariant,

and so also is the induced functor of at”, in 1’; obtained by composition. The induced functor associates with the space pair (X, A) the set H(T’(s(X,A))), which again is abbreviated H(X, A). The

elements of H(X, A) are the cohomology groups Ker T(3q+,)/Im T(3.), and are denoted by H°(X, A). The functor of 3!; in 3?. associates

with the continuous function f:(X, A) —> (Y, B) the set (T’(f#))* which is abbreviated f*. The elements of f* are the R-homomorphisms

(Tl/‘0)" =f*"=H"( Y, B) -—> ”‘0’, A). The indices have been written as superscripts to indicate that the functor of 3?, in Jflis contravariant.

Homotopy Theorem:

Let x,p:(X, A) —> (X x I, A x I) be the

familiar functions )t(x) = (x, 0), p(X) = (x, 1). Formula 5.13 shows

that the induced V and p,* are chain homotopic. By Theorem 10.2, T’(7t*) and T’(,u,*) are also chain homotopic. If T is covariant,

Categories, Functors, and the Singular Theory

117

Theorem 5.4 shows that M, = a,” where h“, p*q:H,(X, A) —-> H,(X X I, A X I). If T is contravariant, Theorem 5.4 shows that

7x” = a“, where N", [i*":H‘1(X X I, A X I)—> H°(X, A). The considerations preceding Theorem 5.4 now show: 11.0. THEOREM:

If f,g:(X, A) —-> (Y, B) are homotopic, con-

tinuausfunctions, then f* = g* andf* = g*.

Excision Theorem:

Let U c A and U c A’. Let i(X— U, A »- U)——>

(X, A) be the injection. The proof of the excision theorem (6.17) contains the construction of a chain homomorphism T(SKA, 3)) L’ T(Sa(X: 3)) —-> T(Sq(X, A) —> 0If T is contravariant, the same sequence is obtained except that the

arrows are in the opposite direction. In the case of homology, Theorem 1.3 yields the exact sequence 1.

I.

3.

' ' '—)Hu(A! B)—_>HG(X’B)—)HQ(X’A)—_>H —1(A: B)_—> ' ' '-

Introduction to Algebraic Topology

118

In the case of cohomology it yields the exact sequence O

5.

~ ~ - Ham, B) 3— 11004 B) (X’, A', B’) be continuous. Each rectangle of the diagram

0 —> SM. B) 4'» S..(X, B) 3;» S..(X, A) ——> o

1..

1»

)v

1

,. ,.. 0 —> Sq(A’, B’) —i——> Sq(X’, B’) ———> SQ(X’, A’) —> 0 is commutative when g#, h#, and k" are the functions induced by f. The commutativity is preserved under an application of T. Now Theorem 1.4 applies to yield, when Tis covariant,

- -—>H.(A,B) 3» Hq(X,B) 1» H.(X,A) 1» H.-.(A,B) —+ - - - --—+H,,(A',B')—"—>Hq(X',B')i>Ha(X',A')—“+H _1(A’,B’)—>- - . in which all rectangles are commutative. A contravariant functor T yields an analogous diagram in which the arrows run from right

to left, and the stars on the functions are written up rather than down. This completes the proof of the fundamental theorems for the homology and cohomology groups of the singular theory defined in terms of T.

The Functors 69 M and flom(

,M) in the

Singular Theory Let S be chosen to be the ring Z of integers. Then the modules Sa(X, A), q20, of the singular theory are free abelian groups generated by the q-simplices a’ for which [0| c X, lal it A. LetR be a preassigned ring with unit, and let M be a preassigned Rmodule.

Categories, Functors, and the Singular Theory

119

Homology For the functor T choose ®z M, that is, if C is a Z-module,

then T(C) = C ®z M. For a Z-homomorphsimf:C —-> D, it follows that Tm=f®zltC®zM-+D®2M.

Since M is an R, Z-module, then T(C) is an R-module and T(f) an R-homomorphism. T is a covariant additive functor of the category of Z-modules in the category of R-modules. With T is associated the covariant functor T’ of the category of Z-chain complexes into the category of R-chain complexes. The space pair (X, A) then

has an associated R—chain complex consisting of the R-modules SAX, A) = T(Sq(X, A» = Sq(Xa A) ®z M,

and the boundary operators an ®z 1. Clearly, Sq(X, A) = 292,, and therefore, by Theorem 9.4, 3",,(X, A) = 29(ZU) ®Z M, where the sums are taken over all

q-simplices a with |a-] c X, |a| q: A. Here, it follows from 8.7 that Z, ’5 Z and (2,) ®Z M ’5 M. The elements of §¢(X, A) can therefore be written as finite sums of the form 2 mao', where mm is defined by mm = a" ®z m, for m, e M. It is clear that computa-

tions are to be made with these sums according to the usual rules for sums and products. The boundaries of such elements are found from (3 ®l2 0' ®zmv) = Z ((317) ®zma) = 2 m,30'

and are in $2-10!, A). Corresponding to a continuous f:(X, A) -—> (Y, B) there is an f“ and a T(f#) =f®z 1. For these, the definition off# yields

(f# ®21X2 me) = E (f*0 (9 ma) = 2 mafia)This proves that the old singular theory of Chapters 4-7 still holds if R is replaced by an R—module M. The choice M = R shows that it would have been sufficient to develop the old theory for Z.

However, no simplifications would have occurred thereby.

Introduction to Algebraic Topology

120

Great care must be exercised in the transition from the results

of the old theory. It is worth heeding the warning: In general, HflX, A) ®z M is not isomorphic to HflX, A). The details can be seen in the book, Homology Theory by S. McLane, under the

heading of ”Universal Coeflicient Theorems.” A review of the proofs in Chapter 7 will yield the following:

11.1. Direct Decomposition Theorem: Let {Xk} be the set ofpathcomponents of X; let A c X and A, = A n Xk. Then, for nonaugmented homology,

ZQHAXM A.) = H,(X, A). 11.2. Computation of Ho(X, A): Let r (0 g r _ D there corresponds T(f) = Homz(f, l): Homz (D, M) —> Homz(C, M) where Homz(f,1)(¢) = d)f. Since M is an R, Z-modulc,

T(C) is an R-module and T(f) is an R-homomorphism. By Chapter 9, T is a contravariant additive functor of the Z-modules into the R-modules. With T is associated the contravariant functor T’ of the Z-chain complexes in the R-cochain complexes. The space pair (X, A) then has an associated cochain complex

--- S4,(X, A) was associated with the injections ik:(X,c, A) —-—> (X, A). Theorem 9.3 shows that there is a chain isomorphism Sq(X, A) —> H S°(X,,, A). From this follows

11.6. THEOREM:

The induced homomorphisms it:H“(X, A) —>

H°(Xk, A) form a projective representation of H“(X, A) as a direct product (in the non-augmented case).

Computation of H°(X, A): Let X be pathwise-connected (for example, let X be a region in the complex plane and let M be the complex numbers). A O-cochain «p is an arbitrary function on the O-simplices with values in M. Its coboundary 3‘1’ is a function on the l-simplices; that is, the paths 0 in X. Let P be the initial point and Q be the end point of 0'. Then

(3'1PXU) = Marv) = MQ) - MP)Since X is pathwise-connected, it follows that «p e Ker 8‘ is equivalent to 1]»(x) = constant.

Introduction to Algebraic Topology

122

In the non-augmented theory, S_,(X) = 0 and therefore, S"(X) = 0. For A 7: 0, it follows that 8°(S"(X)) = 0, and then that H°(X) = Ker 8l/Im 8° ’5 M. Now let A c X, where A at 0. Let P be a point of A. To each point Q e X there is a path fromPto Q, say a'. If «p is a O-cochain, then (81¢)(a) = 1]!(Q) since P e A, and therefore, (P) e S,(A). It follows from 1]" e Ker 8‘ that 1’1‘Q = 0 for each Q e X. Therefore, 1]» = 0 and H°(X, A): 0.

11.7. Cohomology Groups of a Point P: It was just proved that H°(P) ; M in the non-augmented case. In the augmented homology, 80m, = (.) and then (8°¢)(ao) = q5(.). Therefore, 8° is an isomorphism. This means that all O-cochains are coboundaries, and hence, H°(P) = 0. If q < —1, the cohomology groups are zero by

definition. When q = —l, the fact that 8°:S“(P) ———> S°(P) is an isomorphism means that Ker 8° = 0, and hence, that H“(P) = 0.

Now let q > 0. By the proof of 7.5, 8,0, = 0 when q is an odd integer, and aqua = up, when q is even where oqq —>P. When q > 0 is odd, it follows that (804904 2 Mauve) = 0’ and hence, that 84 is the 0-homomorphism. When q > 0 is even, it

follows that (8"¢)(¢r,,) = 4100-1); i.e., that 8" is an isomorphism. When q > 0 is even, it then follows that

H“(P) = (Ker 8"“)/(Im 8)” ”=1; M/M = 0. When q > 0 is odd, it follows that H¢(P) = 0/0 = 0. H°(P) = 0 when q 72 0. 11.8. The Cohomology Groups of S":

Thus,

The proof of 7.12 remains

valid except for the direction of the arrows. This means, in the augmented case, that H"(S") g M and H°(S") = 0 when q at n. In the non-augmented case, H°(S°) ; M + M;

H°(S") ; M and

H"(S") ; M when n > 0; and H¢(S") = 0 whenq ¢ 0, q ¢ n.

Application of Homology to Function Theory The argument of a complex number 2 = r(cos a + i sin a) at 0 is a name of the angle a = arg z. It is only determined up to a multiple of 27:. A line g in the complex plane C defines two half-

Categories, Functors, and the Singular Theory

123

planes H and H1. Choose 2’ e H and z: e H1 so that the vector

2; — z’ is perpendicular to g. The choice of a fixed value B for arg (z: — 2’) permits the determination of the value of arg (zl — 2), 21 6 H1, 2 e H by means of the statement

B — 7r/2 < arg(zl —— z) < B + 7r/2. Then arg(zl — z) is a continuous function of z e H and 21 6 H1.

The Winding Number: Let azl —> X be a (continuous) l-simplex. Let 20 be a point not in la] whose distance from lal is p. Since |a-| is compact, there are numbers {a} such that 0 = to g t, g - - - g t,, = l for which the variation of a'(t) on each subinterval

I, = {t|t e 1, 1,4 g t g t,} is less than p/2. Choose a point 00,) e 1,. Erect on the line segment from 2, to 00,) a perpendicular line g, whose distance from 20 is p/2. Then the points a(t), where t e I, and the points z, for which [2 — zol < p/2, lie in distinct half-planes with respect to g,. Therefore, when t e I, and 2 satisfies [2 — zol < p/2, there is a continuous function of t and 2, say a,(t, 2), whose value at (t, z) is an argument of 0(t) — 2. By addition of a suitable multiple of 27:, it is possible to arrange that a,+,(t,, z) = a,(t,, 2), so that there exists a composite function d(a', t, 2) which is continuous in (t, 2) when t e I and |z — zol < p/2, and whose

value is an argument of 0(t) — 2. For a second function B(0, 1, z) with the same properties,

a(a-, t,z) — B(a, t, z) is an integer multiple of 27:. Since the difference is continuous in t, it must be constant. Consequently, a(a-, t,z)is uniquely determined up to a constant 27m. Denote the uniquely determined

value of d(—, 1, z) — a(—, 0, 2) by V(0', z). This is called the variation in the argument of a- with respect to z, and is a continuous function of z. V(a-, z) is only weakly dependent upon the parametri-

zation of the set 10-]. To see this, let h:I —>I be a continuous, strictly monotone, increasing function. Then V(ah, z) = V(a, 2), because oz(a, h(t), 2) can be taken to be a(ah, t, 2). Let 0 = to < tl < - . - < tn = 1 be a subdivision of I into inter-

vals I, = {tlt e I, t,_l g t g t,}. Choose a continuous, strictly monotone, increasing function h, of I on I, and let a, = «#1,. The simplices a, will be called a subdivision of 0'. For the function a(o-,, t, 2) it is possible to use a(a, h,(t), 2). Therefore, 7|

V(a-, z) = (1(0, 1, z) —- a(o-, 0, z) = 12:]! (a(o-, t,, z) —- a(a-, t,-,, 2))

= ; Von, 2).

Introduction to Algebraic Topology

124

The V(c,, z) are not dependent upon the particular choice of h, since for another h,, say h}, h;‘h§:l ——-> I is a strictly monotone, increasing function. Let A be a closed set disjoint from lol, and let p > 0 be its distance from lal. Let {0-,} be a subdivision of a such that each |a,| has diameter < p/2. If 2 e A, V(c,, 2) depends only upon the points 0(1) and a-(t,-,) and is less than yr. The line segment connecting a(t,) and a(t,_.) is the carrier of a 1-simplex 7,. Clearly, V(cy,,z) = V(a,,z). Hence, if ry is the polygonal line formed by assembling the y,,

V(% Z) = Ell/('71, Z) = 2V0?» Z) = V(c, Z)Thus, the variation in the argument of a" is the same as that of a sufficiently fine inscribed polygon. If c is a l-chain, c = mla, + - - ~ + m,a,, m, e Z, then V(c, z) is defined by V(c, z) = m,V(¢r., z) + - - - + m,V(o-,, 2)

when 2 ¢ |c|. If P. is the initial point, and Q. is the terminal point of 0., then

V(c, Z) = 2m4(arg(Ql — Z) — arg(P: — Z)) up to a multiple of 27:. Now let c be a cycle. Then

30 = t((Qt) — (P0) = 0. Thus, if the weights are counted, each end point is as often a Q, as it is a 1". Hence, V(c, z) = 0 up to a multiple of 27:. If e is a cycle, the integer W(c, z) = 2—17; V(c, z)

is called the winding number of c with respect to 2. If a cycle is replaced by the cycle formed by a sufficiently fine inscribed polygon, then the winding number is not changed. Now let c be a cycle whose carrier consists of s line segments. Each segment defines a line, and, thereby, two open half-planes.

A point which does not lie on any of the lines is in s of the half-

Categories, Functors, and the Singular Theory

125

planes. Their intersection, being an intersection of convex sets, is convex, and is a polygon in case it is bounded. Cut this intersection into triangles and a possible unbounded region by the diagonals from one of the vertices. The plane is then divided into a definite set of triangles and unbounded regions by means of the lines and the chosen diagonals. Let .1? be one of the triangles or unbounded regions. Since .1! is connected, and the integer W(c, z) is a continuous function on .1! , it is constant there. Let .1! be unbounded. When 2 e .1! and {2| is large, then V(a, z) is very small. Therefore,

W(c,z) is very small, and is consequently equal to 0. Let 11,, i = 1, 2, - - ~, r, be the triangular regions for which W(c, z) = w, :fi 0 for z 6 all“ Choose a 2-simp1ex A, for which .1!, is the interior of IA, |. The orientation of A, is to be chosen in such a manner that W(BA,, z) = 1 for z 6 .ll,. Now form the new chain c’ = c —- w,3A, —— - - « —— w,3A,. That W(c’, z) = O for all 2 which are not on one of the boundaries of the

decomposition of the plane can be seen as follows: If 2 lies in none of the lA,|, then W(c, z) = 0 and W(aA,, z) = 0, and therefore W(c', z) = 0. If 2 lies in the interior of IA, I, then W(c, z) = w,, W(aA,, z) = 1, and W(aA,, z) = 0 for j 75 1'. Hence, W(c’, z) = 0.

The carrier of c’ consists of line segments. Subdivision of c’ yields a chain c” such that the interior of each line segment is free of end points of other segments. Let 3,, i = 1, 2, - ~ -, m be the dis-

tinct segments of c”. Let n, be the sum of the weights with which the simplices having 5, as carrier occur. Choose points 21 and 2,

very near an inner point of s, but on opposite sides of .91 in such a manner that the segment connecting them meets only 3, and none of the other 3,. Then V(s,,z,) — V(s,,z,) is very small for i 7: l

and near i27r for i: 1. Therefore, W(cj’z.) — W(ci’z,) is approximately im. It was previously shown that W(ci’ 2,) = 0 = W(cfi’ 2,). Therefore, n, = 0, and similarly, n, = - ~ - = n,,, = 0.

It has therefore been proved that each segment of c” is traversed just as often in the one direction as the other. Now let X be a preassigned open subset of the plane. A chain of X shall mean a chain c for which |c| c X. A cycle c of X is called homologous to 0 (in symbols, c ~ 0) if W(c, z) = 0 for every point z of the complement of X. Suppose that c ~ 0. Then each suf-

ficiently fine polygonal cycle inscribed in c is homologous to zero. If c’ arises from c by subdivision or by a reparametrization, de— scribed before, then c’ ~ 0 also. Now let c be a cycle whose carrier

Introduction to Algebraic Topology

126

consists of line segments and let c ~ 0. Consider the corresponding triangles A“ Let 2 6 MA. If 2 e |c|, then 2 e X; otherwise, W(c, z) is defined and constant in a neighborhood of z. This neighborhood also contains inner points of IA]. Hence, W(c, z) = w, 72 0. From c ~ 0, it follows thatz e X. Therefore, |A,| c X.

Let c be a chain of X with a polygonal carrier and let c ~ 0. Then there is a representation c = c’ + wlaA1 + + w,6A,, where A1, - ~ -, A" and c’ are chains of X. After a subdivision, each segment of c’ is traversed as often in one direction as the other. Hence, c’ is a boundary, and c is also a boundary. Conversely, each

boundary is homologous to zero, as is easy to see. This justifies the definition of the word homologous. Integration:

Let f(z) be a continuous, complex-valued function

defined on X. A l-chain is called rectifiable or a path of integration if and only if the carrier of its simplices is rectifiable. A line integral

I,f(z)dz is associated with each 1-simplex a' of X. If the path of integration is c=2m,a,, the integral c(z)dz is defined by

Icf(z) dz: 2m,]',(f(z) dz. This integral is not changed by subdivision of c or by monotone changes of the parametrization. A path of

integration is called closed if c is a cycle. 11.9. CAUCHY INTEGRAL THEOREM: Let f(z) be analytic in X and let c be a closed path of integration in X that is homologous to zero. Then

fcf(z)dz = 0. Proof: Let A be a 2-simplex whose carrier is a triangle contained in X. Many books on function theory show that jaAf(z)dz = 0. Now suppose that c is a cycle of X, and that c ~ 0. The value of the integral is the limit of the integrals over sufficiently fine polygonal cycles that are inscribed in c. Therefore it is sufficient to assume c to be polygonal and homologous to zero. From the representation c = c' + Swim. and [M‘f(z)dz = 0, it follows that fcf(z)dz = fc,/(z)dz. In a suitable subdivision of c’, all the segments are traversed equally often in both directions. Hence, Ic,f(z)dz = 0. Consider as an example the case f(2) = 1/2. This function is analytic for z 72 0. Let X consist of the plane C from which the origin has been deleted. Let K be the unit circle parametrized by

Categories, Functors, and the Singular Theory

127

cos (2m) + i sin (2m), t e I. The value of d(k, t, 0) can be taken to be 2m, from which it follows that W(K, 0) = 1. Let c be an arbitrary cycle of X and W(c, 0) = w. Then c — wK = c' is a cycle for which W(c’, 0) = 0 and c’ ~ 0. Therefore, fa, (l/z)dz = 0. Hence,

1 1 . [07:12: wjx7dz= 2mW(c,0). Cohomology:

From the de Rham Theorem (not to be proved

here) it follows that the cohomology theory is not affected if considerations are restricted to rectifiable l-simplices. Instead of the set of all complex-valued functions defined on the chains, it is sufficient to deal only with those functions (15(0) that can be written ¢(a-) = j',f(z) dz wherefis continuous on X. The functionfis uniquely determined by ¢. A computation of H1(X) = Ker Sz/Im 8‘ will

now be carried out: The statement, 4) e Ker 82 is equivalent to 8395(7) = ¢(87) = 0 for all triangles 'r with rectifiable sides. This is

equivalent to j,,f(z)dz = 0 and thus to the analyticity of f on X by the Theorem of Morera. Hence, (1) e Ker 32 if and only if f(z) is analytic in X. A «p e S°(X) assigns to each point P of X [actually to each 0-simplex (P)] the number MP). Let a be a rectifiable

l-simplex with initial point P and end point Q. Then (S‘pa) = «14310): 11/(Q)—— MP). Hence, 4) e 3‘(S°(X)) if and only if there is a 1]» e S°(X) such that f,f(z)dz = 111(Q) — MP); that is, if the integral is independent of the path connecting the points P and Q in X. This in turn means that [cf(z)dz = 0 for all closed paths of integration.

Let ¢ 6 Ker 8’; i.e., let f(z) be an analytic function on X. To each closed path of integration c associate a complex number

Res,(c) = (1/27ri)_|'cf(z)dz. The function Res, is called the residue function off. If g is also analytic and Resg = Resf, then Res,-g(c) = 0 for all closed paths c in X, and then 4) = J'f(z)dz and X = J’ g(z)dz differ only by a coboundary 8M1». Conversely, if 95 and x differ only by a coboundary, then Res,(c) = Resg(c) for all c. Therefore, the groupH‘(X) is isomorphic to the set ofresiduefunctions.

Remove a collection of r compact connected subsets from a simply connected open subset of the complex plane. The resulting

region X is said to have r holes. In each hole choose a point, and let xk be the chosen point in the kth hole. Let ck be a closed path of integration in X which has the winding number 1 with respect to

x,, and 0 with respect to x, for j 7': k. Let c be an arbitrary closed

128

Introduction to Algebraic Topology

path of integration in X with the winding number w. about x‘. For c’ = c — w,cI — - - - — w,c, it follows that W(c’, 20) = 0 for each

20 ¢ X. From the Cauchy Integral Theorem it follows that

c(z)dz = w, Ic!f(z)dz + ... + w, J‘c'f(z)dz. Consequently, Res,(c) is completely described by the r numbers (1,, = fc.f(z)dz. Conversely, to each set of r arbitrary complex numbers 01,. there corresponds the function g(z) = (l/27ri)2(o£,,)/(z — xk) which is analytic in X and for which J'ckg(z)dz = ak. Thus, HI(X)

is isomorphic to the direct sum of r summands C. The relations 7.18 say the same about H1(X). The proof at that point can be carried over here, for X is homotopic to a segment with r attached circles, and the cohomology groups of a circle are known by 11.8.

12 Axioms for Homology and Cohomology

In addition to the singular theory, there are other theories for homology and cohomology. The following axioms (of Eilenberg and Steenrod) are weakened theorems of the singular theory which

also hold in other theories. Let .1” be a category of certain space pairs and continuous functions satisfying the following conditions: (1) If (X, A) e 1’, then the pairs (X, a) = X, (A, Q) = A, (X, X), (A, A), and (Q, 25), together with all injections (2)

between them, lie in at”. If (f:(X, A) —-> (Y, B)) E .1”, then .1” contains all the functions induced byf on the pairs mentioned in (l). 129

Introduction to Algebraic Topology

130 (3)

If (X, A) e X, then 1’ contains (X x I, A x I) and the two

(4)

functions defined by Mx) = (x, 0) and p,(x) = (x, l). or contains a space Po consisting of a single point. If X e .91” and P is a singleton space in 1’, then each function P —> X is in Jt’. The space pairs and functions in at” are called admissible. A

space triple (X, A, B) is called admissible if and only if (X, A), (Y, B), and (A, B) are admissible space pairs. Examples:

Until now, consideration has been given to the set of

all space pairs and continuous functions. They satisfy the conditions. Another example is the category of all pairs (X, A), in which X and A are compact hausdorfi‘ spaces, together with the continuous functions between these pairs. A homology (cohomology) theory on .1” associates with each ring R, each space pair (X, A) e .1”, and each integer q an Rmodule H,(X, A) (H"(X, A)). Each continuous function f E at”, f:(X, A) —> (Y, B) has an associated R—homomorphism fl. :Hq(X, A) ——> H,( Y, B) 0* :H”( Y, B)-—> H“(X, A)). Furthermore, there is to each admissible space triple (X, A, B) on R-homomor—

phism 3*:Hq(X, A) ——> H,_,(A, B) (8*:H"“(A, B) —-> H'I(X, A)). For these homomorphisms, the following seven axioms must be satisfied:

1.

1d,, = Id (Id* = Id).

11- (id): = 3M: ((gf)* =f*g*)III.

The first two axioms state that homology theories are covariant functors and cohomology theories are contravariant functors. Iff:(X, A, B) —> (X’, A’, B’), then 8* (8*) commutes with the induced homomorphisms of the homology (cohomology) groups of the space pairs.

IV.

The sequence l.

i.

o

. —+ H.(A, B) —» 11.04 B) —+ H.(X. A) :9 11.01. B)- - l‘

'

o

(- - - C’ and «p’ ——> C be homomorphisms with pi]; = Id. Then «p is a monomorphism, p is an epimorphism, and

Introduction to Algebraic Topology

132

12.2.

C = Ker p 6-) Im «Ir = Coker «p 6-) Im 1]».

Let izKer p ——> C be the injection. If )t’ is the function Proof: given by N = Id — «1»p —-> C, thenph’ = p — Np = 0. Therefore, it is possible to define a homomorphism NC —> Ker p by

)t(x) = 7U(x). Since Nil» = «If — «ppap = 0, it follows that Mp = 0. Hence, the assumptions of Lemmas 10.4 and 10.5 are satisfied for the diagram

his Geri,

Kerp/‘7

‘5

C’

This yields an injective and a projective representation of C as a direct sum of C and Ker p. By Lemma 10.5, the diagonal from C’ to Ker p is exact. Since «I» is a monomorphism and 7t is an epimorphism, it follows that Ker p = C/¢(C’) = Coker «1r. 12.3. THEOREM:

Let p ——> C’ and k:C’ —-> C be chain

homomorphisms with pk = Id. Let A, = Ker pq and B, = kq(C;) be the direct summands of Ca found in Lemma 12.1. Let C be exact with boundary operator d. Then A and B are exact chain complexes under

the restrictions of a. Proof:

Designate the boundary operator of C’ by [3. Since

p(d(Aa)) = B(p(Aq)) = 0, it follows that

01(Aq) c Ker pa.l = A”-.. Hence, a induces a homomorphism azAa—>A,_,. Similarly, it

can be proved that B is a chain complex. Now let C be exact. Then to each a e A, for which 5(a) = 0, and hence, a(a) = 0, there is a c 6 Ca“ such that ot(c) = a. However, c = a’ + b where a’ e A,+1 and b 6 Ba“. Since a(c) = a and a(a’) e A", it follows that a(b) e A, 0 Ba, that a(b) = 0, and then that a(a’) = a(c) = a. Therefore, A is exact. The exactness of B is proved analogously. 12.4. Corollary: lfp:(X, A) —> (Y, B) and k:( Y, B) —> (X, A) are admissible and pk = Id, then the exact homology sequence

Axioms for Homology and Cohomology a.

133

1.

'.

- —> Hq(X, A) —+ H..(A) —+ Hm —’—> Hq(X, A) —> - - splits into two exact sequences: The one consisting of the kernels of p,“ and the other consisting of the images ofk,“ The exact cohomology

sequence splits into two sequences: One consisting of the cokernels of k* and the other of the images ofp*.

Proof: From pk = Id, it follows that p,,k* = Id or k*p* = Id, as the case may be.

Application: Let P e .1" be a singleton space. For each space pair (X, A), the function k:(P, P) ——> (X, A) is admissible by (l) and (2). Now let the function p:(X, A) ——>(P, P) be admissible. Since pk = Id, the corollary can be applied. Since Hq(P, P) = 0 for all q, since H0(P) = M and since Hq(P) = 0 for q 75 0, the exact sequence of Ker p* is the homology sequence of (X, A) except for two locations. For p:X ——> P, let Ker p* = Ho(x). This yields the

so-called reduced homology sequence

- -—>H.(X.A)3'+Ho(A)i>H.(X)—"—>Ho(x,A)i>H-.(A)—>- - -. In cohomology, H°(X) is defined to be Coker k“ where k:P ——> X;

and the corollary yields the reduced cohomology sequence: at

_

4-

_,

r

8‘

- - 17.,( Y) and f*:H°( Y) —-> H°(X), respectively. Proof: Let c e H0(X) = Ker 12*. Then 0 = p*(c) = p;f*(c) and [*(c) 6 Ho( Y). Since f*p’*(M) c p'(M), Lemma 1.1 shows that f*

Introduction to Algebraic Topology

134

induces a homomorphism

of H°( Y) = H°( Y)/p’*(M)

into

H°(X) = H°(X)/p*(M)-

Thus, if the category .1” is restricted to those space pairs for which the function (X, A) —-> (P, P) is admissible, an augmented theory has been obtained for the restricted category. In this theory,

H°(P) = 0 and H°(P) = 0 since, for p = k = Id: P ——+P, both Ker p* = 0 and Coker p* = 0.

In the non-augmented singular theory, it was proved in 7.2 and 11.1 for the path-components X, of X and A, = A r) X), that 2,?H,(Xk, Ax) = H,(X, A). Cohomology has an analogous formula. In the axiomatic theory, only a weaker theorem can be proved.

12.6. THEOREM: Let X = UL; X, be a finite decomposition of X into disjoint, open and closed subspaces Xk. Let A e X and A n Xk = At. The resulting space pairs and functions are admissible. Let i,:(X,,, A,) ——> (X, A) be the injections. Then the (i,),,, yield an injective representation of H,(X, A) as a direct sum of the H,(X,,, A) and the it“ yield a projective representation of H“(X, A) as a direct product of the H“(Xb Ak). In particular,

Ha(X. A) E 26’ 110(s Ale) and

H“(IL A) E 29 H“(X)” Ak)' Proof: (Incorrect for n = 2 in Eilenberg-Steenrod). (1) Let n = 2; that is, X= Xl n X2 and XI n X2 = Q where X1 and X2 are both open (closed) in X. For A c X let A, = A n X,,

y = l, 2. Let ,u. = 3 — 11. Consider the (admissible) injections

hu:(Xu, A.) —> (X, L) A,” A); k,: (X, L) A,, A) —> (X, X, u A). Here, h, is an excision in which A, = X, r) (X, U A,) is cut away. The formula shows that A, is simultaneously open and closed in the subspace X, U A,. The function k, h,:(X,, A,)—> (X, X, U A) is also an excision, since it cuts away the

open and closed subset X,, from X. Consequently, h,* and (k, h,)* are isomorphisms; hence, h,,., and k,* are isomor-

phisms. (The same can be said for h: and k;".) Now consider

Axioms for Homology and Coholomogy

135

the diagram (in which for cohomology the arrows are to be reversed and the symbols q and * raised): Hq(X9X2 U A1)

h.

1:.

£7 Ha(X: X1 U A2)

/TH,,(X, A) \ Ta.

H,(X, u A,, A) *n h“

*9 Hq(X2 u A1, A) h).

Ha(Xh AI)

Hq(X2: A2)

Here i’ and j are the injections and i;* hm = i,,*. The diagonals arise from the exact sequences of the space triple (X, X,, 0 Av, A).

Lemmas 10.4 and 10.5 can be applied since the kw, are isomorphisms and the triangles are commutative (all functions

are injections). Therefore, the 11* yield an injective representation. Since the h”... are isomorphisms, the i“, also yield an

injective representation, and Hu(X9 A) = il*Hq(Xla Al) 6') i2*Hq(A,2, A2)

(2)

Let n > 2, and assume the theorem proved for q = n — 1. Let X’ = UL, X,” A’ = U:=,A n X’, and let i’:(X’, A’) —-> (X, A), i;:(X,c, A) ——> (X’, A’), k 2 2, be the injections. Then i,‘ = i’i; for k 2 2. It has to be proved that a e H,(X, A)

has

only

one

representation

a = 2 ik*(a,,)

with

a,‘ e

H,(X,c, A). From 23:, i,‘*(a,,) = 0, it follows that i,*(a1) +

i; Sag/“(£19 = 0. By case (1), in, and i; form an injective representation of HQ(X, A). Hence, a1 = 0 and 23:, i;*(a,,) = 0. By the inductive hypothesis, the if“, k 2 2 form an injective representation of Hq(X', A’). Therefore, ak = 0 for k 2 2. The proof for cohomology is analogous to the one for homology.

Special Case: Let X be a hausdorff space consisting of a finite set of points, say n, with the discrete topology. Theorem 12.6 yields H,(X) = 0 and H°(X) = 0 for q¢ 0. Furthermore, Ho(X) and H°(X) are isomorphic to a direct sum of n summands M where M = H0(P) is the coefficient module. Let P be one of the points of X.

For the reduced group (in other words, the augmented theory) the reduced homology sequence yields 170(X) ; H0(X, P). It was proved earlier that H0(P, P) = 0, and Theorem 12.6 shows that H0(X, P),

or 11,00, is isomorphic to a direct sum of n — l summands M.

The same holds for I7°(X).

136

Introduction to Algebraic Topology

Those familiar results of the singular theory which were founded upon homotopy theory also hold in the axiomatic theory; for, in the study of the purely topological concepts such as homotopic space pairs, deformation retracts, etc., only the homotopy axiom was needed. This shows that homotopic space pairs have isomorphic homology groups and cohomology groups. In particular, a space which has a single point as deformation retract has the same groups as a point. The computations of the groups of S" and of the graphs carry over to the axiomatic theory with the same results, provided that the functions employed were admissible. The excision theorem

was applied only in the cutting away of open sets in S’l and its consequences are therefore valid.

l3 Mayer-Vietoris Sequence

Preliminary Considerations with Respect to Modules 13.1. THE HEXAGON THEOREM: Let all triangles be commutative in the diagram of R-modules and R-homomorphisms at the

top of the next page. Let the two slanting diagonals be exact, and let kl and k2 be isomorphisms. Then hlkfl II + h2k..."l2 = mlmo. Proof: Lemmas 10.4 and 10.5 show that i,kf‘j1 + izkfi‘v2 = Id. If this is applied to mo(c) for c e C, and the relation j.,mo = II, is used, it follows that m1m0(c) = ml(ilkl—l[l(c) + [ska-112(0) = (hlkl—lll + hzk'z—llzxch 137

Introduction to Algebraic Topology

138

13.2. Corollary:

If mm, = 0, then hlkf'I1 = —h,k;‘1,.

Theorem 13.1 will now be applied to Diagram 13.3:

A

11/ “x i

Y1

13.3.

Y2

/XX "1

Z1

j

i1 k1

22

i2

V B / V\~2

U1

3 5:

“fl

A,,.l

[(2

Uz

n2

Mayer-Victoris Sequence

139

In this diagram, all of the modules and homomorphisms are to have indices q (these have been indicated at the A’s only). All of the triangles are to be commutative. All indicated sequences of at least three modules, including the two outer contours, are to be exact. Furthermore, kl and k2 are to be isomorphisms, and i 1711 = jm1 and i,n2 = jm,.

Additional homomorphisms «pq—> KGB Y2, ¢:Y,@ Y,—-> X, and A:X—> A".1

are defined by Ma) = (hla, — hza), (yny2) = mlyl + ”12?: and A(x) = —s,k“l,(x) where a e Aq, yl 6 Yb y, 6 Y2, and x e X. The Hexagon Theorem (13.1) yields

13.4. THEOREM: A

The sequence V)

1‘

A

. ——>Aa——> Y‘C-B Y2—>X——>Aa_l——>

is exact. Proof:

(1) W0) = mlhx(a) - m2h2(a) = 1.(a) - i(a) = 0(2) (3)

(4)

A¢(yla Y2) = 32k{llzm1(yi) — Slkf'm20’2) = 0: Since 12ml = 0 and 11m2 = 0. «pA(x) = (—h,s,k,“ll(x), —h2s2k;‘l,(x)) = 0, since hlsl = O and his2 = 0.

Let x e X and let A(x) = 0. Then s,,(k,7l ,(x)) = 0 for v = 1, 2. Exactness of the outer contours shows the existence of a

y, 6 Y, such that n,,(y) = k'11,(x). 10.4 and 10.5 yield

For jx e B, Lemmas

jx = (i.k."j1 + i,k{‘j,)jx = i,kf‘I,x + i,k;‘l,x =i1n1y + ianzy =jmi}’1 +jm2}’2-

This shows that j(x — m,yl — mayz) = 0. By exactness, there is an a 6 A4 such that id = x — m,yl — mzy2. Therefore, x = i0 + "10’: + m2}’2 = "11011“ + J’i) 'l‘ m2J’2 = ¢(hi(a) + y“ ya)-

(5) LC! 9501,») = mlyl + mzy2 = 0. An a e A, must be found

for which Ma) = (y,, y,). For this purpose, apply j and use

Introduction to Algebraic Topology

140

jm = inn”. Then, ilnlyl + igngy, = 0. Since i1 and i2 form an injective representation of B as a direct sum, it follows that n1 y1 = 0 and my, = 0. By exactness, there are a, e Y,

such that hvav = y,,, 11 = 1, 2. The original equation yields mlhlal + m2h2a2 = 0, and therefore i(all + a2) = 0. By exact-

ness, there is a b such that :11 + a, = s(b). Let b be written b = ilul + [214, where u, e U”. Then,

a, + a, = sin:‘ + si,u, = slul + sgug.

The desired a is now a = a1 — s,ul = —(a, — s,u,).

This can be seen from

1],»(a) = (h,a1, —h,a,) = (hlal — hlslul, hga, — h,s,u2) =(y1,y,) which holds, since hvs, = 0 and hua, = y for 1/ = 1, 2. If a satisfies «(J-(a)=(h1a,hza)= 0, an x must be found for which A(x) = 0. When u = l, 2, h.,(a) = 0. Exactness shows the existence of u, such that a = spun. From 31', = s", it follows that

(5)

s‘(—i,ul + i,u,) = ——s1ul + 3,14, = —a + a = 0. By exactness, there is an x for which jx = —i Iu1 + i,u,. Then, A(") = —31k1—lli(x) = —slkl_ljljx = "Siki'l—iiui + 1'2“!)

= slkr‘jlixul) = Mu.) = a, because jli, = 03ndj1il = k1Definition: A triad (X, X., X,) consists of a topological space X and an ordered pair of subspaces X., and X,, of X. The triad (1", X,, X,) is to be distinct from (X, X,, X,), provided that X, 7’:

X2. The previously considered space triples were triads (X, A, B) for which A 3 B.

Each triad (X, X., X,) has injections k1:(Xz, X1 0 X2) —’ (X1 U X2: X1) and k2:(A’b X1 0 X2) ——> (X1 U X72, X3).

Mayer-Victoris Sequence

141

The injection k, cuts away X, — (X, m X, , v = 1, 2. The triad (X, X,, X,) is called exact, or a proper triad, if k, and k2 are exci-

sions; that is, if k”, and k“, (for cohomology, ki" and k,*) are isomorphisms in the non-augmented case. It is to be noticed that the concept of a proper triad is a function of the homology theory employed.

Remarks: (1) In checking the exactness of a triad (X, X,, X,), the space X may be replaced by X, U X2 since X does not occur in the definition of exactness.

(2)

To an exact triad (X, X,, X,), there correspond isomorphisms, for instance H,(X,, X, n X,) ; H.106 U X2, X,), which are clearly analogous to the isomorphism theorems of group theory.

Definition:

The complement of a set A in a space X is denoted by

(€11. Examples: (3) In R“1 let X= S", X, = E: and X2 = El'. Then X = X, U X, and S”" = X, n X,. Suppose that these spaces and the injection functions k, :(Ei‘, S"“) —> (S", E2) and k,:(E;', S"") ——> (S", E?) are admissible. The considerations of 7.11 also hold in the axiomatic theory, and show that k,

and k, are excisions. Consequently, (S",E3,',EL‘) is an exact triad in the axiomatic theory.

(b)

Let V, and V, be open subsets of a topological space X. In the singular theory, (X, V,, V,) is an exact triad.

Proof:

Let X = Vl U V,. Let U = ? V2 (the complement of

V,). Then U: V, —— (V, n V,) c V,. Since U is closed and

V, is open, the excision theorem of the singular theory shows that k, :(V,, V, n V,) —> (V, U V,, V,) is an excision. k,, interchange the roles of V, and V,.

(c)

For

Let B,, B2 be compact subspaces of a hausdorfl' space X. Then B, and B2 are closed. By (b), (X, $3,, @B,) is an exact triad in

the singular theory. The theorem for Diagram 13.3 will now be applied to homology and cohomology groups. Let (X, X,, X,) be an exact triad in which X = XI U X2. Let A = X, 0 X,. In the case of homology, consider

Introduction to Algebraic Topology

142

\w/ ”:10” Hq(X2)

\ 13.5

a:

all

Hq(X2:A)

\ v Hq—1(A) m1.

A.

Hq—1(X2)

Hq_1(X1) i2.

‘7

i1.

Hq—1(X)

/1\ Diagram 13.5 which is the same as Diagram 13.3, except for the names of the modules and the two broken arrows. The indicated homomorphisms are boundary homomorphisms or else induced by injections and are assumed admissible. Since the triad (X, X" X,) is exact, k”. and k“. are isomorphisms. The other conditions from Diagram 13.3 follow immediately from Axioms I, II, III, IV. Con-

sequently, Theorem 13.2 can be applied in the case of homology. The broken arrow 5*:HQ(X, X,) —> Ha_l(X,) yields a sequence

Hm —> H.(X, X.) 1'» H.-.(X.) —+ H.-.(X)

Mayer-Vietoris Sequence

143

which is exact by Axiom IV. By Axiom III, the rectangle Ha(X: A) '_) Ha(X, X1)

0.

J.

Ho-1(A) _) Ha-1(Xi)

is commutative. In the case of cohomology, all of the arrows in 13.5 are to be reversed. If the diagram is turned upside-down and the old outer contours are replaced by the broken ones, then a diagram is obtained which coincides with 13.5. Hence, the hypotheses have been checked in the case of cohomology.

This has proved the following theorem: 13.6. MAYER-VIETORIS THEOREM: (a) Homology: Let (X, X., X2) be an exact triad in which X = X, U X2, and A = X1 0 X2. Then the Mayer-Vietoris sequence

- . - —> HMO!) —“> HM) —‘°> Hq(X1)69 H..(X2) 4» Hm —“> He-.(A) —> - - is exact. Here «I», (i), and A are defined by Ma) = (mg, —m,,.,a), (1)061, x2) =i1*x1 + inxa, andA = _a:|:ki;sl* = ankfiisu(b)

Cohomology: Let (X, X,, X,) be an exact triad in which X = X1 U X2, and A = X1 0 X,. Then the Mayer-Vietcris se-

quence

. (— Ha+l(X) mom e 11.04) —‘+ H.(X') —“> 11.- .(A') —+ is commutative, and the analogous result holdsfor cohomology.

Proof:

For all groups which occur in the Mayer-Victoris Diagram

13.5, f defines homomorphisms which commute with the functions of that diagram (injections and boundary operators). Since 1]», ¢>,

and A are expressible in terms of the functions of Diagram 13.5, the corollary follows.

13.8. Corollary (to the Mayer-Vietoris Theorem): Let (X, X,, X,) be an exact triad in which X = X1 U X2. For some q, let H¢+1(X) = 0. Let mv = X U X2—> X, be the injection, 11 = 1, 2. Let

a e Ha(A) andm,*(a) = 0,m,.(A) = 0. Thena = 0. Proof: Since 1.]1‘001) = 0, then a 6 Im Ac“. This is zero since Hq(X) = 0. Hence, a = 0. Corollary 13.8 will be used in the proof of the Jordan-Brouwer Theorem in Chapter 14. Application to Spheres: It has been shown that (8", E2, E!) is an exact triad in the axiomatic theory when n 2 0. Consider the function f:(S", E2, E2) -—> (5", E3}, E2), which is given by f(xo: ' ' ‘1 xn—l, x1!) = (X0, ' ‘ '9 xil-b _xn)'

On A = E3: n E2 = S"“,fis the identity for n 2 0. When n 2 l,

A qt Q. Therefore, the augmented theory can be applied, and Hq(E3,‘) = Hq(E1') = O. The Mayer-Victoris sequence (13.7) now reduces, in the cases n 2 l, to

Mayer-Victoris Sequence

145

o —> H¢(S") —“> HM sn-1)—> o f.

K

o —> H.,(S") 3» H 4(5).-.) —» o. The exactness shows that A is an isomorphism for all q. Hence, the groups H”(S") can be computed and the same results are obtained as in the singular theory (7.12). The groups H‘(S")can be calculated in the analogous way. The commutativity shows that —Af* = A. Hence, f* = —Id is a multiplication by (— 1). That fl. = —Id, even when n = 0, can be seen as follows: Replacefin the case n = 1 by g:(S‘, E1, E1) ——> (S', E1, E1) defined by g(xo, x1) = (—x0, x,). The map g induces the map f(xo) = ——x.,

on S°. Just as before in the case off, the function g now induces a g* such that g* = —Id. Now 13.7 yields the commutative, exact diagram

0 —> Haw) —‘> H.-1(S°) —» o 8:

f.

o —> Ha S‘) 3—» 151..l 5°) —> 0. Therefore, f*A = Ag* = —A and f* = —-Id. That f* = ~Id is proved analogously. Application to Graphs: A graph is the union of a finite set of topological line segments which have at most end points in common.

By the Direct Sum Theorem, it is sufficient to restrict considerations to connected graphs. Such a graph is homotopic to a graph G, which

consists of r circles attached at a point P since the remarks on graphs of Chapter 5 also hold in the axiomatic theory. For r 2 2, G, 3 GH and G, :> S1 in such a manner that Gr—l U ‘5'1 = G,- and S1 n G,_1={P}.

The reader can show as an exercise that (6,, G -1, S‘) is an exact

triad. The Mayer-Victoris sequence for augmented complexes

0 = HAP) 3—» Ha(S‘) («9 mm -.) —‘+ Ha(Gr) —“> Hq—1(P) = 0 shows that 110(61): Ha(S‘) 63 Hq(G,_,) for all q. Induction on r yields

Introduction to Algebraic Topology

146

Hu(G,) = 0 for q q: 1 (in the augmented case)

H1(6,) ; 293M, with r summands M. The result is the same for cohomology. Application to Surfaces of Genus g: Definition:

Let X1 and X2 be topological spaces and let U1 and

U, be homeomorphic n-dimensional euclidean neighborhoods for which U1 :2 X1 and U2 c X,. Let E? be a ball in U1 and let E; be

its homeomorphic image in 0,. Let X be the space obtained by attaching X1 — 12‘? to X, — E; by means of the homeomorphism between E? and Ez'. The space X is said to be obtained by adjoining X1 and X2. If X and a third space X3 are adjoined by the same procedure, the resulting space is said to have been obtained by the successive adjunction of the spaces X., X,, and X3. Definition:

Let F0 be homeomorphic to S2 and let F5, arise from

Fg_l (up to a homeomorphism) by means of adjoining a torus. Each homeomorphic image in R3 of such an F8 is called a surface of genus g. It is referred to as a sphere with g “handles.” (In R3 the handles can be knotted and intertwined.)

Let F5 be a surface of genus g. That such a surface always separates R3 into two disjoint open sets is not always obvious. However,

it is obvious in case Fg is represented in R3 as in Figure 15. Let J be the closure of the bounded complementary domain of Fg and let J’ be the closure of the unbounded complementary domain. J and J’ are referred to as the “inside” and “outside” of F3, respectively. Let R3 be compactified to a homeomorph of S3 by the addition of a single point. This compactifies J’, whose designation will remain the

Figure 15

Mayer-Victoris Sequence

147

same. Then J U J’ = S3 and A = J n J' = F3. The reader should show as an exercise that (S3, J, J’) is an exact triad, that J is homo-

topic to a graph 03, and that J’ is also homotopic to a graph 6‘. Therefore, H.1(J) 6-) H,(J’) = 0 for q at l, and H,(J) @ H,(J’) is isomorphic to a direct sum of 2g summands M (in the augmented case). Since A at Q, the Mayer-Vietoris sequence can be employed for the augmented case (reduced homology groups), and it follows that

H,(F,,) = 0 for q at 1, 2 (in the augmented case).

13.9.

H,(Fg ; M. H,(Fg) ; direct sum of 2g summands M.

For cohomology there is an identical result since the direction of the arrows does not matter here. 13.10. Corollary: IfM is chosen to be afield, itfollows that surfaces of dlflerent genus are not homeomorphic or even homotopic. Remark: In the (augmented) computations just carried out, the symbol H0(Fg) should actually have been used. It differs from the non-reduced group H0(Fg) by only one direct summand. Let (X, X,, X,) be 13.11. The Relative Mayer-Vietoris Sequence: an exact triad, where it is not necessary that X = X, U X2. Let Y =

X, U X2 and let A = X, n X,. Then the relative Mayer-Vietoris sequence in homology

A ,b e ——> H¢(X, A) —> H,(X, X,) G) H,(X, X,) —> A

Ha(X, Y) —> H0'1(X, A) —> and the relative Mayer- Vietcris sequence in cohomology

; H¢(X, A) (L H4(X, X,) ea H"(X, X,) J’— Ha(X, Y) (i Ha"(X, A) (— are exact.

Proof:

Consider the diagram which arises from 13.5 when the

groups of the sequence H.104), H.101), H,(X,), H40"), Hq(X, X2),

148

Introduction to Algebraic Topology

Hq(X, X1), Hq(X, A) are respectively replaced by the groups of the

sequence H.(X, A), HM, X1), H¢(X. X2). H.(X, Y), HAY. X2), Hq( Y, X,), Hq( Y, A), in that order. The functions here are either the

boundary operators or else are induced by the injections (which are assumed admissible). All the necessary hypotheses are satisfied by the new diagram and therefore the theorem is proved. As in 13.7, admissible functions f:(X, X1, X,) —> (X’, X1, X;)

induce functions f,. on the exact sequences for which the rectangles in the diagram are commutative.

14 The Jordan-Brouwer Separation Theorems

This chapter makes use of the singular theory. The triads (X, X” X,) considered consist of the complements of compact sub-

sets of S”. Part (0) of the example preceeding 13.5 establishes the exactness of such triads. In order that the augmented theory be applicable, considerations are restricted to the case X, n X, :1: g. Let X, Y be spaces with X c Yand let izX ——> Y be the injection. If c is a q-chain in X; that is, a linear combination of simplices

01A” ———> X with coefficients in M, then i#c is a linear combination of simplices which arise from 0' by an extension of the image space from X to Y. The notation will be simplified by the omission of

i#, and c will be called a chain in X or in Y depending upon the situation. 149

Introduction to Algebraic Topology

150

The closed ball E’+1 = {x | x: + . - . + x5 3 l}is homeomorphic to a cube I7“, where I is the interval 0 _ 0. Each continuous func-

tion of the compact set I’ in S" is uniformly continuous. Therefore, there is a 8(t) > 0, such that points in 1' whose distance is less than 8(t) have images in S" whose distance

is less than 6,. Now let I’ c I be an open interval of length less than 28(2) and with midpoint I. Let Er(t) be the image of I’ X I"'. Each point of 1737(1) has a distance less than c, from EMU). Therefore, |b.| does not meet the set 1770). Therefore,

c = 8b, in {313" ’(t). The intervals 1’ cover I. Since I is compact there is a p >0 such that each interval of length < p lies in an I’ (Theorem 6.11). Choose a natural number m with (l/m) > p. Let I, be the interval (v/m) g t g (u + 1)/m,

v =0, ---,m —- 1. Let E; be the image of I, x ["1. Then there is a chain b, inVE; for which c = 3b,.

At this point, it is necessary to digress to prove a lemma. 14.2. Lemma:

Let J1 and J2 be closed subintervals of I such that

J. n J, = {t}. Let E’ be the image of J.>< I"1 and let E” be the image ofJ2 x I"'. In VE’ let b' be a (q + l)-chainfor which 6 =

The Jordan-Brouwer Separation Theorems

151

8b’. In ‘65” let b” be a (q~ + l);chain for which c = ab”. Then there is a (q + l)-chain b in? (E’ m E”)for which 6 = 3b.

Proof:

Let X = {sir-Ia), X, = an”, and X, = 3’5"". Then, A=xnn=ewufim

Since 5’ U E” and S" have different homology groups (I? U E” is homotopic to a point), it follows that

t€(E~’ U E”) = A ¢ a.

When X, U X, =%(E~’ n E”) is called X, Example (c) shows that (X, X,, X2) is an exact triad. By inductive hypothesis, Hq+1(X) = Ha+l(gEr—l(t» = 0

The cycle c was a boundary in X, and in X2, and by Corollary 13.8,

c is boundary ofa chain b in A = %’(E’ U E”). Now return to the proof of 14.1: Repeated application, (m — 1) times, of the lemma to the sets E; shows that c is the boundary of a

chain b in @(UL’Q‘ED =€E~K Since c was an arbitrary cycle in {612:7 and had a boundary there, it follows that Ha(%ET) = 0 for all q. This completes the proof of 14.1. Suppose that S", l g r g n, is the homeomorphic image of the

S’ in S". The images of E1, E1, and S"1 will be denoted by 51, EC, and 5"“ (5‘7“ =E'inE‘C). In the case r= n, it will

also be assumed that §”¢S”. Let X=%’S"', X, =‘KEQ and X2 =9§E~L The triad (X, X1, X2) is exact in the singular theory. When r < n, the nth homology group of S" and S" are different. Hence, 3" =/_- S", and

A=X.flX2=‘iS’§’¢® for l g r g n. Consequently, computations in the augmented theory are permissible. Since ET+ and EC are homeomorphic to 1', it

follows from 14.1 that Hq(X,) = Ha(X,) = 0. From this and the Mayer-Vietcris sequence, it follows that Ha+.(X) ; Hq(A). Hence,

Ha+1(%$‘"') ; Hq(?§") for l g r < n and also for r = n when 5‘" at S". Therefore, Hq(%§') ; H,+,(¥S‘°). Since S° consists of two points which can be moved by a homotopy to the north and south poles, respectively, V5” is homotopic to 5"". Therefore,

Introduction to Algebraic Topology

152 ~r g: H

Haws ) _

Sn—l ;_

)

“A

M for q = n — r — l

{0

otherwise.

In the case r = n, this implies that H_.(@S‘") ; MLa contradiction to H.1 = 0 (in the singular theory). Consequently, S" = S". It also follows from this that S 7‘ contains no homeomorphic image of S ’ for

r > n. For such an 5" would have a proper subspace S" which would have to be equal to S" in contradiction to the one-to-one property of homeomorphisms. These results are summarized in the next theorem. When r > n, S" contains no homeomorphic 14.3. THEOREM: image of 5'. When r = n, S’| contains only itself as a homemorphic

image ofS". When 0 g r < n, M forq=n——r~—1

HA? 5") g 0

otherwise.

The augmented homology groups of ‘65" are thus those of an (n — r — l)-dimensional sphere. 14.4. Corollary (The Jordan-Brouwer Separation Theorem): If n 2 l, the complement of 5"" in S " consists ofexactly two pathwiscconnected components, each of which has S‘"" as boundary.

Proof:

(1) Let M be a field Q. Then HJQS‘V‘) = Q in the augmented theory. By 11.2, the number r of path-components is 2.

(2)

Since S‘fl“ is compact, fiS‘r‘ and its path-components KI and K2 are open. The boundary of each component is disjoint from the other component, and is therefore contained in 5"“. Conversely, let x e 5'". It will be proved that x is a boundary point of both K1 and K2; in other words, that each neighborhood U of x contains points of K1 and of K2. Let yl e Kl and y, 6 K2. Let x., e S"‘1 be the inverse image of x. Choose an open disc (a cap in this case) K c 5"" about x0, whose

image K is contained entirely in U. Then S'""—K=E"" is homeomorphic to an 1"". From Theorem 14.1, it follows that

HOWE“) = 0. Therefore, @5"" is pathwise-connected, and

The Jordan-Brouwer Separation Theorems

153

there is a path I‘ in @EP‘ connecting y, to y,. Since yl and

y2 lie in distinct component of g?“ this path must meet 3""; and, indeed, since I‘ 0 §"“ = (25, the path must pass

through K. The set I‘m 3"“ =I‘n K is closed and not empty. Hence, the inverse image of I‘ n 5' is a closed and therefore compact subset of I. Hence, there is a first intersection point s, and a last intersection point 52 of I‘ with K. In each neighborhood of s. e K c U, and therefore in U, there are points of K; for y, is connected with s‘ by a path which, except for the endpoint st, is completely contained in Ki, 1': 1,2.

The following figure shows that the theorem is not trivial even in the case n = 2.

Figure 16

By 7.2, the homology groups of the components of Remarks: ‘65"‘1 are zero for q 2 l. The 0-dimensional group of each component is 0 in the augmented and isomorphic to M in the non-augmented case. Thus, both components have the same homology groups as in in". In the case n = 2, each component can be proved

homeomorphic to E". However, there are counterexamples when n > 3.

Introduction to Algebraic Topology

154

The components of the complement of an S“ 14.5. THEOREM: in S’ are homeomorphic to open discs.

Indication ofproof:

Let S" c S1. Interpret S ’ as the complex plane

in which the point at infinity is one of the points not on 5‘ 1. The bounded component A of the Jordan curve 5“ will be called the inte-

rior of 5". Let f(z) be a regular function that has no zeroes in A. By the Cauchy Integral Theorem (11.9) the function defined by «/ f(2) can be uniquely determined by

/— _ x— i ’ f'(2) f(Zo) exp 2 Iz'fl(z) dz, f(z) ‘—‘ where V f(20) is arbitrarily chosen (but fixed). This is the essential step in the proof of the Riemann Mapping Theorem, according to which A can be mapped conformally, not just homeomorphically, on E l, the interior of the unit disc.

Example for n = 3:

In a paper of E. Artin and R. Fox (Annals

of Mathematics, Vol. 49), the following example is given of an E ‘ whose complement in S3 is not homeomorphic to E".

Figure 17

Here, E' is homeomorphic to a line segment. It is plausible, and can indeed be proved, that the curve V which lies inVE ‘ can not be con-

tracted in ?E ‘ to the point P, which the point held fixed during the contraction. This would be possible in E“.

From this example an S" c S3 can be constructed whose outer component is not homeomorphic to E 3. For this purpose, replace the

The Jordan-Brouwer Separation Theorems

155

curve E1 by a closed tube which gradually becomes more narrow

towards the vertices. The inside of this 5" is homeomorphic to E3 but not the outside. Now make a small hole in the 5‘2 and attach a short tube which lies in the interior of 5‘2. To the free end of the tube attach a smaller copy of the 5'”. In this way one obtains a homeo-

morphic image of S"2 for which neither the inside nor the outside is homeomorphic to E3. There are analogous examples for n > 3.

A homeomorphic image 5“ of S1 in S3 is called aknot. It has

been proved that HI(%§‘) ; M, and Hq(%§l) = 0, for q 7/: 1, in the augmented theory. The homology groups H¢(%§l) are therefore as useless as the groups HAS“) in distinguishing among knots. Definitions:

An open, pathwise-connected (i.e. an open, connect-

ed) subset of R" is called a region. If S‘"“ c Sn and n 2 1, then the Jordan-Brouwer Theorem states that S" — 5"" consists of two disjoint regions. Pick a point

x of one of these regions. Then S" — {x} is homeomorphic to R". The complementary region containing x will be referred to as the outside of 5"”, and the other as the inside. Removing x from its region will disconnect that region when n = l, but not otherwise, for a path joining any two points of a region can be modified slightly

to avoid a given point. This yields the next corollary. Let n 2 2, and let 5“" be a homeomorphic 14.6. Corollary: image of 5"“ in R". Then 5"“ consists of two regions, its path components, each of which has 3"“ as boundary.

14.7. Corollary:

Let n 2 2 and let f:E" —> R" be a continuous,

one-to-one function. Let 5"" = f(S"") and let A be the interior of the region determined by S”. Then f(132") = A. (That this corollary is not trivial should be clear from the Proof: “tube” generated from the Fox-Artin example which was an S" c: 53 for which neither complementary domain was topologically an E3). Since S" is a hausdorff space and E" is compact, it follows that f is a homeomorphism. Let E" = f(E n). Theorem 14.1 states that

Hq(?i “) = 0 in the augmented theory. In particular, this means that VF" = S" — E" is pathwise-connected. Now embed R" in S" (by removing a point of ‘65" from S"). Then, as in the proof of 14.6, R"— E" remains pathwise-connected. The set E" is compact; hence, closed and bounded. Hence, R" — E" is not bounded (contains a

156

Introduction to Algebraic Topology

deleted neighborhood of the removed point). Since Sn‘lci", it

follows that (R" —— E") n 5"” = 25 and that (R" — E") r: B, where B is the outside of 5“". Therefore, 17:" :> Rn — B = 5"" U A. Since f is one-to-one, f(E") and S'"" are disjoint and therefore, f(19") 3 A where A is the inside of Sn". As the continuous image of a pathwise-connected set, f(E7') is pathwise-connected. It was already

known to be disjoint from 5"” and therefore lies entirely in A or else entirely in B. Since A c f(E1'), the only possibility is f(E7') = A. 14.8. Corollary (Invariance of Domain): Let G be a region in R’1 and let f:G —-—> R’l be continuous and one-to-one. Then f is an open

function, f(G) is a region, andf is a homeomorphism of G onto f(G). Proof.

Let U c: G be an open set and y =f(x) ef(U). Let E" c

U be a closed ball about x with positive radius. By 14.7, f(E") is the inside of f(§"") and therefore, an open set in R" which contains y and lies inf(U). Hence, f(U)is open,fisopen, fisahomeomorphism, and finally, as image of a pathwise-connected set, f(G) is a region.

Remark:

Ifs' —> R” is continuous and one-to-one, the image

of an open set does not need to open in f(R‘).

f(U)

Figure 18

The Jordan-Brouwer Separation Theorems

157

14.9. Corollary (Invariance of Dimension): Le! m < n, Then there is no continuous, one—to—one function f sending a region of R" into R'".

Proof:

Embed R“ in R". Then such an f would yield a function

sending a region G in R" into R". By 14.8, f(G) is a region in R",

but, by f(G) c: R’n c R", it follows thatf(G) is not open.

15 Finite Cell Complexes

This chapter describes a process for building more complicated spaces from simple ones by the device of attaching n-cells. The homology theory employed will be the axiomatic one, and all spaces and functions that appear are to be admissible. Let E" be the unit ball in R" and let 5"" be its boundary. Let Xo be a topological space for which Xo n E" = Q , and letf:S"'1 —> X, be continuous. Let X = X, U 3'". Let izXo —> X be the injec-

tion. Now define n" —> X by g(x) = x for x e E" and g(x) = f(x) for x e 5"“. The functions i and f define a composite function flzX0 U E" —> X. A set Vis open in Xif and only iff;‘(V) is open in X0 U E". This defines a topology in X, and f. is then a continuous

function. The space X is said to have been obtained by attaching 158

Finite Cell Complexes

159

the n-cell E" to X, by means of g. The topology defined on X is clearly the smallest topology for which f, is continuous.

If Vis a set in X, then V = (V n X) u (V n E»). 15.1. Denote V n X, by V0 and V n E" by V,. Then Vis open in X if and only if (1) i“(V) = V0 is open in X0.

(2) g“(V) = g"(Vo) U g"(V1) =f"(Vo) U V1

=f‘1(V, flf(S"“)) U V, is an open set in E".

In the special case that V c 1?", it is clear that Vo = :3, V, = V, and g“(V) = V. Hence, the topology defined on X induces the old metric topology on 15'" and B" is an open subspace of X. As complement of E", X, is a closed subspace of X. It will now be proved that the topology induced on X0 is also the original topology, or in other words, that X, is a subspace of X: If V c X is

open, then V.J = V (*1 X0 = i'1(V) is open in X(, by (l) of 15.1. On the other hand, if V0 is an open subset of X,, then f “(V,,) is an open subset of S"" by the continuity off. Hence, f "(Vo) = Wfl S’“1 where W is open in E". Then V, = W n B” is open in E", and W is the disjoint union of V, and f "(V,,). Now let V = Va U V,. Here, V, is open in X0 and f "(Vo) U V, = Wis open in E". Therefore, by 15.1, Vis open in X. Since V, c: B", it follows that V, = V n X,. Hence, V, is open in the topology induced by X on X,. In the special case that X, = S’H and f = Id, then X = E"

and g = Id. 15.2. Lemma: space.

If X0 is a hausdorfl space, then X is a hausdoryf

Let x, and x, be points of X for which x, 72 x,. There are Proof: three cases to be considered: (a) Both x, and x, are in E". Then they

can be separated by neighborhoods contained in E". (b) They lie one in each part of X, say x, e X, and x2 6 E". Then choose I

such that |x,| < t < l and let 0: {x | x e 13'", lxl < t}. Asaneighborhood ofx,, choose X, UB, where B, = {x|x e E", t < Ix] < 1}. Since X0 is open in X0 and f"(Xo) U B, = S’“1 U B, is open in E", it follows that Xo U B, is open in X. (c) Both x, and x2 lie in X0.

Then let V, and V,z be disjoint open subsets of X0 that separate x, and x,. Since V, n V, = Q,f“(V,) flf"‘(V2) = @. Define V; by V; = {x|x e E",x/|x| ef"(V,), |x| > 1}}. Then V,U Vi, i= 1,2, are open subsets of X which separate x, and x,.

Introduction to Algebraic Topology

160

15.3. Lemma:

Proof:

If X, is compact, then X is compact.

The continuity of g and the compactness of E" yield the

compactness of g(E"). The continuity of i guarantees the compactness

of X0 as a subset of X. Hence, X = X. U g(E") is compact. Let the sets B, B’, and B” be defined by B = {x I % < |x| < 1}, B’ = {x|% < |x| < l}, and B” ={x|§-g|x| (X, Xo U B) is an excision

and induces an isomorphism of the homology groups. In the special case that X0 = S'H, the injection (1?" — B’, B — B’) —> (E", 5"“ U B) is also an excision.

The function g induces the following:

g’:(E", s»-1 u B) —» (X, X, u B),

g":(En — B’,

B — 13') ——>

(5" — B’, B — B’), g:(E", 5"“) —> (X, X,) for which the diagram

H,(B» — B’, B ~ B’) 1» 1:1,,(E",S"‘l u B) 9'.’

8c'

ma?» — B’, B — B’) i» Ha(X, X, u B) is commutative by Axiom II. Since g” = Id, it follows that g” = Id, and then that g’ is an isomorphism.

15.4.

X0 is a deformation retract of X0 U B.

Proof: Define F:(Xo U B) X I—> X0 U B by F(x, t) = x for x e X and F(x, t) = g((1 — t)x + tx/lxl) for x e B. Then

F(x, 0) = x since B (X, X.) be the function induced by Proof: g. Then (in the case of augmented complexes), the first part of the lemma follows from the diagram

0 = Hq(E") —> Haas", S"") 4“» Ha-l(S"") —> H.405") = o

it.

Hq(X) i» ,,(X,X.,)

1,.

LHPJXO) in'r,,_,(X)

Introduction to Algebraic Topology

162

in which both a; and g, are isomorphisms. The proof for the cohomology groups is completed by a method analogous to that for the homology groups; it is only necessary to move the indices q and a: up and reverse the arrows. Now replace the term Ha(X, X0) in the homology sequence of

(X, X.,) by Ha_l(S""), replace j, by V = 35.3.7.7,” and replace 3* by f... This yields

15.7. Lemma:

The following sequence is exact:

—> Ham) L» HU(X) L» H._.(Sn-I) —”+ H ”(11, is . Similarly, in the case of cohomology, a replacement of 8* by f, and the definition V = j *§*“ 8"“ yields the exact sequence

0

A0

A A —>P —> 0 be exact. If P is projective or I is injective, then A g P (D 1. Proof: Suppose that P is projective. Then there is an f:P—> A such thatjf= Id. Let a e A. Sincej(a —fi(a)) = 0, there is a b e I

for which i(b) = a —- fj(a). Application of j shows that the representation a = i(b) + fi(a) is direct. The proof can be carried out

similarly in the case that I is injective. Lemma 15.13 together with Equations 15.11 and 15.12 shows that if 11.04,) is injective or Ker f* is projective (in cohomology, if H"0(0) is projective or Cokerf“ is injective), then 15.14.

Hn(X) ; 117.0(0) 6') Kerf: 15.15. Corollary:

(H"(1’) S H”(Xo) (‘3 Cokcrf“).

The isomorphisms of 15.14 are valid when R is a

field. The isomorphism for homology of 15.14 is valid when R is a

principal ideal ring and the coefl‘icient module M is free.

Introduction to Algebraic Topology

164

Proof: (a) If R is a field, the modules that appear are vector spaces and

hence free. Each free module is projective. (b) If R is a principal ideal ring and the coefficient module M is free, then there is a well-known theorem according to which each submodule of M, in particular Ker f,” is free.

Spherical Complexes Definition: Begin with a discrete space consisting of a finite set of points. To this attach one at a time the elements of a finite set of cells. The resulting space is called a spherical complex. Since the discrete space is a compact hausdorff space, Lemmas 15.2 and 15.3 show that each spherical complex is a compact hausdorfi

space. Translator’s Note: Although the concept of a CW complex is not used in this book, it is worth mentioning here that the set of finite CW complexes is a proper subset of the set of spherical complexes. It is true that the finite CW complexes are also obtained by successively attaching cells to a finite set. However, their construction is subject to three further conditions: (1) For each k, all the kcells are attached before any (k + 1)-oells are attached. (2) All the k-cells are attached to the (k — 1)-skeleton by their boundaries. (3) The interiors of the k-cells are disjoint. It is left to the reader to investigate the spherical complex which is obtained as follows: First attach a 2-ce11 to a point P to yield S2 [as in Example (1) below] and

then attach a 2-cell to S2 so that the image of S1 under the attaching function is a curve consisting of infinitely many loops in S2 based at P. The resulting spherical complex is not a finite CW complex. Examples:

(1) Let X0 be a point P and f:S"‘l —> P. Then f is continuous and X = E, U X0 is homeomorphic to S". Hence, S" is a spherical complex.

(2) Let X, be an n-sphere S" and let f:S" —> Xn be the identity. Then X = Xo U E " is E ". Hence, E" is a spherical complex. (3) Let X0 = S‘, P 6 X0, and letf:S° —> Xo be defined by f(S°) = P. Then Xl = X, U E ‘ consists of two circles, a and b, which

are attached at the point P.

Finite Cell Complexes

165

Figure 19 Now take an E2 in the form of a rectangle. Its boundary is S1.

Map S 1 continuously into X1 by sending the vertices onto P, one pair of opposite edges on a and the other pair of opposite

edges on b. This attaches E2 to X. Because of variations of orientation, there are three possiblities for this attaching:

\

P

a

/\b

\

P

P

g

b/\ “b

P

{1

P

P

I

\

P

P

g

bv wb

‘i fi

P

P

P b’

f

P

\

Figure 20 The first yields a torus; the second, the Kline bottle. The third is not a surface (2-manifold) since it has the local appearance

of the vertex of a cone at the point P. (4) The graphs are obtained from a finite set of points by repeated attaching of E ‘. They are, therefore, spherical complexes. (5) The projective spaces and the lens spaces will be studied later. They are spherical complexes. (6) Any space obtained by attaching an n-cell to a spherical complex

is a spherical complex.

15.16. If X is a spherical complex, then Hq(X) = 0 and H"(X) = 0, for q < 0. If m is the maximal dimension of the cells employed in the construction of X, then Hq(X) = 0 and H“(X) = 0 for q > m. Proof:

The first part of the statement follows from 15.8 and the

second from an induction.

Introduction to Algebraic Topology

166

15.17. The following is one way of proving that a given compact hausdorfl' space is a spherical complex: Let I1' be the unit cube in R" (it is a homeomorph of En) and let S"'1 be its boundary. Let h1 :1? —> 1°" be a homeomorphism whose

image space is i" an open set of X. Let hzl1' -——-> X be acontinuous

extension of hl for which i" n h(Sn“) = Q. Denote X — i" by X0 and the restriction of h to 8"“ by f:S"'1 —> X0. Let the space obtained by attaching In to Xo with f be denoted by X’. The function ¢:X’—> X’ which is defined by ¢(x’) = x’ for x’ 6 X0 and by qS(x’) = h(x’) for x’ e I1' is a homeomorphism. Proof:

The function

X“ which the identity on 1““) = X”, and which sends the boundary S"“""1 of 1““ into X“". Let {xk} be a sequence of points in I°“‘“’ = X“ which converges to x e S“‘“"‘. Then the sequence {f(x,.)} = {xk} converges in X“ to f(x)e X“" because f is continuous. In the same way, Y'8 is obtained from Y3“ by attaching an 1“” by a continuous function gzle‘fl’ ——> Y”. Now let {(xk, yk)} be a sequence of points in X“ X YB that converges to a boundary point of

168

Introduction to Algebraic Topology

X, X Y, = jam X idfi) ___ ja(a)+e Ho(X, Xo) (3*: H°(X, Xo) ——> H”(E", S"“)) of 15.1 shows that x(X, Xo) is defined and that

x(X, Xo = x(E") - x(S"") = 1 — (1 + (—1)") = (—1)"Then the proof can be completed by use of Theorem 16.2.

16.5. THEOREM:

Let X be a spherical complex. Let do be the

number ofq-dimensional cells employed in its construction. Then

x(X) = 20: (—1)°a.. Proof:

The construction begins with a hausdorfi‘ space Xo with do

points. By the Direct Sum Theorem, Ho(Xo) = 29R with do summands. Consequently, Bo(Xo) = do; and, of course, Bo(Xo = 0 for

q ¢ 0. Hence, 750%) = (—l)°do. Now the theorem follows from Theorem 16.4 by induction. Remark: For a spherical complex X the euler characteristic x(X) = 2o(— Dado is independent of the homology or cohomology theory employed and always yields the same result. If X is the surface of a convex polyhedron in R’, then X(X) = do — d1+ (1;, where do is the number of vertices, d1 the number of edges, and d2

the number of faces of X. Since X is homeomorphic to S2, it follows that x(X) = 2. This yields do — dl + a2 = 2, which is the wellknown formula of Euler. If the faces are interpreted as countries on a map that is printed on a globe, Euler’s formula gives some

information about the possible numbers of boundary lines and intersection points of boundary lines. In geographical problems on

Introduction to Algebraic Topology

172

a surface F, of characteristic g, the corresponding equation is do—d1+d2=2(l—g).

Examples: (3) Let X1 and X2 be topological spaces containing the balls Er and 15,", respectively. Let X be the space described in Chapter 13 obtained by adjoining Xl to X, by means of the homeomorphism between E" and E2".

The triad (X, X1 — Er, X2 — 1%,") is an exact triad as can be

seen by means of an excision and a homotopy (since E" lies in a larger euclidean neighborhood). By Theorem 16.3, X(Sn_l) + X(X) = (X1 —' £1") + (X2 — Ea")-

On the other hand, X. and X2 are obtained from X1 —— ET and

X2 — E; by attaching cells Br and E2". Hence, Theorem 16.3 yields 1 + (*1)"‘1 + 750’) = x0?) —- (—l)" + X(X2) — (-1)" and then

x(X) = x(Xi) + x(X.) ~ 1 -— (—l)" 01'

X(X) = x(Xl) + x(X,), XIX) = 9609) —- 2,

for n > 0 and odd for n > 0 and even.

(4) The torus T and the Kline bottle K are spherical complexes in which do = 0, a, = 2, and at, = 1. By Theorem 16.5, X(T) = x(K) = 0. The adjoining of g tori to Tyields Fe. Then, Example (3) shows that X(Fg) = 2(l — g).

(5) The real projective plane: The set of all l-dimensional linear subspaces (lines through the origin) of R’ is called the real pro-

jective plane P“. To each x in S2 associate the line through x. This is a function S“ —>P’ which identifies antipodal points x and —x of S“. The function is made continuous by giving P2 the quotient topology: U is open in P2 if and only if its inverse in S" is open. P2 is obtained from the northern hemisphere by identifying antipodal points of the equator. If S‘ is parametrized by means of the angle t and f(x(t)) = x(21) for x = x(l) e S‘,

Betti Numbers and the Euler Characteristic

173

then f(x) = f(—x), f is continuous, and f describes the identifi-

cation of x e S1 with -x e S‘. If E’ is attached to S‘ by means off, then the spherical complex so obtained is precisely P’ with the correct topology. Hence, x(P’) = x(S‘) + (— 1)2 = 1.

Definition: A topological space is called a manifold of dimension n if and only if each of its points has a neighborhood homeomorphic to 13‘" (a euclidean neighborhood). Often there are additional hypotheses of compactness and connectedness (and therefore pathwise connectedness, since each point has a

pathwise-connected neighborhood). The function sending S“ on P2 in Example (5) is a local homeomorphism. Consequently, P2 is a 2-dimensional manifold, which, as continuous image of S”, is compact and connected.

(6) Let (1),. be the surface which is obtained by starting with a projective plane and successively adjoining h — 1 more projective planes. Then 9601)») = X((pn—i) + 1 “ 2 = X( CP" (:5: S" —>P") of (I) to Sm“ (to S") is continuous because SW1 (S") is a subspace of V" and onto because 95(2) = ¢(z/lz|). When 2 and z’ are in 5““ (5"), then |z| = Iz’l = 1; and $(z) = $(z’) if and only if z = e‘“z’ for some real a(z = iz’). Let P and PI be two points of CP" (Pn). Their inverse

images $“(P) and $"(P,) are circles on 52"“ (pairs of points on S"). Consequently, the two inverse images are disjoint closed sets in 52"“ (S"). Since Sm“ (5") is a normal space, the inverses can be separated by open sets V1 and VT The cones from the origin on V1 and V, are disjoint open sets in V*. Hence, ¢( V1) and ¢( V2) are open and disjoint and CP’I (P7‘) is a hausdorfl space. As the continuous

image of a compact set, CP" (P7‘) is compact. 17.1. THEOREM: CP" (P") is a spherical complex obtainedfrom CP"‘l (P"“) by attaching the ball E2” (En). Proof: Embed the n-dimensional vector space V,, in V by associating the vector u = (u,, u,, ~ ~ -, u") with the vector (0, u) = (0, 14,, u,, . - -, 14,.) e V. The vectors u e V" for which lu| g 1 form an

E’" (E") with boundary S’"" (S"“). Define the function g: E“ ——>

S’"“ (g: E" —> S") by g(u) = (M, u). Obviously |g(u)| = 1, g is continuous, and g(u) has a real, non-negative first coordinate. Let f: $g, f: E’" —> CP” (f: E" —>P"). Then fis continuous.

Let P 6 CP" (P e P"). Among all z for which $(z) = P there is precisely one, say (2,, 2,, - ~ -, 2,.) for which 20 is real and non-nega-

tive. Therefore, 20 = V l - [2] and P = $g(z,, ---, 2,). Hence, f is surjective. If u e E“ and v e E’” (u e E" and v e E”), then |u| < l and |v| g 1. From f(u) = f(v), it follows that («/1 — |u|,u) = (v1 — |v|,v) and therefore, («I — lu ,u) = e‘“(«/l — |v|,v) which is possible only for e" = l and u = v. Hence, f(S‘"") n f(E’") = Q (f(S”") n f (En) = (a) and the restriction of f to E“ (E") is one-to-one. Let W be an open set in E‘"; E2” —— W is compact, and therefore has a compact image f(E’" — W). Since f(W) =f(E"') —f(E”' — W), f(W) is open in f(E’") and thus f is an open mapping on ". Hence, the restriction off to E2" is a homeomorphism. It was just proved that f(S""") n f(E’") = Q (f(S"“) nf(1§'")= (25 ). When |u| = 1, then g(u) = (0, u) and f(u) = $(0, u). This shows

Complex and Real Projective Spaces

177

that f(S’"“) is the CP"‘l formed in the V" that was embedded in V (f(S") is the P"'1 formed in the V” that was embedded in V). As a compact subset of a hausdorfl' space, f(SM—1) = CP"'1 is closed in

CF" (f(S"“) = P"“ is closed in P"). Hence, gcpn-l =f(E’") is open in CF" (@P’"l = f(E") is open in P”). Consequently, the conditions of 15.17 are satisfied, and CP" is obtained from C1""'1

(P" is obtained from P"") by attaching E’” (En). The points of the set fa“) (f(£'") have a euclidean neighborhood in this set of dimension 2n (n). Since each point of a projective

space can be sent into any other by a projective transformation, and thus by a homeomorphism, CP" is a 2n-dimensional manifold (P" is an n-dimensional manifold) which, as continuous image of 5““ (Sn), is compact and connected. Exercises:

(a) CF” is a point; P° is a point. (b) P1 is a circle 5". (c) CP1 is a two-sphere S”.

(iroups of the Complex Projective Space CP" If n = 0, CP" is a single point. For the computation of Ha(CP") and H°(CP") use is made of 15.8 to 15.12. Since CPn is built up of cells of dimension g 2n, it follows that Hq(CP") = 0 and H°(CP") = 0 for q > 2n and for q < 0. This proves that Coker f* = 0 and

Ker f* = 0 since H2n_1(CP""); H2"“(CP"") = 0. Formulas 15.9 and 15.10 then yield t (0“) = 0 and H’"“(CP") = 0. Further-

more, Ker f,, = H,,,_I(S’"") and Coker f* = H’“"(S”'“). Hence, 15.11 yields H,,.(CP") ; Han-105““) ; M,

and 15.12 yields

H‘"“(CP”) ; H’"“(S”'") ; M.

When q is different from 2n and Zn —— 1, 15.8 shows that Hq(CP") 1:: Hq(CP"") and H"(CP") ; H”(CP""). This proves for the cases 0 < 2q < 2n that

HM(CP") ; H2q(CP”“) ; - ~ - ; H24(CP") E M and

H,,H(CP") ; HM_I(CP¢) = 0.

Introduction to Algebraic Topology

178

In the same manner, it follows that HWCP") ; M and H“"(CP") = 0. These results can be summarized as

17.2. THEOREM:

For the complex projective space CPn

Ha(CP") __"_’_ H”(CPn) = 0 for q < 0, q > 2n, q odd, q = 0 (augmented),

Ha(CP") ; H“(CP") ; Mfor q even with 0 < q 3 2n, andfor q = 0 (non-augmented). 17.3. Corollary: x(CP") = n + l.

If M = R and M is an integral domain, then

(iroups of the Real Projective Space P" By Theorem 17.1, P" is a spherical complex formed by attaching E" to P"". The attaching function was precisely the canonical

mapping of S"'1 on P"". For the computation of H,,(P") and Hn_,(P"), let a be the antipodal function oz(x) = —x on S". The set EL“ will be designated by E and the set E2 by aE. Consider the diagram (top of p. 179). The indicated homomorphisms are boundary operators, or else are induced by the canonical function f: S"——->P" or by injections. Consequently, all triangles and the two rectangles are commutative. The antipodal function a induces the homomorphisms do: (S", S"“) —> (S", S7“); m: («E 8"“) ——> (E, Sn_l); “2: (Sn, aE) ‘—> (Sn, E)-

The injections on the right side of the diagram have been written in terms of the injections on the left side and a. The symbol * has been omitted from all homomorphisms in the diagram and the following

computation. By definition, the triad (S", E, acE) is exact. Therefore, kl and a2k1al are isomorphisms. The two diagonals are exact because they are parts of homology sequences. By the Hexagon

Theorem (13.1), it follows thatflj = fik‘ ‘1 +fik‘ lIa. By the example to Theorem 7.16 (which also holds in the axiomatic theory), a = (—1)"“. Therefore,

17.4.

h2f=flj = (l + (—1)"“)f,k“l.

Here, the function f2 is the isomorphism g. of 15.5 (except for an

Complex and Real Projective Spaces

179

Hu(¢xE) = 0 H..(S”) I

01210:

H“(S",(XE)

o = H,,_ 1(ads)

j

i1 k

‘

H,‘(S",E)

aailao

H..(S",S"—1) i

H..(E,S"‘ 1)

aoial

71 ’2

azkai

L

H..(aE.S"‘ 1) f2a1

HAP” ‘ ’)—>H..(P")—I2—>H.(P",P" ‘ ‘)—a>H.._ 1(P" ‘ ‘)——>

f

—>H"(S”)_j—>

f1

f3

.(s",&-1)—LH._.(S~-1)—.

isomorphism induced by the homomorphism E —> E"). Consequently, f2 is an isomorphism. The homology sequence of (S",aE) shows, since Hq(aE) = 0, that I also is an isomorphism. Since 1' is the injection, f, = fii. Consequently, fl is an epimorphism. Now consider 17.4. (a) Let n be even. Since P"‘1 is built up of cells of dimension at most n — 1, Hn(P"“) = 0 and j2 is a monomorphism. From 17.4, it then follows that f = 0. (b) Let n be odd. Then I: — 1 is even and f3 = 0 by (a). Therefore, Bf, = 0. Since fl is an epimorphism, it follows that 6 = 0. Hence, j, is an isomorphism. From 17.4, it now follows that

Introduction to Algebraic Topology

180

f = 2j;‘f,k"l = 20, where 0: H,,(S") —+ H,‘(P") is an isomor-

phism. Consequently, f... is a multiplication of M by 2 up to an isomorphism. These results applied tof, yield: (a) Let n be even. Then f3 is the mutliplication of M by 2 (up to an isomorphism). (b) Let n be odd. Then f, = 0. At this point, 15.11 can be applied. In the present situation, X0 is P"“ and f* is f3 (the attaching function). Since H,,(P”“) = 0, it follows from 15.11 that H,‘(P") _=_ Ker f,. Denote by M "’ the kernel of the multiplication of M by 2. There now follow (a) Let n be even. Then H,,(P") ; Kerf3 ; M‘”. (b) Let n be odd. Then H,.(P") ; Kerf3 ; M. By 15.9,

Hn_x(P”) E Hn_x(P"“)/fi(M) = Cokerfi. From this follows (a) Let n be even. Then f,(M) = 2M and Hn_1(P"") ; M. Hence,

H,_l P") ; M/2M. (b) Let n be odd. Then f, = 0 and H,,_,(P") g Hn_,(P"“) g M W. By 15.8, HAP") ; Hq(P"") when 0 < q < n — 1. These results can be summarized in the following theorem. 17.5. THEOREM:

For the realprojective space P"

0, when q < 0, q > n, or (in the augmented case) q = 0. M"), when q is even and l s q g n.

HAP") ;

M/ZM, when q is odd and l s q 3 n — 1.

M, when q = n is odd or when (non-augmented) q = 0.

.

”'6' Corollary '

isodd

.

_ 0,whenn _ ,. _ X(P )— )3 1)“ — {1, when n is even provxded

that M = R and M is an integer domain. The cohomology groups H"(P") are obtained analogously. In

their computation, the direction of the arrows and the order of the products in the diagram need be reversed. This yields the following theorem.

Complex and Real Projective Spaces

17.7. THEOREM:

181

The groups H°(P") are obtained from the list

for Ha(P") by interchanging the role of M ‘2’ and M/2M. The casein which M = Z (the integers) is of interest. In this case,

M ‘2’ = 0 and M/ZM is a group of order2. Thus, the projective spaces P", n > 1, furnish examples of spaces in which some homology groups have torsion; that is, they have an element a ¢ 0 for which there is an integer r such that ra = 0.

18 Maps of S" on S" and Lens Spaces

18-1- Lemma:

Let (X, X1, X2) be an exact triad in which X =

XI U X2 and A = X: 0 X2- Suppose Ihat three functions f, f,, f2,-

(X, A) —-> (Y, B) have thefollowing properties:

130‘) =f(X), When x e X,, f2(x) = f(x), when x e X2, 1300 C B. 1“!"1) C B-

Thenfl. =fn: “Pisa: andf“ =ff +1?-

Proof:

Consider the following diagram (in which the symbol *

has been deleted throughout): 182

Maps of S" on S" and Lens Spaces

183

Hq(Y,B)

f;

f;

”(1(X9X2)

f1 f

f2

HQ(X9X1)

,1

i2

k1

Ha(X,A)

k2

i1

i2

Hq(X,,A)

Hq(X2,A)

By 18.2, the functions f,’ that are induced by f, are well defined and

flip = fy, v = l, 2. The subdiagram H.1(X, X2)

H..(X, A) /’ Ha(X., A) /’

Ha(X, X1)

*\ H,(X,, A)

is a part of the Mayer-Victoris diagram. Since the triad is exact, k1 and k2 are isomorphisms. The diagonals are exact, and the two

triangles are commutative. Therefore, Lemma 10.4 applies, and it follows that Id = i,k;‘jl + izkgljz;

hence, f =filkr1jl +fi2kg‘jz.

Sincefand fl, coincide on X," it follows that fin =fi1iv =flfijllill =flllkll'

Hence,

f=f§jl +f4j, =fi +1":The analogue in cohomology is

f=jlkr1i1f+jzk2“i2f=11fi +J'2fé =fl +12. 18.2a. Corollary:

If Hq(B) = 0 (H“(B) = 0) for all q, and if

f, f1, f2: X —’ Y are the three functions inducedfiy f, f], and f2,

respectively. then f1 = fut + in and f* = f.* + fr.

Introduction to Algebraic Topology

184

Proof:

Since Ha(B) = 0, the diagram for the induced homomor-

phism of the homology sequence becomes

0—>Hq(Y)—j->HG(Y,B)——>0 Tf-fhf:

TfJ'I-f-x

—>Hq(X)—>Ha(X,A)——> is

and therefore, j is an isomorphism. Hence, f =j—lfjs =j_‘(fl +19]: =j_yij3 +j'ff,j3 = ix + f:-

The analogous proof holds for cohomology. Now consider the following situation, which occurs, for example, in the adjoining function for P”: Let Sn be the unit sphere in R’”‘ and letf: S" ——> S" be continuous. Let x0 be a point of S", and suppose that there is an open ball 1?, c S" centered at 9:0 for which the inverse image f"(E0 = 3:, F; is the disjoint union of r open sets

Fi c S" each of which is homeomorphic to E, by a restriction of

f. xo

.— x0

Figure 21

Choose a closed ball E c E. centered at x0. Let F, = f‘1(E) n F,’,. Since fyields a homeomorphism of F5 on E], F, is closed.

18.3. Lemma:

The triads (Sn, F”, (6131,) are exact.

Proof: Choose an open ball U around x0 such that U c 15,. Then (S" — U, F, —— U) ——> (S", F”) is an excision by the Excision Axiom.

Furthermore, (Sn — U, F, — U) is homotopic to ((6131,, F, n (61%,).

Maps of S" on S 7‘ and Lens Spaces

185

Hence, H,(S", F,) ; H,(%F,, F, n glib. The other isomorphism is obtained in the same way. Let y, be the point of S" antipodal to x,. To each 7» for which 0 g A g 1, define a function h,: S”-——> S" as follows: Let [1. be

half of a great circle passing through x and leading from xo to yo (this is unique when x gt x, and x ¢ yo). The image h,(x) will lie on p. Let P be the point of intersection of ,u with the boundary of E. If x lies on the are from P to y,, the distance of h,(x) from yo

shall be 7\, times the distance from x to yo. If x lies on the are from xo to P, then h,(x) shall divide the are from x,J to hA(P) in the same ratio as x divides the are from xo to P. The function h,(x) satisfying these conditions is well defined, and the function H : S" x I —> S" defined by H(x, 7L) = h,(x) is continuous. Obviously, h1 = Id,

hD is a homeomorphism of F on g(yo), h,(%F) = y,, and Id is homotopic to ho. Define g by g 2 110f. Then f and g are homotopic.

If x at UL, F,, then f(x) ¢ E and g(x) = yo. Define g,: S" —> S" by vx =

g()

g(x) if x e F, .

{yorfx egfv‘,

Remember that F, and (€13, are closed and that if x e F, n (615,,

then x #5 U,F, and g(x) = y,. Hence, by Lemma 5.16, g, is continuous. Define «j»: S" ———> S" by

fix) =

g(x) if x e O F, ”=2

y0 if x e g Q21}.

If x 6 UL, F, n? U;=,F,, then x qé ULJS‘, since the F, are disjoint, and g(x) = yo. Then «Ir is continuous by Lemma 5.16. The functions g, g,, and «p induce functions (S", Fl 0 %F,) —> (S", y,) to which Lemma 18.1 will now be applied: By definition, g(x) = g(x) if x 6 F1 and g(x) = y0 for x e %F,. When x e F,,

«[Ax) = y,. When x e %F,, either x ¢ U:=,F, and then x gé UL, F, which in turn implies that «p(x) = y0 = g(x), or else x 6

UL, F, and then «11(x) = g(x). In either case, 11/(x)= g(x) when x e (615}. This means that the hypotheses of Lemma 18.1 are satisfied. Even Corollary 18.2 holds, since H,(yo) = H°(y,) = 0 (in augmented complexes). Hence, g* = g”, + «in, and g* = £1“ + «V.

Introduction to Algebraic Topology

186

An induction on r yields g... = Effigy... and g* =

:-.g;". Since

g and f are homotopic, it follows that

18.4.

f. = E3... andf" = E133“.

Actually, each g. is homotopic to an orthogonal transformation of the sphere, and therefore, g”... = 5;]. However, a proof of this is

beyond the scope of this book. Two simple conditions that imply that g”... = i1 are as follows:

18.5. (1) There are continuous functions fl: S" —> S" which are homo-

topic to an orthogonal transformation of S" and for which both fl,(x) =f(x) when x e F", and fi(x) e ?E when x at I}... Then g = hof, and both g... =1}... = i1 and g” = i1 hold for every coefficient module M in the axiomatic theory.

(2) There are continuous functions f.,: Sn —> S" which are homo— topic to a homeomorphism of S" onto S" and for which both f;(x) =f(x) when x e 13;, and f,,(x) E (613} when 3: ¢ 15'... Again, it follows that g“. = fl... It has been proved that fl... represents an automorphism of HAS") ’5 M. In the case M = Z, it follows thatf” = :tl andfv = i1. Let S1 be interpreted as R/Z (S1 is parametrized by a central

angle) and let n e Z. Multiplication of the elements of R by n induces a homomorphism on: S1 —> S1 which is defined by 0,,(x) = nx. 18.6.

In

the

axiomatic theory, the homomorphism 0,...

is a

multiplication of M by n for any coefl‘icient module M. Similarly, 0,? = n.

Proof:

The conclusion follows readily when n = 0. Since 9-, is an

orthogonal transformation, it follows that 0_.... = —l. The relation 6." = 6-10,. permits consideration to be centered on positive n. Let I be the interval 0 < x < e, where e is sufficiently small. Then

0;‘(I) consists of the disjoint sets 1,, = (l/n)I + (v/n), u = 0, l, - - -, n - 1. From 18.5 (1), it is sufficient to extend the restriction 6n/ll, to a continuous function h: S1 —-> Sl that is homotopic to the

identity. Application of a rotation (which obviously is homotopic to the identity) shows that only Io need be considered. Figure 22 yields such an h and the required homotopy.

Maps of S" on S" and Lens Spaces

187

19..

l

h

Id

0 1..

i

i

L ’1

i

1

Figure 22

Application to Lens Spaces Let X1 = S1 be a circle. Attach E2 to X1 by means of the func-

tion 01.1 S1 —> Xl which multiplies angles by n e Z,. Denote by M ‘1" the kernel of the multiplication by n of M. It was just proved

that Ker 0“ = M ‘n’. Let X2 = X, U E” be the space obtained by the attaching. The exactness of sequence 15.11 shows that 0 '—) H2(X|) —’ H2(Xz) —> Kel' 0,, —> 0

is exact. From H,(X,) = 0, it follows that H2(X,) ; M W. From

15.9, it follows that H,(X) ; M/(nM). From 15.10 and 15.12, it follows that H”(X2) ; M/(nM) and H‘(X2) g M‘"’. The space X2 is homeomorphic to a space Xl which will now be constructed in a somewhat different manner: Let m and n be two relatively prime integers. Let S2 be given coordinates (x, 111‘), where

7t is the product of the geographic longitude by 1/27r and «[r is the latitude. The points of S2 are now divided into equivalence classes such that when a]; > 0, the equivalence class of ()t, 1]») contains only

Introduction to Algebraic Topology

188

the additional point (A, + m/n, —1]/\), and when «p = 0, the class of

(7», 0) consists of all the points (71. + v/n, 0) for u = 0, l, - - -, n — 1. Let Xl be the set of equivalence classes and F: S2 —> Xg be the

function sending points into the class containing them. Let X; be given the identification topology induced by F. Then F is open and continuous and, in particular, F053) is open in Xi. The restriction of F to 13"; is one-to-one, continuous, and open, and hence, is a homeo-

morphism. Furthermore, F(S‘) 0 FOR) = Q. Thus, condition (a) of 15.17 is fulfilled. The restriction of F to S1 is just the multiplication of S1 by n. Hence, Xl is homeomorphic to the space X,. 18.7. THEOREM: For F: S2 —> X,, the induced homomorphisms F* and F* are both zero homamorphisms.

Proof:

The diagram employed here is almost the same as the one

which precedes 17.4 and was employed in the computation of the groups of P". In place of the function a of that diagram, the present

proof makes use of the function a: S2 —> S2 defined by a((7x, «p» = ()1. + m/n, — 1]»). This is an orthogonal transformation whose determinant is —1. Consequently, a* = —l (06" = —1). Let E be E1, then 01E = E3. The functions induced by a and F are distinguished by their subscripts; in particular, 010: (S‘, S ‘) —> (S 1, S ‘), on: (E, S‘) ——> (aE, S1), and a2: (S’, aE) —> (S2, E). The functions 1', j,j1, k, I are injections. As before, the injections on the right side are obtained by means of a from those on the left.

The lower left-hand triangle is trivially commutative. On the lower right-hand side, commutativity is a consequence of Fauna, = F,,.., which is proved as follows: Let x = (7L, 1]») e E; i.e. «Jr _>_ 0. Then

FldoiOL, .1») = F.(7\. + m/n, —.;,) (K’ ‘1’)U (7" + m/"a —‘I")9 if \P‘ > 0

¢=o = g(7\l+v/n,0),if

}=F,(x),

The triad (S", E, dB) is exact. Hence, k. and also a,.k*a;‘ are isomorphisms. All of the hypotheses of the Hexagon Theorem (13.1) are satisfied. This proves that Fuji: = Fulfill: + Fuk:1#d;lSince a. = —1, it follows that Fuj, = 0, and hence, i,t = 0.

Maps of S“ on S" and Lens Spaces

189

HAS”) “2.1.6171

1'.

H2(Sz,ozE)

H2(S‘,E) (12.11.0163

k.

a2.k. 611:1 “0.1.. “1:1

H2(aE,Sl)

H2(E,S‘) F2.

H2(X1) = 0—>H2(X2),—‘>H2(X2,X1) 12.

F.

Fl.

H2(S”).—>H2(32,Sl) I.

Since j,,., is a monomorphism, F,, = 0. In cohomology, an analogous argument shows that F* = 0.

Adjoin E3 to X2 by the function F: S” —> X2. The space that arises is called the lens space L(n, m). The non-zero homology groups of X2 were

Definition :

H1(X2) 2 H206) S M/HM, 1120(2) E H‘(X2) E M”) Ho(X2) ; H°(X,) ; M (in the non-augmented case). From 15.8 to 15.12, there follows, since F, = 0,

Introduction to Algebraic Topology

190

18.8. THEOREM: ofL(n, m) are

The non-zero homology and cohomology groups

Ho(L(n, m)) g M

Ho(L(n, m)) g M

H(L(n, m)) E Ml"M

H‘(Mm m)) g M ""

H,(L(n, m)) g M ""

H2(L(n, m)) E M/nM

H,(L(n, m)) g M

H“(L(n, m)) g M.

All the other homology and cohomology groups, in particular, the augmented groups H0(L(n, m)) and H”(L(n, m)), are zero.

18.9. THEOREM: sional manifold.

Proof:

The lens space L(n, m) is a compact 3-dimen-

L(n, m) is a spherical complex, and hence, a compact

hausdorfl‘ space. To prove that it is a manifold, it is necessary to exhibit a euclidean neighborhood for each point of L(n, m). It is

trivial that each point of E3 has euclidean neighborhood in E". It remains to be proved that each point of the image X2 of E1 under F has a euclidean neighborhood in L(n, m). Pick an open set in E“ which is formed by a small sphere around the north pole. There is a symmetrically placed neighborhood of the south pole. The identification requires that the cap at the north pole (the portion of S“ within the neighborhood) be glued onto the cap at the south pole after the south polar cap has been turned through an angle of m/n. The result of the identification is obviously a euclidean neighborhood. For the investigation of the other points of X,, pick P0, P1,

- - -, P,,_,, equally spaced points on the equator. Together with the north and south poles, the points P, span a double pyramid that is homeomorphic to E3.

It may be assumed that the angle NQS between faces meeting at the equatorial polygon is precisely l/n. The vertices P. (i modulo n) yield a subdivision of the double pyramid in n tetrahedra PJ’M NS. Designate the faces of these tetrahedra as follows: P,P,+,N = at, P,P,+1S = b“ P‘NS = cb PMNS = 11,. Here, d, is identified with em. The lens space will now be reassembled from the tetra-

hedra in such a way as to describe neighborhoods of points of X,. The function F requires that a, be glued onto bum. This is accom-

plished by placing the ith upon the (i + m)th tetrahedron. Since the angle at the equatorial polygon was chosen to be l/n and m is relatively prime to n, the figure closes after all the tetrahedra have been joined. This process has formed a new double pyramid in which the

Maps of S" on S” and Lens Spaces

191

Pi+l

Figure 23 P. form just the north and south poles. The c‘ all lie around one pole, say the north, and the d, all lie at the other. The points in the interior of a, are inner points of the new double pyramid, and therefore have

euclidean neighborhoods. Since the equatorial polygon can be turned, it has been proved that each point of X2 has a euclidean neighborhood and that L(n, m) is a manifold. 18.10. THEOREM:

The lens spaces L(n, m) and L(n, s) are homeo-

morphic when sm E 1 modulo n. Proof: Continue the process of assembling the tetrahedra that was begun in the proof of 18.9 by glueing d‘ to cm. In this fashion, the ith tetrahedron has a face in common with the (i + m)th, this in turn has a face in common with the (i + 2m)th, etc. If s is the smallest solution of the congruence sm E 1 modulo n, then d, is precisely s steps away from Cm in this process. The identification of d; with cm, therefore, yields the lens space L(n, s). In a trivial fashion (by a reflection) it can be seen that L(n, m)

and L(n, —m) are homeomorphic. In general, when n is fixed, certain of the (Mn) (Euler’s 4) function) possible lens spaces are homeomorphic. It was proved by E. J. Brody (Annals of Mathe-

192

Introduction to Algebraic Topology

matics, Vol. 71, (1963)) that the only homeomorphisms between

lens spaces are those just described. (This is difficult to prove.) From the homology groups with integer coefficients, it is possible to prove that n is an invariant. It is easy to see that L(l, 0) is homeomorphic to S3, and that L(2, 1) is homeomorphic to P3. In the case of larger values of n, new spaces appear that are the simplest examples of higher torsion in manifolds. Two such cases are L(7, l) and L(7, 2).

l9 Classification of Surfaces

Let X, and X2 be two topological spaces adjoined as described just before 13.9 to form a space X. Let Y be the space obtained by attaching a ball E7' to X by means of the homeomorphism between the boundaries of E" and EI'. The triad (Y, X1, X2) is exact and XIUX,=Y zYlflX2=En¢Q.

Since Hq(E") = 0 (in the augmented theory), the Mayer-Victoris sequence yields

19.1.

11,,(Y) ; Hq(X,) 69 Ha(X,). 193

Introduction to Algebraic Topology

194

19.2. THEOREM: Let H..(X.) ; H..(X,) = 0. Then Hn(X) = 0. If, in addition, M = R is an integral domain and n 2 2, then Bn—1(X) = Bn—1(Xl) + B —1(X2) + 1-

Proof:

The sequence 15.8 becomes

0 —> II1I(A’)—> [111(Y)—> Hn_l(Sn-l) —) Hn—1(X)

—> Hn_1(Y) —> 0. By 19.1, H(Y) = 0, and consequently H,.(X) = 0. An application of (8,, Kto H,._1(S"") ; Ryields Bn_1(X) = B,,_,(Y) + 1.

19.3. THEOREM: Let (1),. be the surface obtained by the successive adjoining of h projective planes. Let M be Z, the ring of integers. Then H2(,.) = 0 and B,(,,) = h — 1.

Proof: Theorem 17.5 yields H2(1) = H2(P‘) = Z ‘2’ = 0, and H,(1) ; 2/22. This proves that B1(,) = 0. Now suppose that the theorem has been proved for ,,_,. Then it follows for (1),. by letting n = 2, X1 = (Dad, and X, = (I), in the previous theorem.

Remark: It was proved in 13.9 that H2(Fg g2, and now it is known that H,(,.) = 0. Hence, F3 and (1),. are not homeomorphic or homotopic. The euler characteristic distinguished the F3, and similarly, the X is a triangulation of an n-dimensional manifold X, then a proof analogous to that of 19.4 will show that dim X = n. The triangulability of compact connected 3-manifolds was established by E. Moise in 1951, but the question as to which compact

connected manifolds of dimension at least four are triangulable is a difficult unsolved problem. Not even for 3-manifolds is there any approach to a solution to the homeomorphism problem, and no

approach is known to reasonable restrictions of the problem. The subject is exceedingly complicated. The object of this chapter is the classification of surfaces. It was already proved that no two of the surfaces F3 and (1),. are homeo-

morphic. The next theorem will show that there are no other possible surfaces.

19.5. THEOREM: Proof:

Each surface is homomorphic to an Fg or a AIH’U

be the functions defined just prior to Theorem 3.2. Here, in contrast

to Chapter 3, the ranges of the latter two functions have been restricted to A,+q. Definition: To each ordered pair (c’, c”) where c” E S"(X, A; M) and d4 e S°(X, A; N) is associated a p + q—cochain

c" V da 6 S’+°(X, A, P) by means of

0, when p or q is negative

(c’ v door) = {crows -~ -. d.) o warm» - - » dun», when p and q are non-negative.

It is clear that c" v da is zero whenever one of the factors is, and that the operation v defines a bilinear function S"(X, A; M) x S°(X, A; N) —> S“°(X, A; P). In other words, the cup product is distributive and R-homogeneous.

20.1. THEOREM:

8(c" v d“) = 8c" v d0 + (—1)Pc” v 8d“,

where c" e S”(X, A; M) and d“ e S°(X, A; N).

Proof: (1) If p < 0, then 0" v d“ = 0, 8c” = 0, and cp = 0. A similar statement holds for q < 0.

Singular Cup Products

(2)

203

Let p and q be non-negative. It is sufficient to investigate the action of the terms of the equation on an arbitrary p + q + l simplex as follows:

8(c’ v d°)(1-) = (c’ v d¢)(73(A},+q+,)) ' ' ‘a J}, ' ' ' : dp+a+l)) = 11:1 (_1)J(cp V d“X'r(do,

= ”,2 M has a unit element in N. The interesting case is the one in which M = N, M is an associative ring with a unit, and

cup products of pairs of elements of H*(X, A; M) are being considered. Then the results of 13.9, 13.10, and 13.1 1 yield 20.12. THEOREM:

If M is an associative ring with a unit ele-

ment, then H*(X, A; M) is a graded, associative ring with a twosided unit element. The ring in Theorem 13.2 is called the homology ring Definition: of the pair (X, A). M is often taken to be the ring of integers.

Introduction to Algebraic Topology

208

Examples:

(1) H‘(S"; M) is the direct sum of two homogenous submodules: H°(S") and H7'(S") both of which are isomorphic to M; that is, H*(S"; M) has the form Re0 (43 Re". Here, eo is clearly the unit element of H*(S", M). Only the product e,, v en remains to be

investigated; but this product has degree 2n and is therefore zero. The module H *(S", M) is the “module of dual numbers” of Stedy.

(2) The torus and the figure obtained by attaching two disjoint circles to a 2-sphere have isomorphic cohomology groups in all dimensions, but their cohomology rings (with integer coefficients) do not have the same multiplicative structure in dimension 1. The computations here are left to the reader.

Let f: (X, A) —> (Y, B) be continuous. By definition,

f*(6’0) = cl’(f(rr))It follows immediately that

20.13.

f#(c9 V d") = (fatten) V (f#dq)

where c" and d" are cocycles, and that

20-14-

f*(d v .3) = f*(d) Vf*(l9)

where a and ,8 are in cohomology groups of (X, A) with the given coefficients. Thus, f* commutes with cup multiplication.

In case a cohomology ring of (X, A) is under consideration, Equation 20.14 shows that f“ is not just a module homomorphism. It is also a ring homomorphism of degree zero.

21 The Singular Cap Product

This chapter defines a product of a chain by a cochain which

yields a chain, and the corresponding product of the homology by the cohomology group is investigated. Definition: Let M x N —>P be a bilinear function denoted by the product m on. Let c" e SP(X, M) and let ca 6 Sq(X, N). Here ca = 0 ifq < 0, and, if q 2 0, ca = 2,b,a,, where b, e N and a" is a q—simplex. The cap product cp A ca associates with the pair (c’, cu) a (p —— q)-chain of S,_q(X, P) by means of c’Ac,,=0

whenq Ila-AX, P) which is also termed the cap product, and it induces a bilinear function

H*(X, M) x H*(X, N) —> H*(X, P) whereby the homogenous elements ofdegrees (p, q) are sent into homogenous elements of degree q — p.

The Singular Cap Product

213

As in the case of cup products, let M, N, and Q be modules for which (m o n) o q = m o (n o q) and all the products of this form are defined and lie in a module P.

21.8. THEOREM: ea 6 SAX, Q) then

Let

c' e S’(X, M),

d” e S"(X, N),

and

6' A d” Aeq) = (c’v d”) A eq. Proof:

(1) If one of r, p, or q is negative, then both sides of Equation 21.8 are zero. Hence, all three can be assumed to be non-

negative. (2) If p + r > q, then r > q — p, and both sides of 14.8 are zero. Hence, it can be assumed that p + r g q.

(3)

If r, p, and q are non-negative and p + r 3 q, then it is sufficient to prove 14.8 in the special case that ea is the chain be, where b 6 Q and a is a simplex in X. Then

c" A (d’ A 170')

= 0’ A {[d‘"(0(du-m ' ' u dq» ° b](0(do, ' ' 'a d.-p))} = [C’(0(do, ' ' ', da-nxda—p-n - ' ', dq—n» ° d"(¢7(da—m ' ' -, do) ° b]' (”(4, ' ' u da_p)(do, ' ' ', da—p-r»

= [c'o(d,_,_,, ' ' u dq-p) ° flaw..-” ' ' -, dc) ° b]0(dor ' 'a dr-fl-r)‘ 0n the other hand, (6' V ‘1’)" [70' = [(cr V dpx¢7(dq-r—m ' ' ', (1,1))0 b10010: ' ' " da-r—p)-

Here,

(Cr v d’Xd —r—m ' ' ' a dc) = cf(a'(da—r-na ' ' ' ’daxdo’ ' ' ' a dr) ° dp(°'(d -r—p: ' ‘ ‘ a 14a ' ' ‘ 9dr”) = cr(°'(da-r—m ' ‘ 'a (la—9)) °

9(a(dq-p’ ’ ‘ 'a dc)-

This completes the proof of the theorem. If M = N, M is an associative ring and P is an 21.9. Corollary: M—module, then H*(X, P) is an H*(X, M)-module.

Introduction to Algebraic Topology

214

21.10. THEOREM: Let c° be the cocycle whose value is l on each 0-simplex. Then c° is a unit element in H *(X, M) and H*(X, P) is a unitary module. Proof:

0° A be = [6°(tr(dq)) ° blddo, - ‘ -, d.) = be. If M X N —> N is bilinear, M is an associa-

21.11. Corollary:

tive ring with a unit element, and N is a unitary M-module; then H*(X, N) is a unitary H*(X, M)-module. 21.12. THEOREM:

If a e Hq(X), B e H"(Y), and f* and f*

are the homomorphism of the homology and cohomology groups induced by a continuous function f: X —> Y, then

f*(f*3 " a) = ,3 Afwt. Let c" be a cocycle in 3, let a- be a p-simplex in X, and let Proof: b e N. Then f#cp A 170' = [f#c’(0’(dq_p, . . ., dq) 0 510(4, . . ., da—p)

= [c’fdd ' ' 'a do) ° b]‘7(dm ' ' '2 du—p)-

Then f#(f#cp A do

= [C”f0(d -ps ' ‘ '3 do) ° b]f0'(do, ' ' ': dq-u) = cp Aftba's

and this establishes the theorem. Let M = N = R be an associative ring. Let c" be a cochain in S’(X) and let d, be a chain in 3,,(X). Then c" A d, e S0(X). In the module S0(X), the augmented boundary operator 50 associates the

empty simplex with each 0-simplex and associates the product of the coefficient sum and the empty simplex with each 0-chain. If d, = bo, then

c" A d, = (CPU 0 b)(a-(do)) and Jew A be) = 0% o b = c’(ba). By linearity, it follows for arbitrary d, that

«me» A (1,) = cl'(d,,).

The Singular Cap Product

215

Therefore, the question of whether c" A d, is a cycle in the augmented

theory can be settled by the value of c”(d,,). As an exercise, the reader can use the cap product to define the following products for space pairs (X, A): H'(X, A; M) x Hq(X, A; N) —> Hq_,,(X, P) HP(X, M) x Hq(X, A; N) ———> Hq_,(X, A; P) Part of the problem here is that 0(d0, - - -, dad) could have a carrier

in A even when 0(d0, - - -,p,) does not.

22 The Anticommutativity

of the Cup Product

This chapter begins with a consideration of affine simplices. Let (a0, - . -, a.) be an affine simplex A —> E; thus (a0, - - -, au)

6 A(Aq, E). Definitions:

The

homomorphism

pa: A(Aq, E) —> A(Aw, E)

is defined by Pq(aoa ' ' ‘: ac) = 6:1(aq: ' ‘ 's “0),

where ea = (—1)“(°“’”. The homomorphism recursively by

Du: A(Aa, E) —-—> A(Aa, E)

216

is

defined

The Anticommutativity of the Cup Product

217

D_l = 0, D0 = 0, and

Dq(°') = “0(0' _ Pq(a') —‘ Da—l(aao'))a for q 2 1-

Here the statement D_l = 0 is a formality. When q = l,

D1(¢r) = (do, do. a1) — (a... (11,00)22.1. Lemma:

If q 2 1, then aapa + 10.1-18”.

Proof: 3mm. ' ' ', ac) = eaaa(aqs ' ' 's “0) ' ' ‘2 a0), = 1:1 60(_ 1)U‘1(aa, ' ' " fin

where (‘1‘, is located at the (q — i)th position. The lemma follows from

the fact that ea(—— 1)‘1 = eq_1 and €a—l(aq9 ’ ' ‘y (in ' ' " a0) = Pa—1(ao: ' ‘ ‘1 at, ' ‘ '; ac)-

22.2. Lemma:

aqflDq + owed = Id — pa.

Proof: Since D_l = 0 and D0 = 0, the lemma is true when q = 0. Suppose that 22.2 has been proved for q — l; in other words, that

34D".l + D,,_2aa_l = Id — pH. A multiplication on the right by 3., yields aHaq = 3a — p443“ = a, — 3gp“.

By definition of D“,

3ma = aquaofir — p407) - Dq—:(3a0))Since awao + aoaa = Id, it follows that

aaHDaa' = (0' —‘ Pa(‘7) _ D¢_,(3¢0')) _ ao(3q0 _ aaPaa'

— (an — 3aPa)(U)) = a‘ — pqa' — Da_,3¢a'.

This establishes 22.2.

Introduction to Algebraic Topology

218

Notice that the recursion formula 22.2 shows that Doc is a linear combination of (q + 1)-simplices with coefficients that are i1 and that the vertices of these simplices are among a0, - . -, aq. Now let the space E be restricted to the simplex A4, so that pq(A,§) e A(Aq, Ag) and BAA» E A(An+ia Ac)-

22.3. Lemma: Proof:

pad = a'pa(A",) and a' = a'D¢(A}).

The first part of 22.3 follows immediately from the defini-

tions of the symbols involved. For the second part, let v0, - - -, v“,

be integers between 0 and q. Then (an, ' ' '9aa)(dvu ' ‘ 'rdv.) = (duo: ' ' 'sav.)

and (a0, ' ' ': aad ‘ ‘ ‘9 diam) = (aw ' ' ', alien)’

Hence, Dq(Aq'1)aa+l(Ah-H) + aq(A;)Da—I(Ah—l) = aq+q(A4:) + Dq-laq(A:;)

= (Id — pow.) = A: — paw). Now the considerations shift from affine simplices to space pairs and singular homology. Let (X, A) be a space pair, and let S.,(X, A) be formed with coefficients in the ring Z of integers. The reader will

remember that SAX, A) is the factor group Sq(X)/SU(A). Definition:

If a e S.(X), let

a” =

Id

when q g 0

MW.)

when 4 2 0,

and 0

when q s 0

aDq(AI,)

when q 2 0.

Duo =

22.4. Lemma: Equation 22.2 holds under the extended definitions of pa and Do; that is,

The Anticommutativity of the Cup Product

219

60+,Dq + D¢_laa = Id — pa. Proof:

If q 3 0, 3WD” + Dadaaa- = 0 = (Id —- p¢)o-.

If q 2 0, then

8....a + Dq-laqa = U{(Da(A3)3a+1(A&H) + 34(Ah)Da-1(Ah—l)} = am; - MAD) = o — pea.

22.5. Corollary:

The homomorphisms pa and D4 induce homomor-

phisms (of the same name) on the factor group Sq(X)/Sq(A) which also satisfy 22.2. Let M be an R-module. Then the cochain complex Sq(X, A; M) was defined to be Hom (S‘,(X, A), M).

Definition:

The homomorphisms pa and Du induce homomor-

phisms (with reversed direction)

pr: Saar, A; M) ——> Saar, A; M) and

D3‘: S¢“(X, A; M) —> S4(X, A; M) by means of

(93796. = f“(paw and (Df+1f°“)cq =f“1(c0): where ea 6 Sa(X, A) andf0+1 e S"+1 (X, A; M).

22.6. Lemma: Proof:

8D: + D3118 = Id — pa.

Consider the image of the cochain f«+1 under the left-

hand side of 22.6. On the chain ca“, this has the value

Introduction to Algebraic Topology

220

((80? + D3118)!"9%) = (8(q))(€¢) + (Di+n(3f“))(ca) = D?f"(3ca) + 8f"(Daca) =fq(Dq-lac¢ + aDaca)

=f”((Id — pace) = ((Id — P#)f')(€q)Since pf is chain homotopic to the identity by 22.6, it follows at once that p: = Id. This means that if f'1 is a representative of the cohomology class a, then pff*1 is also a representative. Furthermore,

217-

(Pif‘Xo) =f"(pq(a)) = 2«f"(o‘Pa(A3)) = Zuf"(001...

(4))-

The tools needed for the investigation of the anticommutativity

of the cup product are now at hand.

22.8. THEOREM:

If M = N and multiplication M is commuta-

tive, or if a product it o m where n e N and m e M is suitably

defined, then an V Ba = (_ 1)M(Ba V an)

Proof:

Let fP and g" be representatives of the cohomology classes

a and B respectively, where

a e H’(X,A;M) andfil e H"(X,A;N).

Then ptf1' and pig" also serve as representatives, and the cohomology class a v )9 is represented by

(pif’) v (PW)The value of this cochain on the simplex (chain) a is

(P*f’ v p*g‘)( P. This permits the definition of a product N X M —> P by means of nom=mon.

(If N = M, this definition is unambiguous only if multiplication is commutative.) The new multiplication permits the definition of ,8 v a and the computations just carried out prove Theorem 22.8 when

q 2 0 and p 2 0. The theorem is trivial when q < 0 or p < 0. The reader who is familiar with the definition of a grassmannian algebra will see that the graded cohomology ring with coefficients in

a commutative ring is such an algebra.

Index

lndex Additive functor Adjoin 146 Admissible

109

Category 107 Linear— 108

Chain

—functions 130 —pairs 130 ——triples 130 Affine —map 14 —simplex 20 —space 10

Antipodal function

Axioms for—

79

Associative law —for cup product 207 ——for tensor product 87 Attaching of n—oells 158

Augmented complex

29

Axioms —for a category 107 —of Eilenberg and Steenrod

129 ff.

Bilinear

12

169

Boundary —of affine simplex 22 —of prism 41 ——of singular chain 28 4

Cap product 209 Cauchy integral theorem

Carrier

28

—of CP" 1771f. ——of SI 122 —ofP" 180 Cokcrnel 131 Commutative diagram 4 Commutative law —for tensor products Complex Augmented—

19, 86

—operator Brouwer 152

130 ff.

Singular— 121 —ofa point 122

—for cup products

Barycenter 57 Barycentric coordinates

Basis 18 Betti number

27

—complex 4 ——homomorphism 4 ——homotopy 43 Coefficient module 131 Cohomology 120 Cohomology group

126

88

220

27

Chain— 4 CW— 164 Homomorphism of— 4 Simplicial— 194 Spherical—

164

Complex projective n-space Component, path— 71 if. Connected, n—tuply— 81

Contravariant

98

—functor Convex 12

109

———hull 13 Coordinate vector Covariant 98

11

177

Index

226 —functor Cup product Cycle 4

Multilinear— 85 —of a space tuple

108 202

Deformation retract Strong— Degree 81 Diagram 3

Additive— 109 Contravariant— 109 Covariant— 109

51

51

Genus

Commutative— 4 Nine— 92 Diameter 61 Invariance of—

Graded module

120 ——product 17 —sum 17, 70 Divisible module 95 Domain, invariance of— Eilenberg 129 Epimorphism 2 Equator 75 Equivalence Homotopy— 49 —in acategory 108 Exact —sequence Excision

170

74

131

——theorem

Hemisphere 75 Hexagon theorem 137 Homologous to zero 125

Homology 156

—ring 207 ——sequence 6, 130 —sequence of a triad 140 —sequence of a triple 35 Homology group 4 Augmented— 29 Reduced— 29 Singular— 29

——of a chain complex

Five lemma 55 Free module 18 Fox, R. 154

4

Chain— 43 —axiom 131

42

4

—-of a pair 31 —ofa point 73, 120, 131

Homomorphism Chain— 4 Homotopy 42

67,117

Function Affine— 14 Bilinear— 19 Homotopy of a—-

71,

—of CP" 177 —of1"'l 178 —of S" 77

3

141

——axiom

18

205

Graph 53 Homology groups of— 80 Application of Mayer-Victoris theorem to— 145

157

—axiom 131 —of an affine space 12 Direct —decomposition theorem

Euler characteristic

146

Generating system

Dimension

—triad

32

Functor

——equivalence 49 —inverse 49 —of space pairs 49 —-theorem 47, 117 Hull, convex— 13

Index

227

Independent points Injective

—module 163 —representation

12

Nine diagram

Path-component 7111'. Prism construction 39 Product Cap— 209

112

Inverse Homotopy— 49 —ofa morphism 108

Isomorphism

Cup— 202 Direct— 17 Tensor— 86

2

Jordan-Brouwer separation theorem 1 52

—theorem

Divisible—

153 147

106

Real projective n-space Region 155 Representation Injective— 112 Projective— 112 Retract 50

Deformation—

118

205

Homorphism of graded—— 205 Injective— 163 Projective— 163 Quotient— 106

R— l (R,S)— 101 Torsion— 96,103 Torsion free— 103 Monomorphism Morphism 107 Multilinear 85

91

178

51

95

Free— 18 Generating system for—

Graded—

Quotient module

143

—-theorem, relative case Module Basis for— 18 Coefl‘icient— 131

—of functions Projective

—module 163 —plane 172 —representation 112 Projective spaces 175 ff. Euler characteristics of— 180 Homology groups of— 180 Proper triad 141

Kernel 1 Klein bottle 197 Knot 155 Artin-Fox— 154 Lens space 189 homology of— 190 Line 11 Linear category 108 Manifold 173 Mayer-Vietoris —sequence

92

2

Sequence 2 Exact— 3, 5 Mayer-Vietoris— Simplex 14

Affine—

143

20

Empty— 20 Singular— 27 Sphere 75, Chapter 18 Homology groups of—

144 Separation on— Steenrod 129 Subdivision 57 Sum, direct— 17

150 ff.

77,

228

Index

Surface

173

Triangulable space pair

Classification of— —-of genus g 146 Tensor product

86

Torsion 103 Torus 172 Triad 140 Exact— 141 Proper—

14]

195 11‘.

194

Triple, homology sequence of Vector field

Vertex 14 Vieton's 143

79

35

Index of Symbols A v

209 209

H*(X, M) 213

H..(X, A) 30 H¢(X, A)

12

(.) 20

Hom

f: 4

L(n, m) 189 P" 175 Sd 58 S" 75 SAX) 27 d 4, 23 a. 5 x 179

f# 33 f(X,An,--~

31

9Ar)—’(Yr BI, ‘ 3 ' Br)

A0(X, Y) 19 A(X, Y) 19 @(A) 141 C(X, Y) 19 CP" 175 D‘, 65 F, 173

we 39

2, 98

A, 20 A; 24 (1),. 173

229

LITTLE LIST OF TYPOS — The letter R tends to do double duty in the book, book, somesome— stand— times standing for an arbitrary ring while often also stand(and occasionally other things, things, ing for the real numbers (and like a point in a space). space). — Page 3, 3, line 9: 9: In ln condition (2), (2), replace the "subset" "subset" "superset" sign. sign. sign by a "superset"

— Page 5: 5: "homeomorphisms" "homeomorphisms" should be "homomorphisms" "homomorphisms" here. here. — Page 5: 5: "the "the exactness at the qth location" location" should be "the exactness at B_{q-1}". Bi{q—1}". "the

— Page 7, 7, item (3): (3): "Let "Let c_q ciq = : \partial C_{q+1} Ci{q+1} \in H_q(C)" Hiq(C)" "Let c_q ciq + \partial C_{q+1} Ci{q+1} \in H_q(C)". Hiq(C)". should be "Let — Page 11, 11, line 4: 4: "O'' "0" \in E" E" should probably be "O'' "0" \in E'". E'". (Although (Although it isn't very clear why a distinction bebe— E' is made here in the first place.) place.) tween E and E' — Page 11, 11, first displayed equation: equation: "O'(O', "O'(O', Q)" Q)" should be "\phi'(O', "\phi' ——> arrow should be reversed. reversed.

— Page 163, 163, definition of "injective": "injective": In 1n the second diadia— gram on the top, top, the arrow from P to B should be reversed. reversed. Also, Also, the period after this diagram should be removed. removed.

— Page 196: 196: "a "a triangle \Delta_3 \Deltai3 or \gamma" \gamma" should be "a "a triangle \Delta_3 \Deltai3 of \gamma". \gamma". — Page 201: 201: "A "A cochain c^p cAp \in S^p(X, SAp(X, A) A) is" is" should be "A "A cAp \in S^p(X, SAp(X, A; A; M) M) is". is". cochain c^p

— Page 202, 202, Definition: Definition: "pair "pair (c^p, (cAp, c^q)" c)" should be "pair "pair (c^p, (CAP: d^q)". q) " . — Page 202, 202, Definition: Definition: "S^{p+q}(X, "SA{p+q}(X, A, A, P)" P)" should be "S^ "SA {p+q}(X, {p+q}(X, A; A; P)". P)"— Page 203: 203: "when "when j j > P" P" should be "when "when j j > p". p". -

— Page 205, 205, line 6: 6: "H^{p+q}(X, "HA{p+q}(X, A, A, P)" P)" should be "H^ "HA {p+q}(X, {p+q}(X, A; A; P)". P)"— Page 206, 206, the exact triangles from the Mayer-Vietoris Mayer—Vietoris sequence: sequence: The "+" "+" signs should be "\oplus" "\oplus" signs. signs. — Page Page 206, 206, the second second exact triangle on the right: right: The \psi. left arrow should be labelled \psi.

— Page 207, 207, proof of Theorem 20.10: 20.10: In the first display, display, "\sigma(d_1) "\sigma(d71) — (d_0)" (d70)" should be "\sigma(d_1) "\sigma(d71) — \ sigma(d70)". sigma(d_0)". — Page 215, 215, last line: line: The "p_q" "pig" should be a "d_q". "dig". (The (The "p" "p" here seems to be an upside-down upside—down "d".) "d".) — Page 217, 217, Lemma 22.1: 22.1: Replace the "+" "+" sign by an "=" ":" sign. sign. ... and probably many more... more... ...