Homology of Cell Complexes 9781400877751

Table of contents :
Foreword
Contents
Introduction
I. Complexes
II. Homology Groups for Regular Complexes
III. Regular Complexes with Identifications
IV. Compactly Generated Spaces and Product Complexes
V. Homology of Products and Joins Relative Homology
VI. The Invariance Theorem
VII. Singular Homology
VIII. Introductory Homotopy Theory and the Proofs of the Redundant Restrictions
IX. Skeletal Homology

Citation preview

HOMOLOGY OF CELL COMPLEXES BY GEORGE E. COOKE AND ROSS L. PINNEY (Based on Lectures by Norman E. Steenrod)

PRINCETON UNIVERSITY PRESS AND THE UNIVERSITY OP TOKYO PRESS

PRINCETON,

NEW JERSEY

1967

Copyright (£) 1967, by Princeton University Press All Rights Reserved

Published in Japan exclusively by the University of Tokyo Press; in other parts of the world by Princeton University Press

Printed in the United States of America

"Foreword

These are notes based on an introductory course in algebraic topology given by Professor Norman Steenrod in the fall of 1963·

The principal aim of these notes is to

develop efficient techniques for computing homology groups of complexes.

The main object of study is a regular complex:

a CW-complex such that the attaching map for each cell is an embedding of the boundary sphere.

The structure of a

regular complex on a given space requires, in general, far fewer cells than the number of simplioes necessary to realize the space as a simplicial complex. procedure»

orientation —• >chain complex

And yet the

?homology groups

is essentially as effective as in the case of a simplicial complex. In Chapter I we define the notion of CW-complex, due to J. H. C. Whitehead.

(The letters CW stand for a) closure

finite—the closure of each cell is contained in the union of a finite number of (open) cells— and b) weak topology— the topology on the underlying topological space is the weak topology with respect to the closed cells of the complex.) We give several examples of complexes, regular and irregular, and complete the chapter with a section on simplicial complexes. In Chapter II we define orientation of a regular complex, and chain complex and homology groups of an oriented regular

complex.

The definition of orientation of a regular complex

requires certain properties of regular complexes which we call redundant restrictions.

We assume that all regular

complexes satisfy these restrictions, and we prove in a leter chapter (VIII) that the restrictions are indeed redun­ dant.

The main results of the rest of Chapter II are ι a

proof that different orientations on a given regular complex yield isomorphic homology groups, and a proof of the universal coefficient theorem for regular complexes which have finitely many cells in each dimension. In Chapter III we define homology groups of spaces which are obtained from regular complexes by making cellular identifications.

This technique simplifies the computation

of the homology groups of many spaces by reducing the number of cells required.

We compute the homology of 2-manifolds,

certain ^-manifolds called lens spaces, and real and complex projective spaces. Chapter IV provides background for the Kunneth theorem on the homology of the product of two regular complexes. Given regular complexes E and L there is an obvious way to define a cell structure on |K| X [i[ —simply take products of cells in K and in L.

But the product topology on |κ| X |LJ

is in general too coarse to be the weak topology with respect to cloeed cells.

Thus

K X L , with the product

topology, is not in general a regular complex.

To get around

this difficulty we alter the topology on the product.

The pro-

per notion is that of a compactly generated topology.

In order

to provide a proper point of view for this question we include in Chapter IV a discussion of categories and functors. In Chapter V we prove the Kunneth theorem. We also compute the homology of the join of two complexes, and we complete the chapter with a section on relative homology. In Chapter VI we prove the invariance theorem, which states that homeomorphic finite regular complexes have isomorphic homology groups. We also state and prove the seven Eilenberg-Steenrod axioms for cellular homology. In Chapter VII we define singular homology. We state and prove axioms for singular homology theory, and show that if X is the underlying topological space of a regular complex K, then the singular homology groups of X are naturally isomorphic to the cellular homology groups of K. In Chapter VIII we prove Borauk's theorem on sets in S

which separate S and Brouwer's theorem (invariance of

domain) that Rm and R are homeomorphic only if

m - n.

We show that any regular complex satisfies the redundant restrictions stated in Chapter II, and settle a question raised in Chapter I concerning quasi complexes. In Chapter IX we define skeletal decomposition of a space and homology groups of a skeletal decomposition.

We

prove that the homology groups of a skeletal decomposition are isomorphic to the singular homology groups of the underlying space.

Finally, we use skeletal homology to

show that the homology groups we defined in Chapter III of a space X obtained by identification from a regular complex are isomorphic to the singular homology groups of X. We should mention that we sometimes refer to the "homology" of a space without noting which homology theory we are using.

This is because all of the different

definitions of the homology groups that we give agree on their common domains of definition. We also remark that cohomology groups, which we define for a regular complex in Chapter II, are only touched on very lightly throughout these notes. We do not cover the cup or cap products, and we do not define singular cohomology groups. Finally, we wish to express our gratitude to those who have helped us in the preparation of these notes.

First, we

thank Martin Arkowitz for his efforts in our behalf—he painstakingly read the first draft and made many helpful suggestions for revision.

Secondly, we thank the National

Science Foundation for supporting the first-named author during a protion of this work. thank

And finally, we wish to

Elizabeth Epstein, Patricia Clark, Bonnie Kearns,

Barbara SuId, June Clausen, and Joanne Beale for typing and correcting the manuscript.

George E. Cooke Ross L. Finney June 2, 1967

TaTsIe of Contents

Foreword Table of Contents Introduction

i

I. Complexes

1

II. Homology Croups for Regular Complexes

28

III. Begular Complexes with Identifications

66

IV. Compactly Generated Spaces and Product Complexes

86

T. Homology of Products and Joins Relative Homology

106

TI. The Invariance Theorem

138

TII. Singular Homology

177

Till. Introductory Homotopy Theory and the Proofs of the Redundant Restrictions

207

IX. Skeletal Homology

239

IMEODUCTIOIi One of the ways to proceed from geometric to algebraic topology is to associate vith each topological space

X

a sequence of abelian groups

00

CH ( X ) } , and to each continuous map 1 q.=0 Y a sequence of homomorphisms

f

of a space X

into a space

f : H (X) — > H (Y) , q* qr qr one for each

q . The groups are called homology groups of X

as the case may be, and the morphisms

f

or of

Y3,

are called the induced homo-

morphisms of f , or simply induced homomorphisms. Schematically we have

geometry space X map

f: X — >

—^

—->

algebra

homology groups

Y —>

H (X)

induced homomorphisms

f

The transition has these properties, among others: 1. If f: X — > X

is the identity map, then f

is the identity isomorphism for each 2.

If

f: X —S> Y

: H (x) — > H (X)

q .

and g: Y — > Z , then

(gf) 4*

each

q .

= g

·f

1*

That i s , i f the diagram

X

>

Z

gf of spaces and mappings is commutative, then so is the diagram i

for




Hq(Z)

of homology groups and induced homomorphlsms, for each

q. . In modern

parlance, homology theory is a functor from the category of spaces and mappings to the category of abelian groups and morphisms. The correspondence here between geometry and algebra is often crude. Topologically distinct spaces may be made to correspond to the same algebraic objects. For example, a disc and a point have the same homology groups. So do a

1-sphere and a solid torus (the cartesian product of

a one-sphere and a disc.) Despite this, the methods of algebraic topology may be applied to a broad class of problems, such as extension problems. * Suppose we are given a space X , a subspace A h

of A

of A

into a space

Y . If we let i denote the inclusion mapping

into X , then the extension problem is to decide whether there

exists a map of X

of X , and a map

f , indicated by the dashed arrow in the diagram below,

into Y

such that h = fi .

X

*Unless otherwise stated, the word space will mean Hausdorff space. 11

If f

exists, we say that f

lently, that

h

is an extension of h

has been extended to a mapping

f

to X , or, equiva-

of X

into Y .

There are famous solutions of this problem in point-set topology. The Urysohn Lemma is one. The hypotheses are: the space X

is normal,

but otherwise arbitrary; the subspace A = A U A ,

, where A

are disjoint, closed subsets of X ; the space

is the closed unit

interval

[0,1] ; and the map

h: A — > Y

Y

carries A

to 0

and A,

and A.

to" 1 .

Under these assumptions the Urysohn Lemma asserts that there exists a mapping

f: X —S» Y

such that

f = hi : X

>

(A0 U A 1 ) — g

[0,1] = Y

The Tietze Extension Theorem is another solution. Here, X normal, A set Y

is closed in X , and h

is an arbitrary map of A

is to the

of real numbers.

The theorem asserts the existence of an extension

iii

f: X — > Y

of h .

Pathwise connectivity can also be described in terms of extensions. Let X

be the closed unit interval, and let A = {0,1} . A space Y

is pathwise connected if, given two points mapping

h: A — > Y defined by h(0) = y

extended to a mapping

y

and

and

y,

of Y , the

h(l) = y., , can be

f: X —S» Y .

Furthermore, the problem of the existence of a continuous multiplication with a prescribed two-sided unit in a space Y

can be phrased as

an extension problem. We seek a map m:

and an element

in Y

such that m(l,y) = m(y,l) = y

YXY —> Y

for each y

in Y . The require-

ment that there be a two-sided unit determines a function h space A = (l X Y) U (Y X l) = Y V Y

of Y X Y

1

on a sub-

into Y . The existence

of a continuous multiplication with this two-sided unit is now equivalent to the existence of a map m:

YXY —> Y

such that mi = h .

YXY

YVY Corresponding to each geometric extension problem is a homology extension problem, described for each

q by the following diagram:

*

N x

\

φ \

H4(A)

>

Given the groups, the homomorphisms φ:

homomorphism If f φ =ΐ

H g (Y)

i^. and

H (x) — > H (Y) such that

h^ , does there exist a XJ) = h # ?

is a solution of the geometric extension problem, then

is a solution of the algebraic problem for each

other hand, if φ

does not exist, then f

q. . On the

does not exist either. This

latter iniplication is the principal tool in the proofs of many theorems of algebraic topology. Let A

be a subspace of a space X . A map

retraction if fχ = χ stances, A

χ

X —> A

is a

in A . tftider these circum­

is called a retract of X . The Tietze Extension Theorem

implies that if X X

for each point

f:

is a normal space which contains an arc A , then

can be retracted to A . Throughout this and subsequent chapters, IT will denote the closed

unit ball in Euclidean η-space (n-l)-sphere in R n

THEOREM: S "

R

, and

S "

will denote the unit

.

is not a retract of E

11

.

Suppose to the contrary that there exists a map such that

fχ = χ

for each χ

the identity map h:

in S "

. Then

f

f: W — > S is an extension of

ί

v.

,n-1

h

If η = 1 , the fact that E shows that no such

f

A n _i ?* S

is connected, while

S

is not,

can exist.

If η > 2 , a different argument is required. Assume that the homology groups of ET

and of s "

are, as we will later show them

to be, Z

q = 0

0

q> 1

,n-l H (S C " X )%

H (E11) =

If

f

Z 0

exists, then for each

q = 0, η - 1 0 X

which fχ = χ . Using the fact that S

is a

is a point

Sn" χ

onto E31 .

in X for

is not a retract of E

we

can prove the Brouwer Fixed-point Theorem: THEOEEM: Each mapping

f:

E11 — > E11 has a fixed point.

Suppose to the contrary that there exists a mapping

f: E11 — > E11

which has no fixed point. Then we can define a reir&ction g: E11 — > S n ~ For each point which starts at for each χ If χ

χ

of IT we let Rx

denote the directed line segment

fx and passes through

in ET , since

lies in S "

χ . Note that Rx

is defined

f has no fixed point. Let gx = Rx fl(sn" -iix

, then gx = £x} fl S

= χ . The verification

that

g

is continuous is left as an exercise.

Thus

g

is a retractian of E11 onto S n ~ . This contradicts the preceding

theorem, and completes the proof of this theorem. vii

Chapter I COMPEEXES 1. Complexes If A f:

is a subspace of X

and B

(X,A) — > (YiB) of the pair

of X

into Y

that carries A

a subspace of Y , then a map

(X,A) to the pair

(Υ,Β) is a map

into B . One can compose mappings

of pairs. For each pair there is an identity mapping of the pair onto itself. A homeomorphism of X

onto Y -which carries A

(X,A) onto

(Υ,Β) is a homeomorphism of

onto B . A more important concept is

that of a relative homeomorphism. A mapping f|(X-A) maps

f:

(X,A) — S * (YJB)

is a relative homeomorphism if

X-A homeomorphically onto Y-B . A relative homeo­

morphism need not map A

onto B .

Exercise. Let X be a compact space, and suppose that f:

(XJA) — > (Y>B) is a mapping such that

to-one onto Y-B . Show that if Y

f|(x-A) maps X-A one-

is a Hausdorff space then f is

a relative homeomorphism. Exercise. Find an example to show that if Y

is not a Hausdorff

space then f may not be a relative homeomorphism. 1.1. The definition of a complex. A complex K

consists of a Hausdorff space

of subspaces, called skeletons, denoted by which satisfy the following conditions. 1

|κ| and a sequence

|κ | η = -1, 0, 1,...,

1.

is the empty set, and

2. Each

is closed in

3. For

, the components

are open sets in the relative topology of

. They are referred to as

the n-cells of K . 5. For each n-cell

denote that subspace of

|k| which is the closure of and let

in

For each

with the relative topology, there exists a relative hameo-

morphism

which carries

onto

•n c^ . For convenience,

is called a closed n-cell of K , even

though it need not be homeomorphic to E11 . 6. The topology of cells of K: a subset A section 7-

is the weak topology defined by the closed of

|K{ is closed if and only if each inter-

is closed in The relative topology on

by the closed cells of

coincide:

in the relative topology of for each

and the weak topology defined a subset A

of

if and only if

and for each

is closed is closed in

i .

1.2. The structure of a complex. A complex K

is said to be a complex on

, and

is called

the underlying space, or the geometric realization of K . A finite complex is one with finitely many cells. A complex is called infinite if it is not finite. 2

If there exists an integer q > r , then K

r

such that

|κ ,[ = [K [ for each

is said to be of finite dimension. The least such r

is called the dimension of K . If

σ is a cell of K , we will sometimes write

though K

is not defined as a collection of cells.

The subspace

|κ I is a discrete subspace of

which are points by (5), are also open sets of of

σ ε K , even

|κ| . Its components,

|κ I , by (4).

The cells

IK I are called vertices of K . The relative homeomorphisms

f

. are not part of K . To show

n,i

that a candidate for a complex satisfies (5) one need only exhibit some set of

f's . A complex is to remain unchanged when one set of

f*s

is

replaced by another. By (5), an η-cell

σ.

of K , with the relative topology, is

actually homeomorphic to the η-cell IT - S n " not be homeomorphic to S n " morphism 1

,

1

'

f

11

onto "a , then K

n ^ l

'

wise, K

. of E

. If, for each

σ.

6. need

there exists a homeo-

is called regular. Otherill ·Τΐ M M l I

X

is irregular. If

relative homeomorphisms

ση

. Its boundary

is a cell of K

and if none of the

f . is a homeomorphism, then n,x

σ. is called i

irregular. In order to be sure that the topologies in (6) and (7) are welldefined, one must know that

|κ| = q,i U a.

and that

|κ | = i; Uq < n σ?ι

-

These two facts follow, as they should, from conditions (l) through (5). We leave their verification to the reader. Exercise. Estahlish the following elementary facts about "the structure of K .

3

1. 2. 3. Also, although this won't be used immediately, for each

q-cell

of K . (While the first three parts of this

exercise are easily seen to be true, the reader may find this fourth part to require some moments of reflection.) 5. Let X

be a space that is the -union of finitely many disjoint

subsets

Let

let

denote the closure of

in X and

. Suppose that for each i 1. There exists a relative homeomorphism of for some k

k . (We call

the dimension of

2.

a

onto k-cell and call

.)

lies in a union of cells of dimension lower than that of

Then the

are the cells of a complex on X •

6. A function only if

f

from

into a space

is continuous for each cell

7. If K

X

is continuous if and

a of K .

is of.finite dimension, then condition (7) implies

condition (6).

2. Examples A regular complex on S n . For each

identify k

-with the subspace

, and for each

let the

Certainly

is a Hausdorff space,

is closed in The

, each

, and

0-skeleton

consists of a

pair of points. The

(q-l)-skeleton

is the equator in divides

q-skeleton of K be the q-sphere

, and

into two (open) hemi-

spheres,

and

, whose union

is The map

is the homeamorphism defined by vertical

projection:

in

. A reversal of sign gives the hameomorphism for

verification that the

are the cells of a regular complex on

is now trivial by Exercise 5. 2.2. An irregular complex cm Let

. The

denote the point

and let the q-skeleton

5

Conditions (l) through (3) are satisfied, and conditions (k) and (5) are satisfied vacuously for

q. < η . For the one

now define a relative homeomorphism

f:

η-cell,

(S - σ ) , we

(ET, S " ) — > (S , σ ) in two

stages. First, let P = (0,...,0,1) , the point diametrically opposite σ

on

S

, and let E

coordinate set S = S

χ -

be the closed hemisphere of points of S

_ > 0 . Then E

contains P , and has as boundary the

of points of S

with coordinate

a relative homeomorphism

g:

More precisely, for each point

X

1

= 0 . We define

(E,S) — > (S , σ ) by doubling angles.

χ

in B , gx

is that point of S

which is -located on the great circular arc that starts at P through

χ

to

with

σ

and passes

in such a way that

^C (P, origin, gx) = 2 · (ET, S " ) may be taken to

be the identity map. 2Λ.

An irregular complex on E

.

We augment Example 2.2 by taking and by letting

f:

|K | = |'K | = E11 for

q> η ,

(E11, S n ~ ) —S» (E11, S n ~ ) be the identity mapping.

2.5· An irregular complex on the torus T . Let b

θ

and φ lie in the closed interval

[0, 2rr] , let a and

be positive real numbers with a > b , and let T

points of B

whose coordinates

denote the set of

(x,y,z) satisfy the parametric equations

7

χ = (a + b cos φ) sin θ y = (a + b cos φ) cos θ ζ = b sin φ

To obtain an irregular complex K

on T , subdivide T

into the cells

shown in the following picture.

«Mco, (0,0)

φ

2ΤΓ

The cells can be described precisely by using the parametrization of T , which defines a mapping the

f

of the

2-cell τ = [0, 2ΤΓ] X [0, 2ττ] of

θ /-plane onto T . 8

Thus σ = f(0,0) Og = f((0, 2ττ) X 0) α[ = f(0 X (0, 2ΤΓ)) f((0, 2ττ) X (0, 27T)) The mappings

f

. may be taken to be the appropriate restrictions of

n,i

2.6. A regular complex on T . Subdivide as pictured below:

an TT T

21Γ

This subdivision of

τ

induces a subdivision of T

into the cells of

a regular complex:

four vertices, eight edges (l-cells), and four

2-cells, for which

f

2.7·

An irregular complex on real projective A point of P

R

induces the required homeomorphisms.

= (R

n-space, P

is an equivalence class of points of

- the origin) under the relation

where

r

is a real number different from

0 . The topology of

in the quotient space topology obtained from

is

' . Recall that this

topology is defined as follows: Let

be the trans-

formation which sends each point onto its class in

. Then a subset

A

of

is closed if and only if

map

is closed in

is a two-fold covering of

T:

. The . The involution

defined by

is the covering transformation of g . For each

define the

k-skeleton of

to be

Conditions (l), (2), and (3) are satisfied, to real projective

is homeomorphic

k-space, and we can exhibit relative homeomorphisms

to show that each difference

is a cell.

The set

is the union of two open

k-cells, each of which is mapped hameamoiphically by g Let

onto

denote the upper cell, i.e., the one which has its last non-

zero coordinate positive. Let projection of

denote the usual

onto the closed hemisphere

10

Θ k —k

Remark. Mote that P k-i

σ

to P

tion for P σ 1 P P

to P .

k-1 has been obtained from P

with the mapping

g . This suggests an inductive construc­

. Start with a point P

= σ

. Attach a closed 1-cell

by adjoining its end-points to P 1 1 2

Observe that

P

=S

.

by attaching its boundary

To obtain S

by attaching

to P

P

. The resulting space is —2 , adjoin a

2-cell

σ

to

with the double covering map.

This attaching operation amounts to wrapping

S

around P (= S ) twice.

*'

O*

P' P* Given P σ

to P

p· P' , one constructs P

P' by attaching the boundary S

of

by the double covering map g

2.8. A regular complex on P We start with the regular complex on the boundary 11

Γn+1 of the unit

„n+l (n+l)-cube in R '""" . The cells of the complex qn r-dimensional faces

(0 < r < n) of

I

;n+l are the open I"""

. On Γ

we identify points

that are diametrically opposite with respect to the center We thus transform the complex on

ΊΓ

(•§·, 5-,...,2-) .

into a regular complex on P

by

identifying diametrically opposite cells. To illustrate:

P

becomes

with two 1-cells

•

Λ ill.,

2.9·

7

P

becomes

with three 2-cells

A regular complex on R

.

We obtain this complex by considering unit whose vertices are the points of R

η-dimensional cubes

with integral coordinates. The

cubes, with all their lower dimensional faces, are the cells of R

3-

.

Locally-finite Complexes

A complex is locally-finite if each point has a neighborhood that is contained in the union of a finite number of cells of the complex. Such a complex is locally compact. The reticulation of R

described

in Example 2.9 is locally-finite, although the complex is infinite.

12

3·1· A complex that is not locally-finite. Let e. ι

joined to a common vertex ν . For each ο

u

vertex of e's

|K| be the union of countably many regular, closed 1-cells

e. . Give to

and v's . The

e's

i

let v. ι

be the other

|κ| the weak topology with respect to the and v's

determine a regular complex on

|κ|

that is not locally-finite.

Each neighborhood of ν meet each

e.

meets infinitely many of the

in an open subset of e.

because it contains

|κ| of 3·1 cannot be embedded in

Subcomplexes

DEFINITION. A complex L

(in symbols (2)

, and this subset is never empty

for any η .

k. k.1.

it must

ν . ο

Exercise. Show that the space R

e's:

L C K) if (l)

is a subcomplex of a complex

|L| is a closed subspace of

IL I = |L| f! |K I for each

q , and (3) each cell of

of K .

13

K

|κ| , L

is a cell

The proposition below gives a combinatorial characterization of subconrplexes. In particular, it shows that the collection of cells of dimension

< q. of a complex K

subcomplex is denoted by its underlying space is h.2.

PROPOSITION.

collection

C

K

of cells of K

l l } minus

less than

q,-skeleton of K , and

(Characterization of subconrplexes.) The cells of a

K

i2

; it is called the

JK |

and only if, for each cell σ

determines a subcomplex of K . This

are those of a subcomplex L

σ in C , the boundary

of K

if

σ (the closure of

σ) lies in the union of cells of C

of dimension

dim σ .

The proof of this proposition occupies the remainder of this section. k.3·

COROLLARr. The cells of K

are those of a subcomplex of K .

This corollary is an immediate consequence of

k.2.

We now begin the proof of the proposition. Suppose that the cells of

C are those of a subcomplex

L

of

K , let

σ be a

q-cell of

C , and let σ = (the closure of We know from Exercise 2.3 that union of cells of that

L

aC

a IL

in

n

|L|) - σ .

I , and that

of dimension less than

JZi -, f is the

q . It remains to show

a = Sr . But this follows from the fact that |L|

is closed in

|κ| . How suppose that each cell σ lies in the union of cells of Let

a C

of

C

of dimension less than dim σ .

|L| denote the union of the cells of Ik

satisfies the condition that

C with the subspace topology

from

, and let

. Let L

|L| and the subspaces

. We will show that

cells are precisely the cells of C , and that (1)

consist of the space L

is a complex whose is closed in

is a Hausdorff space because it is a subspace of

(2)

, and each

is closed in

, the intersection of

because

with a closed subspace of

(3) W which is the union of the of

q-cells of

C . Each

q-cell is a component

because each is connected and no two have a common

point. Each

q-cell is open in the relative topology of

. This

follows from the fact that

which is closed in

, and hence in

, because

is

closed in It follows from

that

is the union of the cells of C

of dimension (5) For each

q-cell

Now that

, the closure of

is

by hypothesis together with the fact

is the union of cells of , and

in

is the closure of

relative homeomorphism

C

of dimension in

. Therefore,

. Accordingly, the given for K may

be taken as the one required for L . 15

So far we have been able to avoid the question of whether and

JXJ[ are closed in

is closed in lies in

|L |

|κ| . It has been sufficient to know that

|L| , and that if

σ is a cell of

C

[L |

its closure in |κ|

|L| . The question must be settled before proceeding to proper­

ties (6) and (7)· To settle it, and to show several related facts, we prove two lemmas. k.k.

LEMMA.. Let K

be a complex. Each compact subset of

|κ| meets

only finitely many cells of K . Let D

be a compact subset of

|κ| . We show first that D meets

cells of only finitely many different dimensions. Suppose that this is not so. Let

{σ

} be an infinite sequence of cells of strictly ascend-

ing dimension, and let x. be a point of given

IK | contains x.

finite. the then

(Here we use

x. .) If S

only if

, ^i (σ (ID) for each

q. < q , so that

{x.) to denote the entire set

is a subset of

£x.} , and if

i . A

{χ.} Ω |K | is {χ. , x_,...} of

σ is a

q-cell of K ,

(S Π "σ) C ({x·} Π |κ I ) is finite, and therefore closed. Thus S

is closed in

|κ| , and in D

as well.

no limit point in D , because

It follows that

{x.} can have

S = (x.) - {a limit point) would still

be closed. But this contradicts the compactness of D . We show next that in each dimension D cells.

If D

meets infinitely many

an infinite sequence of points is a subset of

£x.} , and if

meets only finitely many

q-cells

£x.) , with

x.

{σ.} , then there exists in

(D Π σ.) . If S

σ is a cell of dim < q , then

SDa

is either empty or consists of a single point. It follows that S is closed in the weak topology of K , therefore, S

is closed in

|K | . By property (7) of the complex |κ| , and hence in D . It follows, 16

as before, that

can have no limit point in D , contradicting the

compactness of D . The lemma is now proved. 4.5. LEMMA. Let

C be a collection of cells of a complex K

lies in the union of cells of C , for each cell

of

such that

C . Let

|C| be the union of the cells of C with the relative topology inherited from

|k| . If A

for each

of

is a subset of C , then A

|c[ , such that

is closed in

is closed in

[k| . In particular,

|c|

is closed in We prove the lemma by showing that if is closed. We assume that By the preceding lemma,

where the first n

we Since andknow therefore, that

is a cell of K,, then , and we argue as follows.

meets only finitely many cells of K ,

of these are the ones that lie in C . Since

IT

so that η

τ η ( U P . ) = τ η |c| .

ι

Accordingly, η

η

Α η τ = Α η τ η |c| = Α η τ η (υ ρ.) = τ η [υ (Α η p.)] .

1

χ

1

Since each A Π p. is closed in p. , their union (finite) is closed in

|K| , so that A Π τ is closed. We now return to the proof of Proposition 14-.2,, and finish showing

that L is a complex. (6) The topology of cells of

|L| is the weak topology with respect to the

[L| . First suppose that A

(any subset of

is closed in

|L|) is closed in that subset of

|L| . Then A Π

|L| . Hence A Π "σ

is closed in "σ for any cell of C . Now suppose that A ( I o is closed in ~5 for each cell is closed in (7) On

σ of C . An application of 4.5 shows that A

|κ[ and hence in

jXJ I the weak topology and the relative topology coincide.

First suppose that A C IL I Then A Π (any subset of AOo

|L| .

i s

closed in the relative topology of

[L |

|L|) is closed in that subset. Therefore,

is closed in ~a for any cell of L . Now suppose that A C |L I

and that A ( I O is closed for each 10 J = IL I , we see that A the relative topology of

a C. IL I · By applying 4.5 with

is closed in

|κ| . Hence A

is closed in

|L | .

A final application of Lemma 4.5 shows that and the proof of Proposition 4.2 is now complete.

18

|L| is closed in

|κ| ,

k.6.

Exercises. 1. Let

σ be a cell of a complex. Show that

σ lies in the union

of finitely many cells of dimension less than dim σ . 2. of

Let K

be a complex and suppose that H

|κ| . Show that H

is a compact subset

lies in the union of finitely many cells of K .

3· Show that each compact subset of a complex K

lies in the union

of the cells of a finite subcomplex of K . k.

Assume that K

is a complex for which each cell boundary

σ

is actually equal to (not merely contained in) the union of a finite number of cells of dimension less than dim σ . This will be seen later to be true whenever K cells of K

is regular. Show that any union of closed

is a closed subset of

|κ| .

An immediate consequence of Exercise 3 above is that if locally compact then K

|κ| is

is locally finite.

A complex is closure-finite if each closed cell lies in the union of the cells of a finite subcomplex of K . The result of Exercise 3 implies that complexes are closure-finite. The complexes that we have defined are the

CW-complexes of J. H. C. Whitehead [I].

5· The Weak Topology for Skeletons The proposition that we established in the preceding section shows that the

q-skeleton of a complex K

is a subcomplex of K . In exam­

ining the proof of the proposition to see how the proposition follows from known properties of K , one sees that property (7) of the defini­ tion of a complex is used in what may be an essential way in the proof 19

of k.k.

It is used there to show that a set S

because it is closed in the weak topology of

is closed in

|κ|

|κ [ . In this section

we consider an example which may throw some light on the extent to which property (T) is really needed. A quasi complex subspaces

|θ | of

Q consists of a space | QJ

| Q|

and a sequence of

satisfying conditions (l) through (6) of the

definition of a complex. There are regular quasi complexes, irregular quasi complexes, finite ones, and so forth. Every complex is a quasi complex, but not every quasi complex is a complex. The quasi complex presented below does not satisfy condition (7).

In Chapter VIII

(Theorem 5.6) we prove that each regular quasi complex satisfies con­ dition (7) (i.e., each regular quasi complex is a complex). We start with the regular complex

K

3 on E

described in 2.1,

and pictured below.

The quasi complex

Q will have

|Q| = |κ| , but

Q will have infinitely

many cells. In dimension zero, there is no difference between Q:

K

and

IQ I = a. U σ_ . In dimensions one and two, new cells are defined

by subdividing

2 σ, . The cell

2 a^ will be included in

Ii | 3 · There is only one

3-cell in

Q:

the cell

cr . By

definition, the topology on each

| Q I is the Euclidean subspace ο

topology, inherited from In

|Q| = |κ| = E

.

|Q| , the Euclidean topology and the weak topology agree. It

is easily seen, however, that in neither

IQ1J nor in

|Q_| does the

Euclidean topology agree with the weak topology. This, in fact, is why the example must be so elaborate as to contain the cannot stop our construction with or

Qp

Q. or with

3-cell

|κ| , defined by

|K| · For two points

α

and

β

φ(α) = a , induces a metric of

|κ| , not necessarily from

the same simplex, d(a,P) = ρ(α,β) · For each simplex function

^T" j s

If K

s of

K , the

is an isometry.

is finite, that is if K

has finitely many simplices, the

weak topology and the metric topology agree. The proof requires showing that each simplex of K K

is closed in the metric topology of

|κ| . If

is infinite, the two topologies may differ, as the following example

shows. Start with the vertices

v.

and the 25

1-simplices

e.

of 2.12.

0(,

Let a.

be the point in

a

Since

i^ v o^ = ! -

d(a., ν ) = s2.fi.

topology for {a.}

e.

y

defined by

1

an·1 Qi1(V1) = l/i .

A

the a.

converge to

ν

|κ| . However, in the weak topology for

is closed. Thus the two topologies for

in the metric |κ| , the set

|κ| do not coincide.

An example of a finite simplicial complex is a simplex with all its faces. We will usually use a single symbol to denote both a simplex and the complex it determines. If the topological boundary of

s

is a simplex, then

s will denote

s as well as the simplicial complex con­

sisting of the proper faces of

s

(those faces of dimension less than

dim s) . 6.2. An alternative definition of finite simplicial complex. A finite simplicial complex K

is a collection of faces of a single

simplex, with the property that each face of a simplex of

K

is likewise

in K . Exercise. Show that 6.2 agrees with 6.1. 6.3.

!THEOKEM. Let

K

be a simplicial complex. The space

the weak topology, together with the subspaces regular complex. 26

|κ| with

JK | , determines a

Here,

|κ | denotes the union of the simplices of K

of dimension

< η .

Conditions (l) and (3) present no difficulty. Also, in

|K| . If

faces of sets of

s

is a simplex of K

s , so that

|κ | Π s

is the union of

is the union of a finite number of closed

|κ| .

The points of

|K | - |K ,| are precisely the interior points of

the n-simplices of of K

|κ| Π s

then

|κ | is closed

K:

functions from some listed set

to the open interval

n-slmplex

(0,l) . The interior

{A,,...,A

-}

s = (s - s) of an

s is connected, and is referred to as an open simplex. Each

open η-simplex is open in IK I . Each point α

of

|κ | because its complement is closed in S lies in

It is easily seen that is a closed

I = s D 1Iin-1I .

is an open η-cell of K

and that s

η-cell of K .

Mappings the standard

s

1IL1-1I · Thus

f:

(ET, S " ) — > (s,s) are provided by isometries with

n-simplex.

Since each simplex of K condition (5) and since

is a closed cell of K

in the sense of

|κ| has the weak topology with respect to the

simplices of K , condition (6) is satisfied. The proof of (7) is left to the reader.

BEFERENCES 1. J. Ξ. C. Whitehead, Combinatorial homotopy I, Bull. Am. Math. Soc. vol. 55 (19^9), pp. 213-245.

27

Chapter II. HOMOLOGY GROUPS FOR REGUIAR COMPLEXES 1. Redundant Restrictions on Regular Complexes. Homology Groups. In this chapter we describe the process of associating with any given regular complex an infinite sequence of groups called the homology groups of the complex. Fundamental to this process are the concepts of chain complex and homology of a chain complex. 1.1. DEFIHITION. A chain complex

C over a ground ring R

(R assumed

to be commutative with unit) is a collection of unitary R-module@

(5 :qeZ}

together with R-homomorphisms

q,

δ : C -> C q

δ 3 = 0 y~x q

. For every q, C q

or the qth chain module, and of C.

We will often write

the boundary operator of C. 1.2. DEFINITION.

If

q

such that for every

is called the R-module of q-chains of C, 9

is called the qth boundary operator

δ for the collection We then write

C = ({C },δ)

complex of cycles of C, written whose qth chain module is Z

.

q™-*-

C

as

Z(c) , is the chain complex

= Ker δ

({Β },θ) , where

For each q, we call Z

({C },δ) .

is a chain complex, then the

B

= Im δ

and

B

B(C) , is the

_ . the R-modules of q-cycles

and q-boundaries, respectively, of C . Note that since we have

({Ζ },θ)

and whose boundary operators

are zero. The complex of boundaries of C , written chain complex

{δ } and call δ

δ δ

= 0 ,

B C Z . q - q

1.3. DEFINITION.

Given a chain complex

module of C is the factor R-module The homology chain complex of C

C = ({C },δ) , the qth homology

Zq'/Bq

and is denoted by J

is the chain complex written

28.

Hq\(c) . Hx(C) .

In this section the ring R will always he the ring

Z of

integers. We will sometimes write "chain complex" for "chain complex over

Z " . If C

is a chain complex over

Z , its chain modules are

abelian groups. Given a regular complex K , we will define homology groups for K by first giving K

an orientation. Any such orientation

gives rise to a chain complex, and then the homology groups for K will be the homology groups of this chain complex. It will be clear that there may be, for a given complex K , many different ways of defining an orientation of K . Thus it will appear that the homology groups of K were not well-defined. We will show, however, that the homology groups of K

are independent of the orientation used to

define them. As we have just remarked, an orientation on a regular complex will induce a chain complex. We will see that only the boundary operators of the chain complexes so induced may vary as we vary orientations. The chain groups, however, are defined independent of orientation in the following way. 1.4.

DEFOTITIOIf.

of K , written

Given a regular complex

K , the group of q-chains

C (K) , is the free abelian group on the q-cells of q

κ. Thus for

q < -1,

C (K) = 0 . In order to define a boundary

operator we will assume that the regular complex K

satisfies cer­

tain restrictions. At this time we will state these restrictions and assume that they are satisfied by all regular complexes. We will prove later that this is the case by using the homology theory of simplicial complexes. Thus we will show now that the redundant

29.

restrictions for regular complexes are satisfied by the subclass of simplicial complexes. Three Redundant Restrictions on a Regular Complex R.R. 1:

If r < q , and

σ

point of the r-cell τ > then

is a q-cell whose closure contains a τ C cr .

To see that R.R. 1 is satisfied if K note that a q-cell of K

K :

is simplicial, we

is then the interior of a simplex of K .

If a point of the interior of an r-s implex lies in a closed q-cell, that is, in a q-simplex, then every vertex of the r-simplex is a vertex of the q-siniplex, and the desired relation holds. 1.5.

DEFHiITIOW.

In a regular complex K , if

closure contains the cell

τ , then

τ

is a face of

τ

is called a proper face of

1.6.

σ , we write

PROPOSITION.

If K

τ

ζ on K , we use Lemma k.Q to

ζ . Then

z'

is determined, up to

an addition of a multiple of 2λ. , so we may set f({z}) equal to the coset of < 2λ. > morphism. ζ

in which

z' lies. As before, f

is a homo­

It is a monomorphlsm because if f({z}) is zero, then

is homologous to a multiple of 2λ

and thus bounds. Since

f

is clearly an epimorphism, it provides the desired isomorphism H n - 1 (K) « Z n - 1 ( L ) / < 2\ > . ^.10.

COROLLAEX.

If K

is a non-orientable n-cireuit,

is isomorphic to a direct sum of Z_

H

χ

(Κ)

and a finite number of copies

of Z . Proof: We may take a basis for

C

all of its coefficients are +1 . C

_(L) which contains Z

1

λ since

(L) is a subgroup of

-,(L) . By a fundamental theorem on finitely generated abelian

groups we can find a basis for the free group

Z

whose elements is a multiple of a basis element of

50.

( L ) each of C

_(L) .

But so

bounds in K

and so is a cycle in L . Thus

itself is a cycle. Therefore

ator of

can be taken to be a gener-

. The corollary now follows.

11. Example.

The projective plane

2-circuit. We triangulate

, a non-orientable

as in the diagram and label the n-

and (n-l)-cells. Note that the chain Thus

so is non-orientable, and

is a free group on one generator,

, so Thus Since

is

connected,

Exercises. Show that the hoitology of l) the torus is , and 2) the Klein bottle is by triangulating them as 2-cireuits and using the methods of this section.

5*

Change of Orientation in a Complex

For this section we will need the notions of chain mapping and chain isomorphism. 5.1. DEFINITION. Given two chain complexes , a chain map homomorphisms

is a collection

such that

51.

and of for

each

q . A chain map is called a chain isomorphism if each of the

φ 's

is an isomorphism. The notion of chain map is important hecause a chain map is

easily seen to carry cycles into cycles and boundaries into boundaries.

Thus a chain map

defined by

φ

φ^: H^(C) -*H^(C)

induces a chain map

(φ#) {ζ} = {φ ζ} . As a matter of convention we shall

usually omit the subscripts whenever referring to the homomorphisms of a chain map or to the boundary operators of a chain complex. Thus we write

φδ = δ'φ

as the defining relation of the chain map φ.

Note that isomorphic chain complexes have isomorphic homology groups and that a chain map

φ: C -» C

and only if there exists a chain map verse of and

C*

φ

such that

respectively.

φφ~

and

is a chain isomorphism if φ" : C* -* C

φ" φ

called the in­

are the identity on C

Here we use the fact that the composition

of two chain maps is a chain map. The basic theorem on induced chain mappings of homology groups is: 5.2.

THEOREM.

If φ: C-* C

and ψ: C -> C"

are chain maps then

Ψ*Φ* = (Ψφ)*: H j C ) ->H*(C") . The proof is easy and will be omitted. This theorem, to­ gether with the fact that the identity chain map induces the identity homomorphisms of homology groups, shows that the assign­ ment to map

C of the chain complex H^(C) and the assignment to a

φ: C -* C

the map

φ^. is a functor from the category of * chain complexes and abelian groups into itself. In the future we

shall usually omit reference to the homology chain complex and talk of the homology groups or regard the collection of homology groups as a graded group. ·* See Chapter IV.

If we are given a regular complex K

with an incidence

function a , we may reverse the orientation on a given cell a K

in the following way. Let

C (κ) be the chain complex for a

given by the incidence function dence function

β

for K . β

incidence numbers involving σ

K

on K . We define a new inci­

is defined to equal a

except for

σ . The β-incidence numbers involving

are defined to be the negative of the ce-incidence numbers involv­

ing a

(we assume that

dim σ > l).

ft complex

Then

β

induces the chain rv

R

GP(K) . We say that

G P ( K ) is obtained from C (K) by

an orientation reversal. We define a chain map as follows.

ψτ = τ

This defines ψ ψ

of

ψ: C (K) -»C (K)

for every cell τ different from σ ; ψσ =-σ .

on all the generators of the

by linearity. It is easy to see that ψ

an Inverse map which reverses the sign on σ

C (K) , and we extend is a chain map and has once again. Thus ψ

is a chain isomorphism and induces isomorphisms of the homology groups of

Ca(K) and

C P ( K ) . And so Η°(κ) » Η^(κ) for every q.

For our general theorem on change of orientation we will need to assume the following lemma: 5.3.

IEMMA.. Any regular complex on the η-sphere

(n > l) is an

orientable n-circuit. The proof of this lemma is given in Chapter VIII, section k. At the present time we may regard the proposition as an additional redundant restriction imposed on a regular complex K : for each q > 1

and each q-cell

(q-l)-circuit. σ

σ of K , the boundary

σ

is an orientable

Notice that E.R.2 asserts that each (q-2)-cell of

is a face of exactly two (q-l)-cells of a , and that R.R.3

implies that

σ

contains the (q-l)-cycle

53.

δσ ,

and hence is. orientable.

Thus we are imposing additionally-

only condition 2 of definition 4.3 on the existence of a path of (q-l)-cells connecting any two (q-l)-cells.

In case K

is simpli-

cial, the condition holds trivially because any two (q-l)-faces of a

have a common (q-2)-face.

5.4. THEOREM. functions

Given a regular complex K

and

and two incidence

on K , then the chain complexes

and

are isomorphic. The isomorphism chosen so that

may be

for each cell

of K .

Proof: We define a chain isomorphism

by starting

at the chain group of dimension zero and working upwards. We let be the identity isomorphism. To specify , let

be a generator with vertices A for

Set

and

on

B . Now

. Similarly,

. Then

Extend

over all of

isomorphism.

by linearity.

Note that

It is clearly an

is induced by a set of independent

orientation reversals: namely, orientation is reversed on each cell a

such that Suppose now that

where

q

is an integer

isomorphism on each and for

has been defined on . Suppose that

for as defined is an

(K) , commutes with the boundary operators

, and satisfies

for each generator of each

. Then we extend

54.

over

so as to have

the same properties: For

σ

a generator of

C (K) we define

ψσ = +σ , where the sign is determined by comparing

ρ

δσ

with ψδ a

in the following way. First note that δ ψδ σ = ψ δ δ σ = 0 , and of cotirse σο

a = 0 . Thus both

subcomplex

σ with the orientation induced by

have coefficients ceding lemma, σ

+1

δ σ

δρσ = εψδ σ

for some

R

q > 2 . So

Z

α

is a

..(σ) « Ζ ,

ε = +1 . We then set σ = εσ . We do C (κ) , and extend by linearity over

the whole group. Then, if σ A

β . Moreover, both

is an orientable (q-l)-circuit, since

this for each generator of

rv

are cycles on the

on the (q-l)-cells of σ . Now by the pre­

regular complex on the (q-l)-sphere, and

and ψδ a

is a generator of

C (K) , we have

ft

ψδ σ = εδ σ = δ εσ = δ ψσ . This relation extends by linearity over/the whole group, and so ψ Clearly ψ σ

commutes with the boundary operators.

is an isomorphism on

a generator of

C (κ) , and ψσ = +σ

C (K) . Note that ψ

of orientation reversals. We extend ψ

for each

is obtained by a collection in this way over all the

chain groups of K , and then the result is the desired chain iso­ morphism 5.5. α

Ca(K) « CP(K) .

COROLLARY.

and

If K

is a regular complex with two orientations

β , then H (K) « Η Π Κ )

for each

q . Homology groups are

independent of orientation. Proof: .Isomorphic chain complexes have isomorphic homology groups. From now on we will write

H(K)

Cu , etc.

55.

to mean H (K) for some

The lemma stated before Theorem 5.k has another important consequence which relates to the possibility of defining an incidence function on a regular complex 5.6· α

K . In fact we have

THEOREM. Given a regular complex K , any Incidence function

defined on a subcomplex

L

of K

can "be extended to an incidence

function of K . Proof: We extend α

Inductively over the subcomplexes

L U L , . . . . Clearly α LUK..

We define OL

of a :

If σ

vertices A

B

over

K

LUK.

to be the following extension

not in L , then there are two

lying in σ , since

closed 1-ball. We define this way α

,

can be regarded as an incidence function for

is a 1-cell of K

and

LUK

[σ:Α] = 1,

σ

is homeomorphic to a

[σ:Β] = -1

arbitrarily. In

is defined on all pairs of cells involving 1-cells of

not in L . Suppose now that we are given an incidence function α q-"J-

defined on the subcomplex

LUK. Si

incidence function α of K

on L U K

q > 2 . We extend a

to an Si"--*-

~™

as follows. For any q-cell σ

not in L we must define incidence numbers for pairs of cells

including

σ . These will be zero unless the second cell of the

pair is a (q-l)-face of σ . Since a σ is a subcomplex of K q-l · oriented by a , by lemma 5.3, Z ~ (σ) « Z . We choose a generator cells of

7 . This generator has coefficients σ . If τ

is a (q-l)-cell of

to the coefficient which

+1 on the (q-l)-

σ , we set

[σ:τ] equal

7 has on τ . Then the only property

of the incidence function α

so defined which we need to verify

is (iii) of Def. 2.1.8. Let ρ be a (q-2)-cell of ά . By E.R.2 there are 2

(q-l)-cells of

τ , τ ρ which are faces of

56.

σ

and have ρ

as a common face. How

α δ q " γ = 0 . So the sum

[σ:τ ][τ2:ρ] , which is the coefficient of zero. Thus (lli) is -verified and a L U K .

dq" γ

on

ρ , must be

is an Incidence function for

We continue this process and extend a By taking

[σ:τ.][τΊ:ρ] +

to all of K .

L to he the empty subcomplex we deri"ve the third

redundant restriction E.R.3.

6.

Homology with General Coefficients. Cohomology.

We have defined

C (K) , for a regular complex K , as the

free abelian group generated by the q-cells of K . An element of C (κ) is then a function mapping the set of q-eells of K

into

the integers which is zero on all but a finite number of q-cells. We generalize the definition of from the set of q-cells of K

C (K) by considering functions

to an arbitrary R-module

G . First

we make the following definition. 6.1. DEFINITION. ring R

Given a regular complex K

and a commutatrve

with unit, the qth dimensional R-module of chains of

with coefficients in R

K

is the set of functions defined on the

collection of q-cells of K

mapping into R

so as to be zero on

all but a finite number of q-cells. The addition and scalar multi­ plication are defined as follows. If functions, r e R , and σ

f

and

g

are two such

a q-cell, then

(f+g)a = ία+SP

and

(rf)o = r(fa) ,

where the addition and scalar multiplication on the right sides are in R . The qth dimensional R-module of chains of K cients in R

is denoted by

with coeffi­

C (KjR) . An incidence function Ct on

57.

K

induces a boundary operator

δ · c (K:R) -»C ,(K:R) because of q q q-1

the unit in R . More precisely, if σ the value 1 on σ

and 0 elsewhere (l and 0 are the unit and zero

CL

of R) we set δ σ = Σ [ σ : τ ] τ , where q

{-1, 0, +1}

Hote that 6.2.

C (K;R)

R .

With t h i s boundary operator the

form a chain complex denoted by

C (KjR) .

C (Kj Z) = C (K) as defined previously.

DEFINITION.

R-module

[σ:τ] lies in the set

T € Λ.

of elements of

chain modules

is the chain which takes

Given a regular complex

G over a commutative ring

of chains of

K with coefficients in

K and a unitary

R , the qth dimensional module G i s the R-module

Cq(K;R) g^ G . This R-module is denoted by C (K;G) . As in Definition 6.1, an orientation α

induces boundary homomorphisms mapping

C (KjG)

into C -(KjG) for each Q . The qth homomorphism is .the mapping a

1 , where

of modules

1

is the identity mapping on

G . The collection

{C (KjG)} together with the boundary homomorphisms

{δ ® 1} form a chain complex which we will write as C (KjG) . 6.3. DEFINITION. The qth dimensional modules of cycles and bound­ aries of the regular complex K with coefficients in the R-module G

are the qth dimensional modules of cycles and boundaries, respec-

tively, of the chain complex

C (K;G) , and are denoted by Z (K;G)

and B (Kj G) , respectively. The qth dimensional homology module H (K;G) of K with coefficients in G Z°(KjG)/Bg(KjG) .

58.

is the factor module

As in the previous section, one can show that all of the chain complexes α,

C (KjG) are isomorphic, and that the homology

χ

groups H (K;G) are all isomorphic. The proof of these facts is easy and will be omitted. Homology with general coefficients is useful in some cases. For example, if G If

G is the group

is the group of rationals, there is no torsion. Z_ , every η-circuit is orientable, because

the boundary of the chain J that for

the case where B

is 2X. = 0 . It turns out, however, is Z , the homology groups with coef­

ficients in an arbitrary abelian group

G

can be calculated from

the homology groups with coefficients in Z . Thus homology with coefficients in a group

G does not give any new information about

the complex K . This is the main result of this section. We need the following definition. 6Λ.

IEFBiITIOIf. If C = ({C },3)

and

C* = ({0'},δ«) are two

chain complexes then the direct sum C Φ C chain complex

of

C

and

C*

is the

({C ®C'},3 Φ δ') whose qth boundary operator is

the mapping defined by (3 ®d')(e,c*) = (3e,3'c·) . This definition can be extended to any finite or infinite number of chain complexes. We remark that the homology of is the direct sum of the homology of C

and of

C Φ C*

C* ; that is, for

each q , H ( C θ C* ) » H (C) Φ H (C ) . 6.5. DEFINITION. An elementary chain complex is a chain complex over

Z of one of the following three types:

59.

i)

(Free) All chain groups are zero except for an Infinite cyclic group in a single dimension.

(All boundary operators

are of course zero.)

ii)

(Acyclic) All chain groups are zero except for infinite cyclic groups in adjacent dimensions,

n

and n-1, for some

n , and all boundary operators are zero except

, which

is an isomorphism onto.

iii)

(Torsion) The complex is the same as in the acyclic case except that than

6.6.

is multiplication by an integer other

0 or

THEOREM.

If G

is a chain complex over

Z

such that each

chain group is free abelian and finitely generated, then C

is the

direct sum of elementary chain complexes. Proof:

From the proof of Theorem II.2.2, we know that

direct summand in only the zero of stricted to

for each

q . We set

. Since

is a cycle, the boundary operator when reis a monomorphism.

The image

of the finitely generated free abelian group choose bases

and

and

there are integers

and

is split into a direct sum over

is a subgroup . Thus we may

for

respectively, such that for such that

is a

.

our first assertion C

q of the chain complexes

60.

... -* O -* D -> Z , -» O -» ... . We now split each of these into a q q-1 direct sum of elementary chain complexes. The elementary chain complexes we obtain are as follows t For every P 1 = 1

we have an

acyclic elementary chain complex with non-zero groups in dimensions q

and

q-1 . For every p. Φ 1 we have an elementary chain com­

plex of the torsion type with the non-zero groups in dimension

q ,

q-1 , and the boundary operator the mapping by p. . For every

z. with

i> k

Θ, which multiplies p i we have free elementary chain

complex with the non-zero group in dimension q. over

q of all these chain complexes is

The direct sum

C .

We will use the theorem above to compute the homology groups of a regular complex group

G . Let α

K with coefficients in an arbitrary abelian

be an orientation on K . The chain complex

CL

C (K) satisfies the hypotheses of Theorem 6.6 as long as K has finitely many cells in each dimension. We will use the result that (ΣΑ.) ® G

is isomorphic to

S(A ® G) . Now by Theorem 6.6,

/γ

C (κ) splits into a direct sum ΣΝ. , where the chain complexes. Then it is clear that

N

are elementary

C (KjG) = C (κ)® G«Z(W. ® G )

where the tensor product notation denotes tensoring in each dimension. For instance where

CT(K) ® G

is the chain complex

( δ ® ΐ ) = 3 ® 1 .

Definition

({C (K) ® G),h ® l) ,

Therefore, by the remark following

6Λ, H*(K;G) « S[H^(K1 ® G)] .

Computation of the homology of K

is then reduced to the compu­

tation of the homology of the chain complexes

6l.

H. ® G .

If

is free, t h e n i s

the chain complex

X

and so

, where

dimension of the single non-trivial chain group If

is acyclic, then

q

is the

G of

is the chain complex

whose groups are all zero except for two adjacent dimensions q , q-1 . The non-zero groups are copies of G , and the boundary operator between them is an isomorphism.

In this case it is easy

to see that Finally, if

is of the torsion type, then

the chain complex

is

where

multiplication by k . Then

is

, the subgroup

of elements of G whose order divides k . This group is written is

, which we write as

We show below that

is isomorphic to

Thus in all three cases

contains

as a

direct summand. The only non-trivial complementary summand occurs in the torsion case. We have therefore proved the following theorem. 6.7. THEOREM.

For each q,

last stan is over all k

where the

for which there is a chain complex of the

torsion type with qth boundary operator

in the direct sum decom-

position of 6.8. LEMMA.. If

G

is isomorphic to

is an arbitrary abelian group then If G

is finitely generated, then

the subgroup of elements whose orders divide k , is isomorphic to , where

T

is the torsion subgroup of G .

62.

,

Proof: To prove the first assertion, we define setting

for all

contains kG , and so

•by-

. The kernel of

then

induces a hcanomorphism

Define

= coset containing ng , for all It is easily verified that

and

are in-

verses of each other. Thus By the fundamental theorem for finitely generated abelian groups, together with the relation

, the

second assertion reduces to the case where If G is infinite cyclic then Suppose that is of order

G is a cyclic group.

and

are both zero.

G is of order n , generated by

. Then

(n,k) . Let r be an integer. divides kr divides r .

Thus ^G

is generated by

and so has order

(n,k) . So

and the proof is complete. 6.9. COROLLARY. Let G be a finitely generated abelian group. Then, for each q ,

where

is the torsion subgroup of

and T

is the

torsion subgroup of G .

6.10. PROPOSITION. Let G be a torsion free abelian group. Then

63. Proof: For then

for all k .

Exercise. plane

Compute the homology groups of the projective

with coefficients in

groups with coefficients in Z

. A tabulation of the homology

and in

gives:

The oth homology group on the left, ]Z, gives rise to the 0th group on the right; and the 1st homology group on the left,

,

gives rise to both the 1st and 2nd groups on the right. Given a regular complex K

we define the cohomology groups

of K by first associating with K

a cochain complex.

6.11. DEFINITION. A cochain complex

C over a ground ring

(commutative with unit) is a sequence with R-homomorphisms . of

E

of R-modules together such that for each q,

The R-module

C , and the homomorphism

is called the module of q-cochains is called the qth coboundary

operator of C . The modules of q-cocycles and q-coboundaries, denoted by pectively.

are the R-modules Ker The qth cohomology module of

and is denoted of

C

and

Im

res-

is the factor R-module

. The collection of cohomology modules

C together with the zero boundary operators forms a cochain

complex called the cohomology chain complex of C

6.12 DEFINITION.

and denoted

The qth dimensional module of cochains of a

regular complex K with coefficients in the R-module

6k.

G , written

q

C (K;G) , is the R-module

Hom(C (KJR),G) . The cochain complex

associated with an oriented regular complex -K is the cochain com­ plex

-

q

C (K) = ({C (K{G)},6}), where

8 q = (-l)%om(oq,l) . The

modules of eocycles and coboundaries of this cochain complex form the modules of eocycles and cohoundaries, respectively, of K with coefficients in G . These modules are written Zq(K;G),

Bq(K;G) .

The qth homology module of K with coefficients in G is the factor module If u

Zq(K;G)/Bq(K;G) , and is written Hq(KjG) . is a cochain on K

of dimension q, and c is a

q-chain on K , then for the value of u on c we write Note that

(u,c) .

(Su,e) = (-l)q(u,dc) .

Exercise. 1. State and prove a theorem about decomposing cochain complexes over

Z whose cochain groups are free and

finitely generated into a direct sum of elementary cochain com­ plexes. Deduce from the theorem that the cohomology of a finite regular complex with coefficients in an arbitrary abelian group G is determined by the cohomology with coefficients in Z . 2.

Let C = ({Cq},5) and D = ({Dq},5') be two cochain

complexes. Define a cochain map φ: C -»D to be a collection of homomorphisms Show that

φ

φ : C -»D

q

such that

Φα+1δ

= δ'φ

induces a cohomology homomorphism φ*: H*(C) -*H*(D) .

Prove an analog of Theorem 5.2.

65-

for each

q .

CHAPTEK III REGULAR COMPLEXES WITH IDENTIFICATIONS 1. The identifications Let K he a regular complex, and let a of K of equal dimension. A homeomorphism called an identification on K f|a

carries

and τ he cells

f of σ onto τ is

σ < σ the restriction ο onto a (closed) face of τ of the same dimension

σ

if whenever

as σ . ο 1.1. DEFINITION. A collection F of identifications on K family of identifications on K 1. For each σ

in K

is a

if each of the following holds: the identity homeomorphism

σ —S» σ

lies in F . 2.

For each

f in F , f"

3·

If f: P — > σ and g: σ — > τ gf: ρ — > τ

h.

is in F . are in F , then

is in F .

If f: σ — > σ is in F , f is the identity homeo­ morphism.

5.

If f: σ — > τ is in F and σ < σ , then in

f\~ 1

ο

is

O

F .

A complex with identifications i s a p a i r

(K,F) with

K a regular

complex and F a family of identifications on K . If

(K,F)

i s a complex with identifications, the r e l a t i o n σ ~ τ

if

f: σ —> τ

i s in F

is an equivalence relation on the cells of K . This follows from

66.

properties (1)-(3) for clude that if

f: σ—>τ

That is, if σ ~ χ onto

g: τ — > a

and

f: σ — > τ

Accordingly, if

σ

F . From properties (3) and (4) we con­

and h: a — > τ

are in

F

are in

then F

g=f

then

.

h=f.

then there exists one and only one map in F

of

τ . This fact will he used to construct an incidence

function in the proof of 2.k. The family points of

F

also determines an equivalence relation for

K : χ ~ y

We let

K/F

if a map in F carries χ to y .

denote the set of equivalence classes of points of

The natural function

S:|K| — > K/F

defines a topology for K/F K/F

(where

sx

is the class of x )

in the familiar way: a subset X

is closed if and only if s~TC

|κ| .

of

is closed in K . Throughout

this chapter we will omit absolute value bars whenever we can to simplify notation. We define the q-skeleton space

s(K ) . The space

complex.

K/F

(K/F) of

K/F

to be the sub-

and the skeletons

( K / F ) form a

The proof of this fact is routine, andwe taketime here to

point out only a few of the more important considerations. Each open cell of K/F

is the homeomorphic image, under

least one open cell of K . If that

s maps onto u

so = u , then the cells of K

are precisely the cells equivalent to

Property (k) listed for cell of K

s , of at

F

σ .

shows that no two points of the same

are ever identified by a map of

F . Property (5)

guarantees that if two cells are identified then so are their boundaries.

67.

The complex K/F

need not be regular, but relative homeo-

morphisms may be obtained for the cells of κ/F

from any set of

homeomorphisms given for K by composition with s : 11

1

(^,S - ) —±-> (σ,σ) ·>._ homeomorphism ^"*v.

(s|F)h

s|o

^ \ ^

J (υ, ύ)

The map

(s|a)h

is a relative homeomorphism because

s|a

is a

relative homeomorphism.

2. Let K/F

The homology of K/F

(K,F) be a complex with identifications. Although

is not necessarily regular it is possible to define a chain

complex based on the cells of K/F whose homology is isomorphic to the homology of the space K/F . We mean that K/F

l) the space

does carry a regular complex, so that its homology groups are

well defined (see page 33 of Chapter Il),and

2) these homology

groups are isomorphic to the homology groups of the chain complex Ca(K/F) that we are about to define. Proofs of (l) and (2) appear in section 3 of Chapter DC. The justification for presenting the chain complex at this stage is the ease with which its homology (and hence that of K/F) may sometimes be computed. 2.1. DEFHHTIOIi. An incidence function α

on K

under

and

F

if whenever

[ρ:τ]α = Ifp: ίτ!α

f: ρ — > σ

is in F

.

68.

is invariant

τ < P , then

2.2. EXAMPLE. The torus from a disk by identification. Let K he a regular complex on the unit square with four vertices, four 1-cells and one 2-cell:

Define

f: P 1 — > p_ by

f(x,0) = (x,l) , and define

by e(0,y) = (l,y) · Then take 1. the identity map on 2.

f, g, f"1

and

F

g: O 1 — > O 2

to consist of

τ

g"1

3. the restrictions of these mappings to bounding cells. An incidence function invariant under

F

is given by the diagram

below (following the arrow-notation introduced on page 35 of Chapter II):

2.3. EXAMPLE. Real projective η-space

F11

from S n

by identi­

fications. Let K be the regular complex on S n We take

F

described in 1.2.1.

to be the collection of maps obtained by restricting

69.

the involution complex

T

(see 1.2.7) to the closed cells of K . The

P^ = K/F has one cell in each dimension. The incidence

function given on pages 36 and 37 of Chapter II is invariant under P. 2.4.

IiMMA.

Let F be a family of identifications for a regular

complex K . Then there exists an incidence function on K is invariant under

that

F .

We construct the function α

by induction on successive

skeletons of K . The argument resembles the proof of II.5«6. For of 1-cells.

η = 1 , choose a 1-cell from each equivalence class Gall the vertices of σ A

equivalent to σ , let τ

to

σ . (Note that we allow

identity.) Define A and

and B . For each τ σ σ f be the unique map in F that carries

[τ: B ] = I .

τ

= f" A

σ

τ = σ , in which case

and B = f" B , and set τ σ

is the [τ:Α ] = -1 τ

All other incidence numbers involving cells of

K. we set equal to zero. It is easy to check that α is an incidence function on K_ Suppose now that α incidence function on K to showing that a

f

as defined

which is invariant under

F .

has been defined as an invariant

. . We devote the next few paragraphs

can be extended to an invariant function on K .

Choose a preferred cell from each equivalence class of q-cells. The union of K plex L Let

and the preferred cells is a subcom-

of K , and so by 11.5-6

a

can be extended over L .

σ be a q-cell of K-L . Let τ be the unique preferred cell

equivalent to If

.

ρ

σ , and let

f: σ — > τ be the unique identification.

is a face of σ , define

Γσ:ρ3α = [τ:ίρ3α . If ρ

70.

is not

a face of

, set

. We assert that the function

thus defined is an incidence function on

invariant under

F .

Properties (i) and (ii) of II.1.8 are clearly satisfied by Before verifying property (iii) we show that a F . As

is invariant on cells of

is invariant under

by the inductive hypoth-

esis, we need only check incidence numbers involving q-cells of K. Let

and

be q-cells of K with an identification The unique preferred cell

also equivalent to

equivalent to

is

, and so we have identifications By the remark on page 67 , we have

Let

be a face of

. Then by construction since by construction.

Thus

is invariant under To show that

F .

satisfies property (iii) of II.1.8, let a

be a q-cell of K-L , and let preferred cell faces of and

. Suppose

identify and

are (q-l)-dimensional

with a common (q-2)-dimensional face are (q-l)-dimensional faces of fa =

(q-2)-dimensional face

with a

. Then with the common

. Thus

because ce is invariant under

F . The second expression is zero

71.

•because a

is an incidence function for L , and

is a cell of

L . The proof of 2.3 is complete. Let

(K,F) be a complex with identifications, and let

be an incidence function on K each integer

that is invariant under

q we define

F . For

to be the free abelian group

on the q-cells of For each q , the map

induces a homo-

morphism

defined by

A boundary operator

To show that

is defined by

is well-defined, let

be in F .

Then

The last equality holds because because

for

- is invariant under . As

varies over the faces of

F , and

varies over the faces of , so the last expression

equals Oneeasily verifies that

Thus

72.

is

a chain complex, which we shall denote by

. Note that

is a chain map. The q-th homology group of

is denoted by

In the examples which follow, we will anticipate the •verifications in Chapter 3X and refer to

as the q-th

homology group of the space K/F .

3«

The homology of a torus

We begin where Example 2.1 left off, with the identifications and orientation exhibited in the diagram below:

The complex K/F has four cells:

so, sp

and

these cells is a cy-le in the chain complex

Thus

is trivial, so that

and

73.

ST . Each of

The homology and cohomology of We start with Example 2.2. cell

The complex

in each dimension, so that

The boundary operator

for

is determined by

Schematically the chain complex

where

is multiplication by 2.

The homology sequence dimension 0 for 1

has one

2 is 3

can be written

Accordingly 4

5

...

n

with It follows coefficients dimension from 11.6.7 in0 that 1isthe 2 7k. sequence 3 4of... homology n groups of F11

The cochaln complex obtained from dimension

0

1

2

is 3

... n

Here, odd Horn even Thus,

for odd

_

for even for even n for odd n

Exercise.

Compute the cohomology groups of

with

coefficients in

5. The homology of the Klein bottle We construct the Klein bottle from a disk, with the identifications and orientation shown in the following diagram:

The non-trivial chain groups are

75-

The boundary operator is given by

Therefore, 6.

and Compact 2-manifolds without boundary

We will assume that the reader is familiar with the fact that each compact 2-manifold M without boundary can be represented as a 2-sphere with , then M by subdividing

handles and

crosscaps.

If

can, in fact, be obtained from a closed 2-cell into an even number of edges and by identi-

fying these edges in pairs in an appropriate fashion.

Such a sub-

division determines a regular complex on

, and the identifica-

tions are carried out in such a way that M

is a complex with

identifications. A good description of the identification process may be found in Chapter 6 of Seifert and Threlfall's Lehrbuch der Topologie. In the process described there, each sequence of four consecutive edges, oriented and identified as in the following diagram,

76.

produces a handle. Each pair of consecutive edges oriented and identified as in the following diagram

produces a crosscap. Each M h

is obtained by subdividing

handle-producing sequences and k

(for some h , and some

into

crosscap-producing sequences

).

We label the edges of the i-th handle-producing sequence with the letters

, and we label the edges of the

i-th crosscap-producing sequence with the letters denote the projection. 2h+k

1-cells, and one vertex.

indicatrix to

Then M

. Let

has one 2-cell,

Note that by attaching a circular

, and using the arrows already given, we obtain

an incidence function on

which is invariant under the

identifications. Computation of Case l):

Also,

k = 0 . We have

Thus

morphic to a direct sum of 2h Case 2):

and copies of Z .

k = 1 . We have

77-

is iso-

Thus

. As in Case 1, every l-cell of M

The boundaries are generated by

is a cycle.

, and so

morphic to the direct sum of 2h

copies of Z

is iso-

and a single copy

of Case 3:

Again

k = 2 . We have

As before, every one-cell of M

Thus

is a cycle.

is free abelian on the 2h+2 generators

Another system of generators for

Thus

is

is generated by 2h+2 elements and has a single

relation, that twice

is zero. So

is isomorphic

to the direct sum of 2h+l copies of Z and a single copy of Zg .

7-

Lens spaces

In this section we define, for each pair

(p,q) of rela-

tively prime positive integers, a 3-manifold L(p,q) called a lens space.

We compute the homology of lens spaces, using a

family of identifications on the 3-sphere. Let X be a topological space and let G be a group of homeomorphisms of X

onto itself. Then G determines an equiva-

lence relation on the points of X , as follows: x

78.

and x' in

X

are G-equivalent if there exists a g

in G so that g(x)=x'

It is easy to verify that G-equivalence is an equivalence relation. The identification space whose points are G-equivalence classes is said to be obtained from X

by collapsing under the action of G ,

and is denoted by X/G . Let K be a regular complex. A homeomorphism onto itself is called a (cellular) isomorphism of K every cell of K

if f maps

onto a cell of K of the same dimension.

clear that an isomorphism preserves the face relation: then fσ < fτ .

f of |κ|

A group

cellular if each g

in G

G of homeomorphisms of

It is

if σ < τ ,

|κ| is called

is an isomorphism.

Suppose that a cellular group

G on

|κ| satisfies the

following property: (*)

If σ

is a cell of K , and g maps a

that

g

in G

onto σ , then g

is such

is the

identity. Then for each g

in G , and each

σ

in K , g|a

is an identi­

fication on K , and the collection of these identifications forms a family F of identifications as is easily verified. The identification space K/F

is the space

We specialize. Let S in the complex plane

|K|/G .

be represented as the unit sphere

2 C . That is,

3

2

S = ((Z0JB1) e C I z 0 I 0 + V Let

ρ

\ = e2lFi'V

and

q

1

= D ·

be r e l a t i v e l y prime positive integers.

, and define

T: S 3 —> S 3

79-

by

Let

Then T and its iterates

form a cyclic group

G of order p . The collapsed space

is called the lens

space

mod p then

and

In the special case p = q = 1 , we have X = 1 and

and

then

is the antipodal map. Thus We compute the homology of L(p,q) by constructing a regular complex on

for which G is cellular and satisfies

condition Let p

and q he fixed. Set

and n = 0,1,...,p-1 ,

To show that the are the cells of a regular complex on

with dim

we

shall appeal to Exercise 5 of 1.1.2. It is clear that the are disjoint and that their union is

Also, the

with

k = 0 and k = 1 clearly form a regular complex on the circle zQ = 0 . The closed cell

is the intersection of

set

80.

with the

If we set

and regard

then P is the half-2-space

as Euclidean 4-space, . The intersection

of the whole 3-space y^ = 0 with ._ is a closed 2-disk, and

is a 2-sphere. Thus is a regular 2-cell. A

similar argument"holds for

with n = 1,...,p-l .

The subset

of

is a hemisphere and thus a closed disk. There is an

obvious map in E

then

defined as follows. If with

. Define

Then f is a homeomorphism of boundary sphere onto

lies

onto

Thus

which carries the , and similarly each

for n = 1,...,p-l , is a regular 3-cell. Finally, it is an easy matter to show that each

lies

in a union of cells of dimension,less than k . Ery the conclusion of LI,, Exercise 5, then, the

are the cells of a complex on S .

The complex is regular because the cells are regular. Since

it follows that G is a group

of isomorphisms of the cellular structure. Moreover T V

if and only if m = 0 mod p . Hence G satisfies condition * above. An incidence function CC that is invariant under G is given by:

81.

(All other incidence numbers are set equal to zero.) The collapsed space

has one cell in each dimension.

If s denotes the projection of aries in

onto

then the bound-

are given by

Thus the homology of L(p,q) is given by and Note that H ^ L(p, q)) is independent of q . For an alternative description of lens spaces, and for more information, the reader should consult Hilton and Wylie's Homology Theory, page 223.

8. Complex projective n-space In this section we anticipate later chapters in order to state a result which allows us to compute the homology groups of certain spaces very quickly. In Chapter VII, we define homology groups for arbitrary spaces in such a way that whenever a space X

82.

carries a regular

complex the homology groups H (X) defined for X agree with the cellular homology groups of the complex. But the homology groups of X need not be computed from a regular complex on X . They may in certain cases be computed from an irregular complex on X . 5·1. THEOREM. Let X be a space that carries a complex K with the property that for each q the topological boundary of each q-cell is contained in K „ . Then for each q , H (X) is isoq_2 ^ qv morphic to the free abelian group on the q-cells of K . For example, suppose that K is the irregular complex on S

given in 1.2.2. Then 5-1 gives the homology of S immediately. The proof of 5·1 appears in Chapter IX. Let C

denote complex η-space, and let S

sented as the unit sphere in C an automorphism of S \(zv The space S

be repre­

. Each λ in S C C defines

given by scalar multiplication ,...,ζ ) = (\z , ...,λ.ζ ) . ο' ' η v ο' η

/S is called complex projective n-space, and is

denoted by CP . The space CP consists of a single point; CP

is a 2-sphere. The projection map is denoted by 2 n + 1

B:

S

_ > CP* . For k < η , we identify

c/ r, n \ , {(z , . . . , z , 0 , . . . , 0 ) } O K-

and the inclusion S

j , „n+l of C .

CS

k+1 C „ Then

with the subspace a

2k+l S

_i+l n „2n+l = CT Π S

,

is consistent with the action

of S . Thus we have a diagram of inclusions and projections:

83.

For

the subspace is real

is a closed 2k-cell with boundary is a relative homeomorphism of

We claim that onto

It is sufficient to show that for each point k in the open cell

contains exactly one point equivalent

to x under the action of

and this we show as follows: Since we have

which lies in

Further,

and

is real and non-negative in

then

is real and non-negative and hence equal

to 1 . So

and x is the unique point in

equivalent to x . Now let

and for each k > 0 let is a 2k-cell, and each

For each

8b.

point CPn lies in precisely one

Furthermore, since is a relative homeomorphism

the cells a k

satisfy the conditions of Exercise 5 of 1.1.2.

They are, therefore, the cells of an irregular complex K on with skeletons q even q odd. We may now apply 5-1 "to obtain

85.

CHAPTER IV COMPACTLY GEKERATED SPACES AKD PRODUCT COMPLEXES In this chapter we introduce the concepts of compactly generated spaces and of the product of complexes.

Sections 1-6, 8 are

based upon notes prepared by Martin Arkowitz for lectures he gave in Professor Steenrod's course in the fall of 1963· 1.

Categories

1.1. DEFIKITION. A category ζ

is a non-empty class of objects,

together with a set M(A, B) for every two objects A (^ . For each triple A,B,C from the cartesian product If

f ε M(A,B) and

f X g and

is denoted by

and

of objects there exists a function

M(A, B) XM(B,C)

to the set M(A, C) .

g ε M(B,C) , then the image in M(A,C) of gef , and is called the composition of

f

g . Two conditions are imposed: 1.

fo(goh) = (fog)eh .

2.

For each object A 1.

in M(A, A)

f

in M(A, B) .

in ^

such that

, there exists an element f„l A = Ig, f = f

for each

The set M(A, B) is called the set of morphisms from A We write

f: A — > B

to indicate that

f

to

B .

is a morphism in M(A, B).

When no confusion is likely to result, we shall write 1.2.

B in

gf

for

gef.

Examples of C a t e g o r i e s The category X

of sets and functions between them. The

objects are the sets, the morphisms are the functions, and the

86.

composition of two morphisms is the usual composition of functions. The category J

of all topological spaces (objects) and

continuous maps (morphisms) between them.

The composition of

morphisms is the usual composition of maps. The category (y. of abelian groups (objects) and homomorphisms. The composition of homomorphisms is the usual one. The category JH

of groups and homomorphisms, with the

usual composition of homomorphisms. The category^ , in which the objects

(X,x ) are the

topological spaces with base point, and the morphisms are homotopy ( χ > χ 0 ) — ^ ^'Vr)

classes of maps

2. 2.1. DEFINITION. from C

to Jj

an object

FA

'

Functors

Let £> and Ju be categories. A functor

is a function that assigns to each object A in Jj

and to each morphism

f: A — > B

F in

G

a morphism

Ff , so as to satisfy one of the following two sets of conditions: 1.

F(lA) = 1ρ Α ,

Ff: FA — > FB

and

F(feg) = FfnFg , or

2.

F(1A) = 1 ^ ,

Ff: FA — > FB

and

F(f.g) = Fg0Ff .

If F

satisfies (l) then F

satisfies (2) then

F

is a covariant functor.

If F

is a contravariant functor.

2.2. Examples of functors The functor

Horn . Let A

and

G be abelian groups,

and let Hom(A,G) denote the abelian group consisting of all homo­ morphisms

μ: A —S» G , with addition

87.

μ +μ

defined by

If a: A' — A

and 7: G — > G' are homomorphisms, then the

correspondence

a homomorphism Horn (a, 7): is Hom(A,G) > Hom(A',G' ) .

It is easy to see that 1. Homl

is the identity map of Hom(A, G) .

2. If

and are all homomorphisms, then

Horn(a'a,rr') = Hom(a,r)Eom(a'

) .

Therefore, for a fixed G the correspondence A (a: A' — > A)

Hom(A,G) > (Hom(a,lJ: Hom(A, G) — H o m ( A ' ,G)) u

is a contravariant functor from

For fixed A , the

correspondence G (7: G — > G')

> Hom(A, G) > (Hom(lA,7): Hom(A,G) — > Hom(A,G'))

is a covariant functor from Q

to Q .

The homology functor. For each space X in

let

H

n(

x

)

he the nth singular homology group of X with respect to some fixed group G. For each map f: X — > Y let homomorphism

be the induced

. The properties of homology groups

(established later in these notes, and independently of this example)

88.

then guarantee H n

to he a covariant functor from ^J to ^

The cohomology functor Hn . For each space X

.

in J

let

HnX) be the nth singular cohomology group of X with respect to some fixed group

G . For each map

f: X — > Y , let

the induced homomorphism

be

. The properties to

be established later about cohomology groups then guarantee H11 to be a contravariant functor from j to a . The fundamental group functor. For each pair define

(X,Xo) in

to be the fundamental group of X

For each homotopy class

{f} of maps from

to

at Xq . ,

define n^f to be the homomorphism induced by

f . It is known that

of f , that fore

depends only on the class

id^ is the identity and that

. There-

is a covariant functor from Exercises. 1.

If G is a group, let

[G, G] denote the commutator

subgroup. Show that the assignment to each group group

G/[G,G] induces a functor from

G of the abelian

to Q_ .

2. Make up a functor using ® .

3. Products in a category 3.1. DEFINITION. Let category ^ the A a

be a collection of objects from a

. An object P of ^

is said to be the product of

if there exists a collection

(called projections) with the following property:

of morphisms Given any object

X

in Q

and any collection of morphisms

exists a unique morphism each

f: X —S» P

{f' : X — > A } , there

such that ρ f = f

, for

Ci .

3.2. DEFINITION. A category ζ the product of any family

is a category with products if

{A } of objects of Q

exists in

ζ, .

If the product of any finite family of objects exists in Q7, then Q

is called a category with finite products. In a category ζ, a morphism

equivalence if there exists and B

f' f = 1. . If equivalent.

f: A —>· B

f' : B —S* A

f: A — > B

in G

is called an such that

ff =1.

is an equivalence we call A

B

and

In J) , for example, "equivalent" means "homeo-

morphic." Exercise. Show that the product (if it exists) of a collec­ tion

{A } of objects in a category is unique up to equivalence. The uniqueness of products up to equivalence allows us to

write EA a

for

P .

Exercise.

Give examples of products, using categories

defined in 1.2. Show that J P

is a category with products, with

the cartesian product with the product topology.

90.

4.

Compactly generated spaces

4.1. DEFINITION. A space X following property: a set A is closed in H

is compactly generated if it has the is closed in X

for each compact subset H

if and only if A Π H

of X .

Clearly, "closed" may be replaced by "open" in 4.1. Exercises. 1. generated. 2.

Show that if K

is a complex then K

is compactly

(Hint: use 1.4.4.) Show that a function from one compactly generated space

into another is continuous if and only if its restriction to each compact set is continuous. 4.2.

PROPOSITION. Each locally compact topological space is com­

pactly * gener ated. Proof.

Let X

closed and H C X

be a topological space. If F C X

is compact, then F D H

is

is closed in H by

definition of the relative topology of H . To complete the proof, we show that if F C X then there exists a compact subset H Let χ be a limit point of F Because X H

of X

is not closed,

with H Π F not closed.

that does not lie in F .

is locally compact there exists a compact neighborhood

of χ . The set F Π H

is not closed in H because it does

not contain its limit point χ . Remarks. For Hausdorff spaces, 4.1 can be stated in a slightly simpler form because "closed in H " can be replaced by

91-

"closed" (meaning "closed in X").

With compactly generated space

defined as it is in 4.1, Proposition k.2 holds for any X

in J

.

To preserve this generality we shall assume in sections 5 and 6 is any space in ^J . The results of these

which follow that X

sections are of course applicable to Hausdorff spaces. In later sections of this chapter, as well as in the other chapters of these notes, we shall revert to our assumption that spaces are Hausdorff spaces. 4.3. PROPOSITION.

If X

is any topological space satisfying the

first axiom of countability, (for example, if X then X

is compactly generated. Proof. Let A

pact subset H

be a subset of X

of X , A Π H

limit point of A . Since X

converges to sequence

χ . The set Y

sequence

{x } of points of A - {x} which consisting of the points of the is a compact subset of X .

is closed in Y . But A

{x } , whose closure in Y

and A

be a

satisfies the first axiom of counta­

{x } together with χ

Therefore A Π Y

such that for any com­

is closed in H . Let χ

bility, there exists a sequence

in A

is a metric space)

already contains the

includes χ . Thus χ

is

is closed.

5. For each space J is the underlying set X A

The functor k in J

, let k(x) denote the space that

with the topology defined by

is closed (open) if and only if A Π H

closed (open) in H H

is

for each compact subspace

of X .

92.

The topology of k(x) is called the weak topology with respect to compact subsets of X .

The identity map

k(x) — > X

tinuous because each set that is closed in X If X 5.1.

is closed in k(x) .

is a Hausdorff space, then so is k(x) . LEMMA.. X Proof.

X

is con­

and k(x) have the same compact• sets. If H

is compact in k(x) then H

is compact in

because the topology of k(x) is finer than that of X . Let H be a set that is compact in X , and let

a covering of H

by sets open in k(X) . By definition of the

topology of k(x) , each U D H many of the U

{U } be

ΠH

cover

is open in H . Thus finitely

H , so that H

is compact in k(x) .

The following two corollaries are immediate. 5.2.

COROLLARY. For each X

in

3 >

k

(x)

is compactly

generated. 5.3.

COROLLARY. X

is compactly generated if and only if X

and

k(x) are homeomorphic. Let X to

Y

then

f: X — > Y

f

and Y be spaces. If

f

is a function from X

is also a function from k(x) to k(Y) . If

is continuous, then

as follows: Since

f

f: k(x) — > k(Y) is continuous,

is continuous on X , the restriction of

to each compact subset of X

f

is continuous. Thus by 5.1 the

restriction of f to each compact subset of k(x) is continuous, and by Exercise 2 of the preceding section continuous.

93-

f: k(x) — > k(Y) is

We shall denote by

the category whose objects are

compactly generated spaces and whose morphisms are contiguous maps. 5.4. DEFINITION. The functor

is the

functor that assigns to each X the space k(x) f: X — > Y the map f:

k(x)

and to each map

— > k(Y) .

6. Products in 6.1. PROPOSITION. The category Proof. If lection of objects of Xi

in

is a category with products.

(i an arbitrary index set) is a col, then

, the product of the

, is defined to be the space

is the product of the X^ in product in ^J^ .

, where

. We show t h a t i s are morphisms

a

then

the f^ are also morphisms from Z to X^ in ^

. Since

is a category with products, there exists a unique morphism for which p^ are the projections

for each i in I , where the . The functor k then

furnishes us with morphisms unique morphism

Thus,

and with a ! such that

is the product of the X.. in

.

If X

and Y are compactly generated spaces then their

cartesian product X X Y in ^

is not necessarily compactly

generated, as an example of C. H. Dowker [l] shows (see Section 8 of this chapter). In contrast, their product in 3

k

is (by definition) always compactly generated. It will

be important to distinguish between the two kinds of product when we consider products of complexes.

7.

The product of two complexes

Let K and L be complexes. We define the product K X L of K and L to be the space

together with the sub-

spaces

In this section we show that K X L is a con?)lex. In Section 8 we discuss conditions on K sian product

and L which insure that the carte-

be a complex. We also given an example

of two complexes K and L whose cartesian product is not compactly generated, and thus not a complex. 7.1. LEMMA. The space

and the skeletons

satisfy conditions l) through 5) of Definition I.1.1. Proof. The spaces

clearly form an ascending

sequence of closed subsets of

whose union is

we have

95-

Thus the components of , where a i

form

are products of the is an i-cell of K and

(n-i)-cell of L . To show that each cell

is an is open in

, we observe that

The right hand side is the union of finitely many closed sets and so is closed. Relative hameomorphisms for the cells of K X L are easily obtained. If

and

are given, then

Since the boundary

is the required relative homeomorphism (see the exercise below). This completes the proof of 7.1. Exercise.

Show that the product of two relative homeomor-

phisms is a relative homeomorphism. 7.2. LEMMA. If K and L are complexes, then each compact subset of

lies in a union of finitely many closed cells of

K XL. Rroof. Let H be a compact subset of and q be the projections of

onto

pectively. Then pH is compact in

. Let p |K| and

|K| . By 1.4.4, pH

|L| resis con-

tained in a union of finitely many cells of K , and hence in the

96.

union of their closures: finite union

. Similarly, qH lies in a

of closed cells of L . Thus H lies in the

finite union 7-3-

of closed cells of K X L .

COROLLARY. Each contact subset of

lies in a union

of finitely many closed cells of Proof. Let H he a compact subset of'

. Since

is a Hausdorff space, H is closed. Set .

Then H^ is closed in H and so is a

compact subset of

By I. ^.3* K^ and

plexes. Thus 7.2 implies that IL

are com-

is contained in a union of

finitely many closed cells of

. Thus

is

contained in a union of finitely many closed cells of Recall that a quasi complex Q consists of a Hausdorff space

|q| and a sequence of subspaces Q^ satisfying the first

six conditions of I.1.1. 7.LEMMA.

Suppose that the Hausdorff space

|Q| together with

the, subspaces Q.^ , i = 0,1,..., satisfies conditions l) through 5) of 1.1.1. Suppose that

| Q| is compactly generated and that,

in addition, each compact subset of

|Q| is contained in a finite

union of closed cells. Then Q is a quasi complex. Proof. Let A be a subset of

|QJ which meets each

closed cell of Q in a closed set. We have to show that A is closed. That is, since

|Q| is compactly generated we have to

show that A meets each compact set in a closed set. Let H be

97-

a compact subset of c e l l of

Q.

|Q| .

Then

HC

U c1 i=o

where each

σ

is a

Then A n H = A O H n ( U σ 1 ) = H η ( U (A η σ 1 )) . i=o i=o

Since A Π σ

is closed for each

is closed in

i , A(IH

is closed. Thus A

|Q| , which completes the proof.

7.5. LEMMA. A closed subspace of a compactly generated space is compactly generated. Proof. Let X be compactly generated, and let A closed subset of X . Suppose that B that B Π C

is closed in C

show that B Then H Π A

is a subset of A

be a such

for every compact set C C A .

We

is closed in X . Let H be a compact subset of X . is closed in H , and so is a compact subset of A .

By hypothesis, H Π B = (H η A) Π B is closed in H Π A . Since A

is closed, it follows that H Π B

meets every compact subset of X Since X B

in a relatively closed set.

is compactly generated,

is closed in A

7.6. THEOREM.

and so A

If K

and L

is closed in H . Thus B

B

is closed in X . Therefore

is compactly generated. are complexes, then K X L

is a

complex. Proof. By 7.2 and 7 Λ ,

By 7-1 we need only verify conditions 6) and 7)· KXL

fies condition 6).

is a quasi complex. That is, K X L

satis­

Note that condition 7) merely states that each

skeleton of a complex has the weak topology with respect to closed cells. Since

(K X L)

is closed in

98.

|κ| X J L | , we know by 7.5

that

(Κ X L)

(K X I·)

is compactly generated. Finally, by 7·3 and 7.4

is a quasi complex. That is, (K X L)

has the weak

topology with respect to closed cells. Thus K X L

satisfies 7)

and the proof is complete.

8. The cartesian product |κ| X |L| It follows from Exercise 1 of Section h and Theorem 7.6 that the cartesian product

|K| X |L|

is a complex if and only if

|κ| X |L|

of two complexes K

pactly generated for a large class of pairs

plexes, then

(Milnor [2]) |κ| X |L|

8.2. THEOREM.

If K

|κ| X |L|

is com­

(K,L) .

and L

are countable com­

is compactly generated.

(J. H. C. Whitehead [4])

complexes and L

L

is a compactly generated

space. We now give two results which show that

8.1. THEOREM.

and

is locally finite, then

If K

and L are

JKj X |L|

is compactly

generated. For a definition of "locally finite" see 1.3· If X sets of X

is a topological space, a collection Qi

is called a base for compact subsets if each compact

subset of X

is contained in some member of Lu .

8.3. LEMMA. If K only if K

of compact

is a complex, then K

is countable if and

has a countablefrasefor compact sets.

Proof. Let K be countable. Sy Ί.ΛΛ, for compact sets of K of closed cells of K .

a countable base

is given by the collection of finite unions

Conversely, suppose K has a countable base compact sets- Every point of K

for

is contained in some member of

• By 1.4.4, each member of

is contained in a finite union

of cells of K . Thus K is a countable union of finite unions of cells, and is therefore countable. The following theorem thus implies 8.1. 8.4. THEOREM. (Weingram [3])

If X and Y are Hausdorff

compactly generated spaces each having a countable base for compact subsets, then X X Y is compactly generated. Proof. The proof is a restatement of Milnor's proof of 8.1: We know that the identity function is continuous. To show that f is a homeomorphism, and hence that X X Y is compactly generated, we need only show that if U is open in

then f(u) = U Suppose that x X y

is open in X X Y .

is a point of U . We will find a

neighborhood V of x , and a neighborhood W of y, such that

We may assume that X = U A^ and that Y = U B. , with and

.. all compact.

We may also assume that {A^} and {B^ x e Aq

are bases, and that

and y e BQ . Now,

is open in

, and contains

and BQ are compact Hausdorff spaces and hence regular. Thus there exists a set VQ open in A Q 100.

and containing x , and a

set W Q

open in BQ

and containing y , such that

Assume, by induction, that

and

, open in A q and

B^ , contain x and y , respectively, and that The set

is closed in

as in A^ , and is therefore compact in

. Similarly,

compact in

is open in

*

. Also,

_

There exist , therefore, sets , and open in

Let

and

and

as well

containing

is

and

, such that

and

, Then

Furthermore, V is open in X and W

is open in Y , since they

meet each set of a base for compact sets in an open set. This concludes the proof of 8A. A complex K is called locally countable if each point of K has a neighborhood contained in a countable sub complex of K. 8.5.

COROLLARY. If K and L are locally countable complexes,

then

is compactly generated.

See the theorem of Wallace on p. lbs of J. L. Eelley's General Topology.

101.

Proof.

Let A

compact subset of

he a subset of

|κ| X |L|

|κ| X |L|

meeting each

in a (relatively) open set. Let

(x,y) be a point of A . Then χ ε U C |K'| , where U in

|K| and K'

is a countable subcomplex of K . Similarly,

y ε V C IL' I , with

V

complex of L . Now

open in

|L| and

L'

a countable sub-

IK' I X IL' I is compactly generated, and A

meets every compact subset of

|K*| X ]L*| in a relatively open

set. Therefore, A Π (|Κ'| X |L'|) is open in In particular, A Π (U X V) of

(x,y) in

is open

|κ·| X |L'| .

is open, so that A

is a neighborhood

|κ| X |L| . It follows that A , being a neighbor­

hood of each of its points, is open in

|κ| X |L| .

8.6.

is locally compact, and Y

THEOREM.

(Weingram [3])

If X

is compactly generated, then X X Y Proof.

Let U

be a subset of X X Y

compact subset of X X Y that U X

is compactly generated. which meets each

in a (relatively) open set. We must show

is open. Suppose that

(xQ,y0)

is a point of U . Since

is locally compact, x. lies in an open set V whose closure

is compact. The set V X y . (V X y n ) Π U

is open in V X y n . Therefore, there exists a

neighborhood W

of x„

in X

The compact Hausdorff space V neighborhood W

of x Q

of all points y

in Y

is open in Y . Since to show that

is a compact subset of X X Y , so

such that W

Xy

C U

and W 1 C V.

is regular, so there exists a

with W . C w . . Let Z be the collection such that W Y

X yC U .

We assert that Z

is compactly generated, it will suffice

Z meets each compact subset of Y

102.

in a (relatively)

open set. Let H C Y Then, since V 1

be compact. Let y

belongs to

sects the compact set

Z , W

VXH

he a point of H Π Z .

X y C U . The set U

inter­

in an open set, by hypothesis.

Wallace's theorem there exists A

open in V

and

By

B open in H

with w"2 x Y 1 C

AXBCUD(VXH)

But this implies immediately that Thus x_

Z

B C Z / and so H Π Z

is open in Y . By construction, W

lies in W

and y

.

lies in Z , U

X ZC U .

is open. Since

is a neighborhood of

(x Q ,y 0 ) and so is open. This completes the proof. We conclude this section by describing an example due to C. H. Dowker [l] of two complexes M

and

Ν

for which

|M| X |N|

is not compactly generated, and thus not a complex. The complex M 1-cells, where

I

is a collection

{A.|i ε 1} of closed

is an index set with the power of the continuum.

The A. have a common vertex, u~ . The complex N erable collection a common vertex,

is a denum-

{B.|j = 1,2,...} of closed 1-cells, all having v. .

We suppose that each A. val 0 < x. < 1 , with x. = 0

is parametrized as a unit inter­

at u~ . Likewise, we suppose

each B. to be parametrized as a unit interval 0 < y < 1 , with J J y j

= 0

at V 0 . We now identify

I with the set of all sequences

(i ,i_,...) of positive integers. Thus, if

103.

(i, j) is a pair of

Indices, with j an integer and with

a sequence

of positive integers, we may define p.. to he the point with coordinates Then

is closed. If H is a compact subset of

then

is

finite because H lies in a union of finitely many closed cells of M X N (T.2). In particular, P meets each closed cell of M X N in a finite, and hence closed, set. Therefore, P is closed in

We show now that P is not closed in For each i in I , and for each positive integer j , let a. and h. be positive real numbers. Let U be the neighJ borhood of u Q

in

that is given by

be the neighborhood of v^ in

, and let V

that is given by

Then U X V is a neighborhood of

in

We shall now choose a pair

of indices for which

lies in U X V . This will imply that of P in

. Since

is a limit point is not a point of P , it

will follow that P is not closed in is chosen so that is,

. The pair

is a point of

is smaller than either

we pick a sequence

. That or

so that for each j=l,2,...

both

. We then choose

larger than

. To accomplish this

. With these choices,

, so that

lies in U X V .

104.

to be an integer and

Thus P

is not closed in

|M| X |N| , and

|M| X |N| is

not compactly generated. Exercise. If K |K| X ILJ

and

L

are regular complexes, then

is compactly generated if and only if one of the follow­

ing conditions holds: a) One of K,L b) Both K

and

is locally finite. L

are locally countable.

(HIHT: The sufficiency of these conditions is given by 8.5 and 8.6.

To show necessity, suppose neither condition obtains. Then

we may assume that K and that

L

fails to be locally countable at a point χ

fails to be locally finite at a point y . Embed

Dowker's example

|M| X |li| as a closed subset of

based on the point The closed subset

(x,y) . Suppose

[κ| X |L|

|κ| X |L| ,

compactly generated.

|M| X |N| would be compactly generated by 7·5>

and this we know to be false. The contradiction establishes necessity.)

BIBLIOGRAPHY [l] Bowker, C. H., Topology of metric spaces, Am. J. Math., vol. 7^ (1952). [2] Milnor, J.,

Construction of universal bundles I, Ann. Math·,

vol. 63 (1956). [3] Weingram, S.,

Bundles, realizations and k-spaces, doctoral

dissertation, Princeton University, 1962. [h]

Whitehead, J. H. C ,

Combinatorial Homotopy I,

Math. Soc, vol. 55 (19^9)·

105.

BuI. Am.

Chapter 5 THE HOMOLOGY OP PRODUCTS AMD JOINS RELATIVE HOMOLOGY 1. The homology of K x L Let K and L be regular complexes. In section IV.7 we defined the product complex K X L. The main result of this section (see 1.5) is the computation of

in terms of

I and H^(L) for

regular complexes K and L that sure finite in each dimension. For the moment, however, we allow K anil L to be arbitrary regular complexes. Since K and L are regular, K X L is regular. The m-cells of K x L are the components of

The products on the right, as well as all other operations considered in this chapter, are carried out in the category of compactly generated spaces. where

(See Chapter IV.) Each m-cell has the form

0 is a cell of K, T is a cell of L and dim a + dim T = m.

If A is an arbitrary set, let F(A) denote the free abelian group generated by the elements of A. It is easy to show that for any sets A and B

From this it follows that

-IO6-

where the

isomorphism

is defined "by the correspondence

If K and L are oriented "by incidence functions we may orient of

and

in the following way. A face of a cell

is a cell

with

and

Accordingly, we set

All other incidence numbers are defined to he zero. We call the incidence function so defined the product incidence function. To verify that the product incidence function satisfies property (iii) of II. 1.8, let A face

of

"be an m-cell of

with dim

of dimension m-2 is of one of three

forms: a) b) c) In case a), we have

-107-

In case b) we have

Case c) is similar to case a). Properties (i) and (ii) are easily verified. Thus we have associated with each pair of oriented complexes K and L the oriented complex

We denote the boundary operator

associated with the product incidence function by define

and we

to be the chain complex

1.1. DEFINITION. Given two chain complexes

and

over a ground ring R, the tensor product the chain complex whose mth chain group is

and whose boundary operator is defined by

It is easy to see that 1.2. THEOREM. The correspondence isomorphism of chain complexes:

-108-

induces an

is

Proof: We know that the correspondence an isomorphism

defines

of the mth chain groups of

and

Thus we need only show that these isomorphisms commute with the boundary operators. Let

be a generator of

Then

-109Under the isomorphism maps to algebraic and in rhus terms the Weand wish proof now of setting: turn to isdefine complete. to , Because the given andaproblem hamamorphism chain of 1.2 complexes of we computing may consider C and D, the compute problem in terms in an of

Let

and

be cycles in C and D respectively. Then in

we set

First we verify that this definition is independent of the choice of cycles:

But this latter cycle is the "boundary of

so that In order to complete the verification that

is well-defined, we

need only note that and similarly that

is linear in the second component. Thus

is a well-defined hamamorphism We extend

to

preserving homomorphism

by linearity, and obtain a degree-110-

We now assume that C and D

are chain complexes over Z

that are free and finitely generated in each dimension. We apply Theorem II.6.6 and write C and D each as a sum of elementary chain complexes of types i (free), ii (acyclic) and iii (torsion):

Using the fact that the tensor product of direct sums of chain complexes is the direct sum of the tensor product of chain complexes, we have

By the remark following 11.6.b,

Also,

For each pair

(i,j) we have the hamomorphism

which extends by linearity to give -111the hamomorphism

1.3. THEOREM. The following diagram is commutative:

This theorem will allow us to replace Proof of 1.3. Let

and

by the hqmamorphism

be cycles of C and D respectively.

We have

where each

lies in

and each

(1) Thus

-112-

lies in

Then

and the proof is complete. We remark in passing that

is a natural transformation of

functors. That is, given chain complexes f:

and g:

and maps

then the diagram "below is

commutative

The proof is easy and is left to the reader. We now investigate the hamomorphism

This hcmamorphism

is completely determined by the individual

and each

in turn depends only on the type of

shall see that and

and

We

is an isomorphism except in the ease that

are both of the torsion type with torsion numbers which are

not relatively prime. In this case we shall prove that

is a

monamorphism. To reduce nine cases to four, we say that an elementary :hain complex is of type Case

if it is of type

, Let M and N be free elementary chain complexes.

That is, for some p, q. we have

-113-

generated by a, if i = p

0

if i

p

Z, generated by b, if i = q

if Then

is also a free elementary chain complex, with Z, generated by a ® b, if k = p+q.

0 The boundary operators in M, N and generated by

are all zero, and so if i = p

0 Z, generated by

if

0=0.

.0 Z, generated by

if k

which maps

Thus is an isomorphism.

-111).-

Case

Let M be a free elementary chain complex with

one generator a in dimension p. Let N be a chain complex with two generators, c in dimension with

= nd for some

and d in dimension q, We show that is an isomorphism.

We have Z, generated by a if i = p

generated by

Thus

has a single generator,

ension

erated by

is elementary and of type

in dimension

and

with of dimension

which maps

gen-

in dimension

. Thus

generator,

has a single

and of order n. So is an isomorphism.

is like case

Case a

of dim-

and of order n.

The chain complex

Case

{d} if i = 1

Let M be a chain complex with two generators,

in dimension p+1 and b

in dimension p, with

Let

be a chain complex with two generators, c in dimension

and

in dimension

with

Then

-115-

generated by

if i = p

generated by

if j = 1

is cyclic and of order

The tensor product

the

greatest common divisor of m and n. Thus single generator

of dimension

The chain complex ension

of dimension

of dimand

The boundary operator is defined by

We first compute are generated by over the integers, Thus the

Since

and of order 0.

has four generators, and

of dimension

has a

The boundaries in dimension and

As x and y vary describes the subgroup generated by

-boundaries are generated by

generates the cycles, we have generated by -116-

Next we compute

In order that

-117-

maps Thus is isomorphism The Since Dividing such it an isthat The the isomorphism necessary hamomorphism by because and -boundaries is and to -cycles in aboth sufficient monamorphism. dimension are are the are relatively of domain generated in course that p+q. dimension and In generated prime, In generated by dimension range dimension there of by bya Thus is are an integer 0. a is an k

We are now ready to prove 1.4. THEOREM. If C and D axe chain complexes of abelian groups that are free and finitely generated in each dimension, then a is a, monomorphism of

onto a direct summand of

The complementary sirmmand, in dimension m, is isomorphic to where

denotes the torsion subgroup of

the qth homology group of C. Proof. By 1.3 it is sufficient to show that the theorem is true if C

and D are elementary chain complexes. In cases and

we have shown that a is an isomor-

phism. The statement about the complementary summand is true because in each case at least one of the chain complexes has torsion-free homology groups. In case

we know that

is a mono-

morphism onto a direct summand whose complementary summand is

The only torsion in in

is

is

in dimension p. The only torsion

in dimension q.. Thus

The proof of the theorem is complete. 1.5. COROLLARY (the

relations). If K and L are regular

complexes with finitely many cells in each dimension, then

-118-

where

denotes the torsion subgroup of

Proof. This follows immediately from 1.1 and 1.4. 1.6. COROLLARY. If K and L are as abovef and G is either for a prime p or the group of rationals, then

Proof. Apply 1.5, 11.6.9, and II.6.10. Exercises. 1. Compute the homology of 2.

relations for cohomology. Using the results

of the exercise at the end of chapter II, and ideas similar to those used in the proofs of 1.3 and l.k obtain a formula relating and

-119-

2. Joins of Complexes. Let X and Y be compactly generated spaces of X and Y is the quotient space of identifications: for all

The join under the following

and and

The projection map tion i:

defined by

a subspace of

If

is denoted by p. The funcembeds X as

Similarly, we cam regard Y as embedded in

and

then the line segment from x to y in

is the subset

Each point of

with

lies on a unique

The mapping cylinder of a map f:

is the subspace of

that includes all line segments with the points of

The join

together is hameamorphic "to the space

obtained by identifying the two copies of union of the mapping cylinders of the projections

in the disjoint and i

Recall that throughout this chapter we work in the category of compactly generated spaces.

-120-

The join of a point and a space X is called the cone on X, and is written CX. The natural inclusion

embeds X as the

base of the cone. A space is contractible if there exists a map F:

with

the identity map and

constant. The

cone on any space X is contractible, and provides the simplest way of embedding X in a contractible space. 2.1. LEMMA.. The join of a closed p-cell and a closed q-eell is a closed (p+q+l)-cell. The join of a (p-1)-sphere .and a closed q-cell is a closed

(p+q)-cell. The join of a

(p-1)-sphere and a

(q-l)-sphere is a (p+q.-l)-sphere. Proof; Let Let

and

be simplexes.

be the simplex on the vertices

We define a map

for f

by setting

and

and Thus

and

a continuous map one and onto. Since

then induces

Furthermore, is compact and -121-

is one-

Hausdorff,

is a homeomorphism.

This proves the first statement of the

Let

denote the unit ball in

define a map

We

by setting

for

then Thus

a continuous map

A straightforward cal-

culation shows that onto

is actually a hameomorphism of

The restriction of

onto

induces

to

is a homeomorphism

This completes the proof of the lemma. The reader should note that the homeomorphism

Exercise.

Show that the join of two spaces is arcwise connected.

We now introduce a gimmick that will simplify the computation of the homology of the join of two complexes.

Let K be a regular

complex. We adjoin to the collection of cells of K of dimension of K.

-1. We stipulate that

Since every

an ideal cell

be a face of every cell

1-cell has precisely two vertices the redundant

restrictions are all satisfied, Given an incidence function on K, we define additional incidence numbers involving

-122-

by

We define the augmented chain complex K

"by setting

of the oriented complex

the free abelian group on the

and by defining

q.-cells of K

for any q-cell

and is generated

toy

Thus

To show that

is a chain

complex, we observe that

in dimensions

and if c

is a zero-chain, then

where

defined in II.1.11. We remarked following

In e

is the index of c,

II.1.11 that

for any 1-chain c. The homology groups of

are denoted by

and are called the reduced homology groups of K. chain map

In

The obvious

induces an isomorphism

for

isamorphically onto a direct summand of A generator for the complementary summand is given by the homo-

logy class of v, where v

is any vertex in K.

Thus

For a categorical definition of reduced homology, see the exercise at the end of VT.1*. Let K

and L be regular complexes, with ideal cells

of dimension

-1. We have inclusions

i:

and and

For notational convenience we write and

Thus we have

an inclusion of K

We define the join complex K ° L

and L to be the space

together with the subspaces

(1) We leave to the reader the job of verifying that K ° L thus defined is a complex. We write cells of K ° L

for the ideal cell of K o L. The

are then given as follows.

ibly ideal) cells of K

and

Let

a and

be (poss-

L, respectively, of dimensions m -123-

and

ii respectively.

Then by 2.1,

is a closed cell of K « 1

of dimension

with interior

open cells of K ° L

are not joins of open cells.

write

Thus the

to denote the (open) cell

following 2.1, the •boundary of so

We will, however, By the remark

is the union

is a face of

only when either

and

and

and

Given incidence functions an incidence function

a

on K

and

on

on

K » L "by setting

and

of

L, we define

(2)

for all cells

of K

and

Thus

then

extends the incidence functions

complexes K

L. Note that if we set

and

L

of K ° L.

a

and

on the sub-

The verification that

is

indeed an incidence function is routine and is left to the reader. It follows from (l) that

The boundary operator in the chain complex If

a is a generator of

and

then

is given by (2).

is a generator of Here we write

-12!*-

for the chain

in the summand

Thus the computation of braic problem:

is reduced to the following alge-

Given two chain complexes define a new chain complex

ator

and C ° D, with boundary oper-

by-

Describe the homology of C ° D

in terms of the homology of C and

of D. If

is a chain complex, define the suspension of

C, written

2.2,

sC, with boundary operator

by

LEMMA.. The function

defined by is a chain isomorphism.

Proof:

It is clearly sufficient to show that

Let

-125-

is a chain map.

2.3.

COROLLARY.

Let K

and

L be regular complexes with finitely-

many cells In each dimension. Then

Proof:

This follows from l.U and 2.2.

2.h. DEFINITION. A regular complex K Equivalently, any cycle in

2.5.

PROPOSITION.

let K

Then K ° L, the cone on K,

Proof:

If K

Let

z he a cycle on K ® L. Since

of K

must bound in

0

such that L,

0

z

Let K he is a finite

L, there exists a finite sub-

z lies on K' ® L. Applying 2.3, z

and therefore in K •> L.

Alternate proof of 2.5 which does not use 2.3: Then

L he a

is acyclic.

is finite, 2.5 is a corollary of 2.3.

linear combination of cells of K complex

hounds in K.

he a regular complex, and 3et

point.

arbitrary.

is called acyclic if

where

Suppose We have

-126-

Thus

and

It then follows that

boundary of the

z

is the

and so K ° L is

acyclic.

3. The homology of SK The join of a complex K suspension of K,

3.1. THEOREM.

If K

with a

and is denoted by

If K

O-sphere is called the SK.

In this section we prove

is a regular complex then

is finite in each dimension this result follows from

Corollary 2.3. To give a proof for general K

we define a chain map

that raises dimension by an isomorphism of homology groups.

1, and that induces

For each chain

c

on K we

define

where

A

and B

are the vertices of the

is joined. Thus if

we have

-127-

O-sphere with which K If dim

then

Thus

induces a homamorphism

To show that with element of the

is a monamorphism, let Each chain

d

in

is a

so that

c

the ideal

is a boundary.

is onto, suppose that

q-cycle on SK. Then

Also

q-chain on K

is a sum e Accordingly

This implies that To show that

c be a

so that

This says that

so that is a cycle. Therefore

so that

is onto. Thus

is

an isomorphism, as desired. As an application we show that it is possible to construct a complex with prescribed finitely generated homology groups, provided that the given If K wedge

and

is free abelian of rank L

are complexes, we shall use

read

to denote a complex obtained from identifying a vertex -128-

of K with a vertex of L.

In general

depends on the choice

of vertices, but it is easy to show that if K regular then

Let -ry of a

and

L

are both

is regular and

be a regular complex obtained from wrapping the bound2-cell p

times around

than one. For example,

where p

is an integer greater

33aen the homology groups of

are given as follows:

If

denotes the k-th suspension of

ively by

Let

(defined induct-

then

G be a finitely generated abelian group. Then where

F

is free abelian.

Let Y

denote the space where the

number of spheres equals the rank.of F. -129-

It follows that

Given a sequence

of finitely generated abelian groups,

set

where for each

has

been constructed so that Then

and

abelian group of rank

we can construct a complex with homo-

logy

If

is a free

by forming the disjoint union of W with

r-1 points.

Relative Homology. A pair

(K,L) of regular complexes is a regular complex K

and a subcomplex

L

of K.

by an incidence function,

The pair is oriented if K say, and

L

is oriented

is oriented by the restric-

tion of For each pair i: the pair

(K,L), we have an inclusion homomorphism We define the group of relative

(K,L), written

q.-chains of

to be the quotient group

-130-

If c

is a chain on K, we denote by

set of Let

[c] the co-

which contains c. (K,L) be an oriented pair.

operator both in L

Let

denote the boundary

and in K. We define a homomorphism by setting

The map

for any

is well defined because

maps

to

Thus the collection forms a chain complex, written ij-.l. DEFINITION. The pair

(K,L), written

chain complex C(K,L). the pair b.2.

qth (relative) homology group of the oriented is the

qth homology group of the

If we wish to stress the orientation

of

(K,L), we shall write

LEMMA.. The inclusion map

is a chain map.

The projection map

is a chain map.

Proof: The first assertion is precisely the statement that

L is

a subcomplex oriented by restriction of the incidence function on K. To prove the second assertion, let

Then

We thus have the following commutative diagram:

-131-

By 4.2, i and

j

induce homomorphisms

and

We wish to define a hamamorphism Then sane

for

We have

is a chain,

in fact a cycle, on L. We set chain on L, we vary The map

If we vary c by a

by a boundary, and so

is well-defined.

is called the boundary or connecting hamamorphism for

the oriented pair

(K,L).

U.3• THEOREM. The sequence of groups and homamorphisms

is exact. Proof;

(i) Exactness at

Let

z = [c],

Then ker

Thus

Suppose

for same

Then

Also,

It is then obvious that

Thus ker

(ii) Exactness at So ker

Then

Let

such that

Then

We have so

z

is homologous to a cycle on L. Thus

and

ker (iii) Exactness at

Let

Then

-132Thus ker be such that

then

bounds a

chain

4

on L. Thus

and

shown that ker

We have

The proof of 4.3 is complete.

The sequence (l) is called the relative homology sequence of the oriented pair

(K,L).

Suppose that

are orientations of the pair

(K,L). Let

he the chain isomorphism of II.5.h. carries chains on L to chains on L

Then

and so induces

is a chain isomorphism, and induces an isomorphism

Relative homology

groups are independent of orientation.

Furthermore, the triple

induces an isomorphism of the exact homology sequences associated with the orientations

and

That is to say, the

diagram below is commutative:

The proof that this diagram is commutative is routine and is left to the reader. Remark; The exact sequence

-133-

is of course split exact. Thus we have maps and k: h

such that

restricts chains on K

he defined as follows.

hi = 1 ,

jk = 1. The map

to the cells of L. We may take k to

If c

is a chain on L

h:

is a chain on K, then

where

d

of K

not in L. Then k([c]) = c'. Using the splittings h

k, we could

and

c'

is a chain involving only cells and

define a map

(See diagram P» 131) A tedious calculation shows that Thus

carries cycles to cycles and boundaries to boundaries, and

induces a map

It is easy to prove that

Note that the splitting maps h Exercise:

Let K

and L be complexes, v

vertex of L. The complex K v l , vertices v

and v'

wedge of K

and

K x v* U v x L

and k

are not chain maps.

a vertex of K, v*

a

obtained by identifying the

of the disjoint union K U L,

is called the

L. The wedge can be regarded as the subcomplex

of K x L. Using the Kiioneth relations for,the homo-

logy of a product, compute

Show that

using the results in section 2 concerning In defining relative homology groups we had occasion to refer to the exact sequence

associated with any pair

(K,L). More generally, we may define a

-13-

short exact sequence of chain complexes to be a sequence

(1) of chain complexes and chain maps such that for each

q the sequence

(2) is exact. We may proceed as above to define a connecting homamorphism j

for each q.

is onto, we have

z = jc

for same

chain map,

since

Since

By the exactness of

for same monamorphism,

If j

(2),

and since d

is a cycle. We set

is well-defined, we vary c by

i

= id is a

To show that

id' for some Thus

is a

Then

d varies by a boundary. We

receive a sequence

which is called the homology sequence of the exact sequence (l). One proves just as in U.3 that the homology seuqence (3) is exact. is another short exact sequence of chain complexes, then a homamorphism of short exact sequences of chain complexes is a triple that the following diagram is commutative:

-135-

of chain maps such

It.h. PROPOSITION.

A homomorphism of short exact sequences of chain

complexes induces a hamamorphism of the associated homology sequences. In other words, the following diagram is commutative..

The third square is more complicated. for some

Let

Then

for some

"by definition. We have

z = jc

Then and so

Thus Exercise: group

(Relative homology with coefficients in an arbitrary abelian

G.) Verify that the following constructions are valid: Let

G be an abelian group, and let be an exact sequence of free chain complexes. Then

is exact, and, as in the case

we have a relative homology

sequence which is exact. For an arbitrary oriented complex pair -136-

(K,L) set

Then we may define the relative homology group with coefficients in G, to be the group

Exercise: (Relative cohomology with coefficients in G.) Let G be an abelian group, and let

be

an exact sequence of free chain complexes. Prove that the associated sequence

is exact. Define coboundaries in these cochain complexes as in Chapter II and verify that

and

are chain maps. Finally, define a

connecting homomorphism

Obtain a se-

quence

Prove that this sequence is exact. If (K,L) is a pair of complexes, set The group

is called the qth relative cohomology

group of (K,L) with coefficients in G and is written It follows from the exactness of the sequence above that we get an exact sequence:

This sequence is called the exact cohomology sequence with coefficients in G of the pair (K,L).

-137-

Chapter VI. THE OTVAIANCE THEOREM 1. Remarks on the Proof of Topological Invariance. Let K and K' be regular complexes. Suppose that f: is a continuous mapping with the following property: for each cell 0 of K, there is a cell

of

of the same dimension such that

f 0 = . (We say that f maps cells onto cells.) Let

he an inci-

dence function for K1. Define an incidence function

where ify that

and

are arbitrary cells of K. Then it is easy to ver-

is an incidence function for K. We define a chain map as follows: if

then

Then

and

for K by

is a (j-chain of K,

is a homomorphism. Moreover, if

are the boundary operators associated with

respectively, and if

is a generator of

By linearity, this relation holds over all of chain map. Consequently,

and

then

Thus

is a

induces hamamorphisms

for every q. We then say that the hamamorphisms

are induced by the mapping f and write them as

Examples of mappings satisfying the condition given above are a) the inclusion of a subcamplex in a complex and b) the projection of -138-

a complex K

to the complex K/F

(as long as K/F

obtained from K

by identification

is regular).

In this chapter we extend this definition of induced homomorphism. First, in section 2, we extend it to mappings

f which have the prop­

erty that the minimal Subcomplex containing the

f-image of any cell

is acyclic . Such mappings we call proper mappings. Then, in sec­ tions 3 and k, using the notion of subdivision, we show that an arbi­ trary mapping

f: K

>• K'

of finite complexes

can be factored

into proper mappings. We define the homology homomorphisms induced by

f to be the composition of the homomorphisms induced by the fac­

tors of f. Returning to our first topic, we note that if f: the identity mapping then fjj.: H (K)

so does

> K

is

f maps cells onto cells, and

> Ξ (K) is the identity isomorphism for each q. Also, > K»

if f: K

K

gf, and

in homology theory.

and g: K«

>• K" map cells onto cells, then

(gf )# - g^f^.. These two properties are fundamental In section k we show that they hold for the induced

homomorphisms of arbitrary continuous mappings of finite regular com­ plexes. The invariance theorem (k.9) is a corollary of this result. We emphasize that in this chapter we prove the invariance theorem only for finite regular complexes. 2.

Chain Homotopy, Carriers, and Proper Mappings.

Suppose we are given two topological spaces X continuous mappings

f

and

f., from

See V. 2.k. for a definition. -139-

X

and Y,

and two

to Y. We say that

f

is_

hamotopic to

written

F:

such that

if there is a continuous mapping and

We

can reformulate this condition as follows. We define mappings and

from X into

Then two mappings

and and

frcm X to Y are hamotopic if and

only if there exists a mapping

such that FpQ = f

and We define in a similar way the notion of chain hamotopy. Suppose that we are given two chain complexes C and and

from C to

and two chain maps

Let I be the cell complex for the

closed unit interval consisting of a 1-cell I and two vertices and

Let I also represent the chain complex for I with the

boundary operator plex

defined by

Then in the chain com-

we have

and

for c a chain of C. We define chain maps to

and

from C

by

2.1. DEFINITION. Two chain maps

are chain

hamotopic if there exists a chain map F: and and we write

such that

F is called a chain homotopy of to indicate that -140-

and

and

,

are chain hamotopic.

2.2. LEMMA.. Let and

and

be

chain maps from C to

are chain homotopic if and only if there are homamorphisms such that

where

. Then

and

for each q,

are the boundary operators of C and C' respect-

ively. Proof: If

and F is a chain homotopy of

and

set

Then

If

is given satisfying the conditions of the Lemma, define

a chain map F:

by and

for c a q-ehain of C. Then F clearly maps q.-chains of into q.-chains of

, and

Thus F is a chain map and provides the desired chain homotopy. The collection deformation for

of homamorphisms and

is called a chain

, and is said to exhibit a chain homotopy -Ikl-

of

and

2.3. THEOREM. Let and

and

be chain maps from C to

. If

are chain hamotopic, then

Proof: Let

be a chain deformation exhibiting a chain homo-

topy Analogously, of and let, Let and z be a be q.-cycle of complexes,and C. Then cochain let be cochain maps. A collection homamorphisms

of

satisfying

is called a cochain deformation. One proves as in 2.3 that if a cochain deformation for the cochain maps

Now let K and

is then

be oriented regular complexes, and suppose

that

are chain maps. Let G be an abelian

group. Then

and

induce chain maps and cochain maps

Ham

Ham are chain homotopic. If

. Suppose that

and

is a chain deformation exhibiting a -1^2-

chain homotopy of for

and

and

then

is a chain deformation

Furthermore,

cochain deformation for Hcsm

is a

and Ham

It follows that

and

A carrier from a regular complex K to a regular complex K' is a function X which assigns to each cell u of K a non-empty subcamplex

of

in such a way that if

then

is a subcomplex of

A carrier X from a pair of complexes

(K,L) to a pair

is a carrier from K to

which when

regarded as a function on the cells of L is a carrier from L to

If

is a continuous function then a carrier for

f

is a carrier X

such that for each cell

X

is said to carry f. Carriers for a map always exist — set for each

assigning to each

of K,

The minimal carrier for f is the carrier in K the smallest subcomplex in

contain-

ing A carrier X from K to the subcamplex

of

is acyclic if for each a in K

is acyclic. A map f:

is

called proper if the minimal carrier for f is acyclic. We may relativize these notions in the obvious way, noting that if is given, then the minimal carrier for f |K is automatically a carrier from

, The concept of a -143-

proper mapping is fundamental in cellular homology theory because a proper mapping from one regular complex to another can be shown to induce homomorphisms of homology groups. Finally, we call a chain map

proper if

for each O-chain c on K, In 2.b. LEMMA. Let K and

In c.

be oriented regular complexes. If X

is an acyclic carrier frcan K to map

then there exists a proper chain

such that for each cell

a chain cm

, (We say X carries

of K

is

) Furthermore, if

and

cp^ are any two proper chain maps carried by X, then there exists a chain deformation

and

exhibiting this chain hamotopy

which is also carried by X. Proof: We construct

by induction on dimension. For each O-cell

A of K, we select a vertex B of the non-empty subcamplex X(A) of

We then set

and extend over all of

by

linearity. Then for c a O-chain of K we have In

= In c.

Given a 1-cell

and

In c

= In on

of K,

is a O-chain on

Since

is acyclic, there exists a 1-chain

such that

1-chain on

We set

and

so that

the boundary operators

and

, We extend

Then

is a

commutes with over

by

linearity. Suppose that

has been defined on

for all

so that it commutes with the boundary operators and is carried by X. Let

be an n-cell of K. Then -Ikk-

is an (n-1)-chain on

since

is a chain on

Also,

is a cycle on

Thus by the acyclicity of n-chain c on

such that

by linearity over

because

we may choose an We set

and then

We extend

commutes with the boundary op-

erators and is carried by X. In this manner we define To prove the second assertion of the Lemma, let two proper chain naps of C(K) into

for all n. and

both carried by the

acyclic carrier X. We construct a chain deformation a chain homotopy of

and

is a vertex of K, X(A). We set

has index zero, and thus bounds in equal to a 1-chain on x(A) having

O-chain on K,

n

exhibiting

by induction as follows. If A

as boundary and extend by linearity over

Suppose that

be

Then if c is a since the last term is zero.

a

has been defined for all dimensions less than

so that it is carried by X and satisfies

Let

be an n-cell of K. Then

is a chain on

We have

Then since

is acyclic we can choose an (n+l)-chaln c on

whose boundary is tend by linearity over

We set

and ex-

We continue in this manner for all n.

defined is carried by X and exhibits the desired chain homotopy of

and

The proof of the Lemma is complete.

Wow suppose that X

is an acyclic carrier from

(K,L) to

By Lemma 2.4, X, as an acyclic carrier from K to carries a proper chain map

The restriction of

cp to C(L) is a proper chain map

which is

carried by X as an acyclic carrier from L to

Thus if we set

we obtain a well-defined homamorphism Since

is a chain

map. Furthermore, let

be proper chain maps carried by X.

Then by Z.k, there exists a chain deformation homotopy of i

and

and we may suppose

Thus if

exhibiting a chain is carried by X.

It follows that if we set for

then

is well-defined.

Thus

and

are chain hamotopic. In other words, X induces

the triple of chain maps

defined uniquely up to a chain

homotopy. 2.5. COROLLARY. An acyclic carrier X from

(K,L) to

induces a unique set of homology and cohomology hamamorphisms

-1^6-

for any coefficient group G. The triple of hamomorphisms defines a homomorphism of the relative homology sequences of the pairs (K,L) and

In other words, the diagram below is

commutative;

A similar result holds for cohomology. Proof: Everything but the caramutativity of the diagram has been proved, and that follows from V.k.k. 2.6. COEOLLABY. If and

is proper, and if

are any two acyclic carriers for f, then the associated homo-

logy and cohomology hamamorphisms for

and

coincide. We say

that f induces these hamamorphisms and write them as

and similarly for cohomology.

-li+7-

Proof:

x

Χ(σ) Q then X,

Let 0

X be the minimal carrier for 1

(°) Π \(v)'

φ

SiBiS, if φ

is also carried by

as carriers from K

and

X-

2.7.

are

to

X

Then for each

X,; here ve are regarding

COEOLLARY. If

X

is an acyclic carrier from K

σ of K,

the dimension of

K1

to

X. Suppose that

ψ

is also carried by

a collection of homomorphisms Jj

So

Jj

σ

is a q.-cell of K, then

σ,

X. φ

carried

X. Then by the Lemma there is

carried by

σ, which is a (q.+l)-chain on

is zero and

X

such that

Χ(σ) is at most that of

Proof: By the Lemma, there is at least one proper chain map

If

X, "X , and

The unique homomorphisms induced by

then there is exactly one proper chain map carried by

φ - ψ.

X,

φ # , φ , φ # , etc.

for each cell

by

σ of K,

is a proper chain map carried by

and

K1.

f.

X which gives a chain homotpj

Χ(σ) is of dimension at most

q.

Χ(σ), must be zero. Thus each Jfi

φ = ψ.

Examples of proper mappings: We assume throughout that a regular complex on a closed cell is acyclic. This will be proved in section Ij- of this chapter. (a) Inclusion of a subcomplex: Let complexes, and let carrier for and so

i

i

Ϊ: L

> K

(K,L) be a pair of regular

denote the inclusion map. The minimal

assigns to each cell

σ

of

L

the subcomplex

is proper. The induced homomorphisms

the homology and cohomology sequences of the pair

1½

i^

and

(K,L).

* i

a

of K,

are those of

b) Automorphism of a complex: Exercise. Let T: K x L

> LxK

for x e K ,

is proper, and the minimal carrier

y e L. Then T

be defined by T(x,y) = (y,x)

preserves dimension, so, by 2.7, there is a unique chain map φ in­ duced by T. Show that cp(c d) = (-I)1^d ® c) for c e C (K), d e C (L). Note that the mapping ψ: C(K) ® C(L)

> C(L) R

be given by a diagonal matrix,

with k minus ones starting from the upper left, and then n-k+1 plus ones. Compute T„* in H η(S ). c) Projection of a product onto its factors: Let f: K x L X

> K be the projection mapping. The minimal carrier

for f satisfies Χ(σ X Ό

= σ· Thus f is proper and induces

a unique chain map φ, by 2.7· Note that r

φ(σ ® τ) = ,

0 if dim τ > 0 a

if T is a vertex.

d) Simplicial mappings: Let K and L be simplicial complexes. Then f: K

> L is a simplicial mapping if f carries the vertices

of each simplex of K into vertices of a simplex of L and is linear in terms of baryeentric coordinates. A mapping of the vertices of K into the vertices of L can be extended to a simplicial mapping if and only if it maps vertices spanning a simplex of K spanning a simplex of L. If opposite a vertex A, then

onto vertices

σ is a simplex, and τ σ is the cone on τ. Thus

is the face a is acyclic

by V.2.5. It follows that a simplicial mapping induces a unique chain map φ: C(K)

> C(L). -1^9-

3. 3.1.

DEFINITION.

Subdivision of a Regular Complex Let K he a regular complex. The (first iterated)

subdivision of K, written

Sd K,

vertices axe the cells of K

is the simplicial complex whose

and whose simplexes are defined as follows:

A finite collection of cells of K Sd K

form the vertices of a simplex of

if and only if the cells of the collection can he arranged in

order so that each is a proper face of the next. It is ohvious from the definition that We topologize

is a simplicial complex.

|Sd K| by giving it the weak topology with respect to

closed simplices. Note that if L

is a subcomplex of K,

is a simplicial subcomplex of Sd K. is not equal to 3.2. THEOREM. to

Sd K

then

Sd L

Note also that in general Sd(K )

(Sd K) . If K

is a regular complex then

|κ| is homeomorphic

|SdK|. Ih the proof of this theorem we will use the following lemma.

3.3·

a

LEMMA.. If

a complex is the join of the vertex Proof: A cell τ

of Sd a

a with the subcomplex

is a collection of faces of

he arranged in an increasing order:

σ < σ-, < ··· < α .

OP. = σ or T

is a cell of Sd Jr. If

of the vertex

σ with the cell

k =0

then

τ

Sd a,

is a cell of a regular complex, then

is the vertex

of the join of the vertex

a. = a,

then

a < . · .K'

=

f« r = > SdV

In this diagram the mappings g* division homeomorphisms.

and g are the iterations of sub­

(See comments before 3·^-·) The mapping

f

is defined by the relation f = g*f'g. We say that f induces f * S

or that f · is the mapping f as a mapping of Sd K

I*

into Sd K 1 .

We shall sometimes write f for f'. Before stating our theorem we prove the following lemma from general topology. 3.8.

LEMMA.. (Lebesgue Covering Lemma).

space and let

Let X be a compact metric

{U.} be an arbitrarily indexed open covering of X.

Then there exists a number δ > 0, called a Lebesgue number of the covering

{U.), such that every set of X of diameter less than

δ

is contained in same U.. Proof: For each χ e X and every i, let d(x,i) be the radius of the largest open ball around χ contained in UJ. Define d(x) = i.u.b. d(x,i). i Then d(x) is finite for each χ since X is bounded. We show that d; X have

> R is continuous. Letting (d(x) - d(y)| < e whenever

ρ be the metric on X, we

p(x,y) < €. Next we note that d(x",

is positive for each x, since the U. cover X. Thus d maps X

-156-

into a compact subset of the positive real numbers. So d(x) is bounded away from zero by some

e > 0. Choose any δ such that

0 < δ < e. Let W be a set of X of diameter < δ. Choose χ ε W. W

is contained in the open δ-ball V about x. But

δ < e < d(x).

Thus for some i, W C V C U.. So δ is a Lebesgue number of the . covering {U.}. 3.9. THEOEEM. Let K

and K* be finite regular complexes, f a

continuous function mapping K Then there exists an integer N f

is proper as a map of Sd K

into K 1 ,

and r a positive integer.

such that for each s > IJ, the map into Sd K*.

£

Proof: The complex K is covered by the collection £j of open sets —1 "X" r f (St A.) , where A. varies over the vertices of Sd K». Since K 3

'

3

is a compact metric space, we may by 3-8 choose a Lebesgue number δ / H for the covering O . Let Ν be such that the mesh of Sd K is less than δ. Given s > K, we show that the minimal carrier for s f: Sd K

r > Sd K'

is acyclic. We note first that the smallest sub-

complex of Sd K' containing a given point set X is the union of the closures of those cells of Sd K* which have a non-trivial inter­ section with X. Let

α be a cell (open simplex) of Sd K. Then the diameter of /— -1 r

a is less than

δ so that

σ L· f StA for some vertex A of Sd K'.

Thus fa C StA. If f a meets a cell τ of Sd1K', then A is a vertex of τ. We can thus write τ as the join of A with the face τ

of τ opposite A: τ = A · τ.. Thus the smallest subcomplex of

SdrK·

containing

ta

is

UT = U A ο τ = A ο υ τ . Since the A fonir# τ τ A If A is a vertex of a regular complex K, the (open) star of A, written St A, is the union of all (open) cells of K having A as a vertex. -157-

join of a vertex and a complex is acyclic, by V.2.5, the smallest subcomplex containing

is acyclic for each

of

minimal carrier for f is acyclic, and f is proper.

-158-

Thus the

k. The Induced Homomorphisms and the Invariance Theorem. In this section we describe the process of associating with each continuous mapping of one finite regular complex into another a sequence of homomorphisms of the homology groups of the complexes, and similarly for cohomology. We have already shown that a proper mapping induces such homomorphisms. Thus Theorem 3«9> which associates with an arbitrary mapping of a finite regular complex a proper mapping, will play an essential role in all that follows. 4.1. THEOREM. Each continuous mapping f:

of

pairs of oriented finite regular complexes induces hamomorphisms and for each q , such that the following conditions hold: a) If

is the identity mapping, then, for

each H^K,!} G) are the identity isomorphisms. b) If

are arbitrary continuous mappings, then, for each q ,

and Proof: Let

be the statement that the theorem holds for all

finite complexes of dimension at most n . Let statement: If dimension of

be the following

is a cell of an oriented regular complex and the is at most n , then 159

is an acyclic complex.

Then

and. Let

dim

are all obvious. be a cell of a regular complex and let

Then the dimension of

homeomorphism f:

, where

is

. We have a

is the regular complex

on the (m-l) sphere given in Section 1.2. Then by are induced homomorphi sms

, there and

for each q. . Prom b) and a) we derive that

Similarly,

Thus

is an isomorphism so that

160 Consider the following diagram: We Here and cell have secondly of therefore We is that This The generated first shown So cycle implies derive is actually that zero by first some has for generates .that is preliminary coefficient We acyclic. have results, by so the that on sowhich above each that diagram. follow (m-l)

from

B

η

.

4.2. IiMMA.. If the dimension of the oriented regular complex K is at most η

and if g:

(SdK, SdL) • • > (K,L) is a subdivision

homeomorphism, then a) g

and

g"

b) g^. and induced by g

g

are proper maps. , the homology and cohomology homomorphisms

given by 2.5, are independent of the choice of g .

c) Sx, and

g

are isomorphisms.

Proof: By 3·3 and V.2.5, if

σ is a cell of a regular complex,

then Sda is acyclic. The minimal carrier cell

σ of K

into the subcomplex Sdff of

proper. Next we show that g SdK , then

for g"

maps a

SdK . Thus g

is also proper. If

is

σ is a cell of

σ is a collection of cells of K which can be arranged

in an increasing sequence: σ is mapped by g mapped into

σ. < σ, < ... < σ

into an interior point of

. The vertex σ

. All of

σ , so the smallest subcomplex of K

is "σ . Since B

X.

that ~5

σ

g

of

σ is

containing

go

is a cell of dimension at most η , we have from

is acyclic. Thus the minimal carrier

acyclic and so

σ

X»

for g

is

is proper. We have proved a ) .

Since the minimal carriers X,

and

X. for

g

and g are

independent of the choice of g , it follows that the homology and cohomology homomorphisms induced by of the choice of Now let X»

φ,

respectively.

g

and by

g

are independent

g . This proves b ) . and Let

φ?

be proper chain maps carried by

σ be a cell of K . Then l6l

φ,σ

X, and

is a chain

For each cell closed cell contained in The carrier

X

from

acyclic by

K

. X

1 . Since

X

Thus to

K

is a is a chain on

sending each cell a to

carries

is

and also the identity chain map

does not increase dimension, we have from Corollary

2.7 the fact that

Let

of

This of course implies that

be a cell of

SdK . If

of cells of is a chain on

K

0

is the collection

then we have

. For each cell

.

of

Consequentlyj

Thus , we have

is a chain on

is then carried by the acyclic carrier which sends each cell of

SdK

map, and so

into

. But so is the identity chain

by 2.4. .

Thus

It follows that

and

and

are isomorphisms and the lemma

is proved. 4.3. K

COROLLARY.

If

K

is a regular complex and

of dimension at most

Proof:

n , then

is acyclic by

is acyclic for every

. For

by Lemma 4.2.

NOTATION.

a is a cell of

, and q arbitrary, Thus

is acyclic for

For each pair of nonnegative integers

r

and

s

we have a homeomorphism obtained by composing subdivision homeamorphisms. 4-5-

LEMMA. If

for all

K

r, s . If

is of dimension at most , then

162

n , then

is proper

If

r > s , then

Proof:

Suppose that

minimal carrier

X

r < s , and let

for

acyclic by it-.3. Thus

sends

a be a cell of a to

Is proper.

map carried by the minimal carrier then

is a chain on

If

The

and so is is a proper chain

for By Corollary 2.5>

and by Theorem II.5-2,

Also, by Corollary 2.5 and the exercise at the end of Chapter II, we may derive analogously that

Now suppose that

Each

r > s . We have

, maps cells into (open) cells since

163

it is a subdivision hameomorphism. and so is proper by

Thus

maps cells into cells

. More explicitly, let

There are uniquely determined cells with

be a cell of in

and

The minimal carrier

for

sends each face of

sends

is a chain on

that

, and

c

. Thus if is a chain on

• From this it follows immediately

is a chain on

for

. Also,

to a subconrplex of

is a proper chain map carried by then

to

. Thus the minimal carrier

carries the proper chain map

relations involving

k.6.

COROLLARY.

and

The desired follow immediately.

Given nonnegative integers

r,s,

and

t ,

and

Proof:

This corollary follows immediately from Lemma k.5 and the

fact that

, proved in Lemma h.2.

We proceed with the proof that

. Let

f:

be an arbitrary mapping of pairs of finite regular complexes of dimension

Let

be a positive integer. 16k 3-9) there exists an integer N such that for as a mapping of the following diagram:

r

into

Then by Theorem f

is proper

. Thus we have

In this diagram

and

morphisms and

are iterated subdivision homeoThe mappings

are

all proper, and we make the folio-wing definition. ^.7.

DEFINITION.

If

f:

is a mapping of finite

regular complexes of dimension at most cohomology homomorphisms induced by

n , then the homology and

f , written

and

respectively, are the compositions

and

In order to justify this definition, we prove:

k.8. of

LEMMA.. The induced hcanomorphisms

r

Proof:

and

are independent

s .

Let

such that

and

r

be given and suppose

f , as a mapping of both

into are induced by

, is proper.

s

and

t , with and

In the diagram below,

f:

165

, are

and

By b.6,

, and similarly for cohomology.

Thus we need prove only that and that

Let

be the minimal carriers for

and let •

be proper chain maps carried by

If

is a cell of

is a chain on

If

, then

, and

r

and

chosen so that

f

is a chain

. The desired equalities follow. r1

be given, with

is proper as a mapping of

and into and

So

. Thus

and so Let

, and

is a chain on

, then

on

, and

are induced by

, and suppose

s

is

into In the diagram below,

f:

166

By k.6, we need only prove that

and Let

be a cell of

for

and

Then

Let

be the minimal carriers

proper chain maps carried by

is a chain on

is a subcomplex containing

is a subcomplex of

containing

Thus

Thus if

is a chain map carried by the minimal carrier for

is a chain on equalities follow.

. Consequently

and the desired

The proof of the lemma is complete.

We show now that the induced hamomoiphism satisfies properties a) and b) of the theorem. is proper by

. If

f:

To prove a) we note that the identity map

167

is an arbitrary proper

mapping of finite regular complexes of dimension

, then by an

argument similar to the proof of Lemma k.Q, the induced homomorphisms

and

given by Definition U.7 coincide with those previously

defined (Corollary 2.6) for

f

as a proper mapping.

Property a)

follows as a special case. To prove b), let

f:

and

g:

be mappings of pairs of finite regular complexes of dimension Let

r be a positive integer. We subdivide

finely that if A of

is a vertex of

so

, then there is a vertex

such that

Fox, if

B

is a Lebesgue

number of the covering

a vertex of

then it suffices to take

S

so large that Mesh

order to ensure that stars of vertices of also require that is a cell of fcrl f1

. Then we take

are acyclic, we t

, there is a vertex A

g'

are the mapping induced by

so large that if

of

, as in the proof of Theorem 3-9and

In

f

such that

In the diagram below, and

g

respectively:

Arguing as in Theorem 3-9, we show easily that

, and

are all proper.

, and

respectively. a cell

Let their minimal carriers be

Then each

of

carries a proper chain map A

, and for some vertex

B

for some vertex A

is a chain on of

of

, . Given . Thus

. But

. This implies that for each cell

168

r

of

St B

of Theorem 3«9>.

and so

. As in the proof

> which is the minimal subcanrplex containing

, can be written in the form plex of

depending on

where

is a subcom-

T . Thus

is an acyclic subcanrplex of

We define

If

then

so

is an acyclic carrier.

Since

is a chain on

So

Thus

is a chain on is an acyclic carrier for

But

so

carries and so we have

. Thus

and

are both carried by

Consequently, and

Prom Lemma k.5 and the proof of Lemma If.2,

Thus 169

A similar argument shows that

Thus

We have thus shown that

is true for all

the processes used for different

n

morphisms lead to the same result. f:

n . We note that

in defining the induced homoMore precisely, if

is a mapping of finite regular complexes,

then for

n

and

n'

both greater than the dimensions of the com-

plexes, the induced homomorphisms given by

and

, coincide.

Also, the properties a) and b) follow since in any one Instance we may take

n

large enough and apply

. Theorem 4.1 is proved.

It is important for later considerations to note that in our proof we do not assume Lemma II.5-3that

K

and

K'

We only assume, in effect,

satisfy the redundant restrictions.

way, no matter how we orient

If and

K1

Stated another

our proof yields the fact

(see below) that their homology groups are isomorphic.

This is one

way to prove, incidentally, that homology groups are independent of orientation!

4.9.

(See Chapter VIII, Section 4.)

COROLLARY.

(The Invariance Theorem).

If

f:

is a homeomorphlsm of oriented finite regular complexes, then for each

q. we have

Thus the homology and cohomology groups are topological invariants.

Proof: Thus

Similarly, is an isomorphism.

By similar reasoning,

170

is also.

If we are given a map regular complexes, then

f:

f

of oriented finite

as a mapping of

homomorphisms of the homology groups of we shall write

K

K

into

K1

induces

into those of

K'

which

Similarly, we have homomorphisms

It is easy to see that the triple

defines a

homomorphism of the exact homology sequences associated with and

, using Corollary 2.5 and the definition of the induced

homomorphism.

if. 10.

Thus we have

C0R0UARY.

The exact homology and cohomology sequences of a

pair are topological invariants.

Exercise.

(Reduced homology groups.) For each complex

unique map

f:

K • • > point induces homomorphisms .

Set

. Show that if

is a continuous map, then . if

L

is a subcomplex of

Here

is induced by

g

induces a map

Show that the sequence below is exact, K :

i:

is given by the composition , where

and

K , the

is the inclusion,

is induced by the boundary homomorphism of the ordinary

homology sequence.

One must show that the composition (point)

More generally, let

is zero.

be a category with a maximal object

satisfying (i) For each object

, there is one and only one

morphism

171

P ,

(ii) If A

and B

are objects of

, and

,

then Let F:

be a covariant functor from

to the category

of ab.ellan groups and homomorphisms. Define the reduced functor F

by setting

If

, since into

for to

, it follows that F(f) maps

. Thus we may define Verify that

is a covariant functor from

' . Carry out an analogous definition for a contravariant

functor F:

, using cokernels.

172

5· The Properties of Cellular Homology Theory. Cellular homology theory assigns to each pair of regular complexes a sequence of homology groups and to each continuous mapping of pairs of regular complexes induced homomorphisms of the homology groups. In this section we state the seven basic properties of cellular homology theory. It turns out that these properties characterize cellular homology theory. A description of the precise sense in which this is true and a proof of the characterization may be found in Foundations of Algebraic Topology, by Eilenberg and Steenrod. We keep the coefficient group G I.

If

f:

fixed throughout this section.

(K,L) — > (K,L) is the identity, then

f^

is the

identity isomorphism in each dimension. II.

If

(K,L) — > (K',L') and

f:

arbitrary continuous mappings, then III. If

Let δ#

f:

and

(K,L)—>(K',L')

g:

(Κ',Ι/) — > (K",L") are

(gf) # = g^f^ ·

be an arbitrary continuous mapping.

o\£ are the boundary homomorphisms of the exact homology

sequences associated with

(K,L) and

(Κ',Ι·1) ι respectively, then

· (X,A) the

qth homology group

and the collection of these groups will form a direct

spectrum of groups. The limit group of this direct spectrum will be by definition the 1.1. DEFINITIOH.

qth singular homology group

of the pair

(X,A).

A directed set (class) is a set (class) A to­

gether with a binary relation
(X,A) and

(K,L)

g:

S(X,A) as follows. If (K',L')

> (X,A) are given, then Jjf.

f < g h:

if and only if there is an isomorphic embedding > (K«,L·) such that

(K,L)

f = gh. The ordering
A),

(K",L") f < f U g

S(X,A),

(K',Lr) respectively, into

(X,A), be given. Let (K*,L').

is

(K,L)

induce a mapping and using the obvious embeddings we

and g < f U g.

Then

(S(X,A), (X>A)

( i . e . , f o r each

f e S(X,A)), l e t H f (K,L;G) be a l a b e l e d copy of H (K,L;G). Technically, H (K,L;G) is a group isomorphic to H (K,L$G) together y.

y.

with a definite isomorphism u e H (K,LJG), in

we write

H (K,L;G)
H (K,L;G).

If

[u,f] for the element corresponding to

u

HJ(K,LJG).

An isomorphic embedding is an embedding which maps cells onto cells. We shall often use "embedding" to mean "isomorphic embedding".

183

We wish to show that the triple together with appropriate admissible homamorphisms, is a direct spectrum.

To define the admissible hamamorphisms, we suppose that and that

Thus

and

and there is at least one embedding satisfying morphism mapping

. The admissible hamo-

to

are then the homology

hamamorphisms induced by embeddings precise, if

h

h

such that

To be

is such an embedding, then the mapping is an admissible hamamorphism.

Note that

there may be more than one admissible hamomorphlsm between a given pair of groups.

2.1.

PROPOSITION.

For each

q., the triple

with the admissible hamamorphisms defined above, is a direct spectrum of abelian groups, written

Proof:

We verify the properties of a direct spectrum, referring to

Definition 1.2.

Property (i) is clear.

Property II of cellular homology theory.

Property (ii) follows from To prove (iii), let

and that

and

be given.

are embeddings of

(K,L) in

and let

satisfying

We find an admissible

hcmcmorphism carrying ment.

Suppose

and

We define a complex

into the same ele-

as follows,

on the unit interval with vertices at

and

then the complex obtained from the disjoint -union

184

let

I be a complex

1.

is

by the identifications Let

for all he the obvious inclusion.

Define

by setting

Then

g'

is well .defined because

that

g'

is continuous.

It is easy to see

Also,

by construction, so the map is an admissible hamamorphism.

I i

I

Since in

hh^

and we have

map

(K,L) onto opposite ends of Thus

erty V of cellular homology theory.

Thus

and the proof of the Lemma is complete.

185

by Prop-

2.2.

DEFINITION.

coefficients in

The

qth singular homology group of

G, written

is

(X,A) with the

limit group of the direct spectrum

Notation:

If

then we write

is given and for the equivalence class of

taining

186

•a. con-

3. The Properties of Singular Homology Theory In cellular theory we found it relatively easy to compute homology groups and difficult to prove that they are topological invariants. The reverse is true of singular theory. Given a continuous mapping

of pairs of

topological spaces, we define the induced homomorphism as follows. If we associate with each mapping g of a pair of complexes

(K,L) into (X,A) the mapping

we obtain an order-preserving function from S(X,A) to S(Y,B). For each g

S(X,A), define by setting

collection

The

{cp } is clearly a homomorphism of direct spectra (1.5), S

and is said to be induced by the mapping

The

resulting hamamorphism of limit groups is called the homology homomorphism induced by f and is written I. If

is the identity, then is the identity for each q.

II. If

and

are

arbitrary continuous mappings, then It follows that the singular homology groups are topological invariants. If (X,A) is given, we define a boundary homomorphism for each q. as follows. If we associate to each mapping

the mapping

we obtain an order-preserving function from

18?

S(X,A) to S(A). For each f e S(X,A), define by setting where

is the boundary liomamorphism of the relative homology

sequence of the pair

(K,L). To show that the collection

preserves the successor relation and thus is a h anamorphism of direct spectra, let

be a mapping such that there

is an embedding

with then

Let

is a successor of

By

property III of cellular homology (VI.5), where morphism for

Thus

is the boundary homois a successor of

in the direct spectrum

. Therefore

is a hamomorphism of direct spectrum. We define to be the hamamorphism of the limit groups of

and

induced by

III. If

is a continuous mapping of pairs of

spaces, then Proof of III: Consider the following diagram

Here

is the hamcmorphism of direct spectra induced by

induced by

and

and

is

are defined as in the discussion

188?

before III. The composition

is a hamcmorphism of direct spectra;

the hamamorphism of limits induced by h anamorphisms of limits induced by

is the composition of the and. by

thus, to show that

Similarly for

it suffices to show that

the above diagram is commutative. Let g:

be given. Then if

we have

On the other hand

Thus

and the proof is complete.

IV. If i

is a pair of spaces, let

and

denote the inclusion mappings. Then the sequence below, called the homology sequence of the pair (X,A), is exact:

Proof of IV: We prove exactness at

and leave the rest as

an exercise to the reader. a)

Suppose that denote the inclusion. Define

189?

is given. Let

so that the diagram below is commutative:

If

then

Since

an embedding,

but

b) Ker

so

Suppose that

is given and that

satisfies

Then there exists a map

and an embedding that

and

is

such

We denote the embedding

and we have inclusions

by

Then

By the exactness of the relative homology sequence of the pair

for some

We show

that

Now

is represented by and

and

, it follows that

have the common successor

spectrum

Since

Thus i

in the direct and ker

Note that II and III imply that the homamorphisms of singular

190?

homology groups induced by a map

commute with

the hamamorphisms of the exact homology sequences of the pairs (X,A) and

(Y,B). Thus f induces a hamamorphism of exact sequences.

V. Ihvariance under homotopy: If into (Y,B) such that

and

are maps of (X,A)

then

Proof of V; Let

be given, and suppose that Let

connecting

and

be a homotopy Define

for

and

for

by and

Since

Define

Then property V of cellular homology

theory implies that

Thus

have the common successor

and

in the direct spectrum

(See diagram below.) It follows that

3.1. COROLLARY. A homotopy equivalence between two pairs of spaces induces isomorphisms of their singular homology groups.

191?

3.2. DEFINITION. An Inclusion mapping excision if

is called an

The excision is called proper if the

closure of Y - X is contained in the interior of B. VI. If

is a proper excision, then the induced homo-

morphism

is an isomorphism for each q.

We give the proof of VI in section VII. If X is a point then

VTI follows from the fact that a map factored through an acyclic complex.

192?

can be

h.

The excision property.

1)-.1. A counterexample for improper excision: Let X be the closed region of the xy-plane that lies below the curve y = sin —, above the line y = -2, and between the lines χ = 0 and χ = 2/ττ.

X is

homotopically equivalent to a point, so H?(X) = HT(X) = 0. Let A be the boundary of X. By the exactness of the homology sequence for

(X,A),

oj

ΐξ(Χ,Α) « ΐξ(Α).

y = -2

We show f i r s t t h a t

HT(A) = 0.

We claim that i t i s enough t o show

that (l) L

If into

f:

L

> A i s a mapping of the f i n i t e regular complex

A,

then

f

is homotopic t o a constant.

Suppose t h a t we have proved ( l ) . and suppose t h a t of

[u,f] e H 1 (L).

Let

f:

L

> A be given,

We have the inclusion

L as the base of the cone on

L.

193

Since

h:

f - constant,

L C CL there

exists a map the class

> A

g: CL

gh = f.

[hgUjg] as a successor. But

so 1IxU = 0. That is,

{[u,f]) = 0

Proof of (l): Suppose that that part of

a sequence

f:

CL

[u,f ] has

is acyclic (V.2.5),

and so (l) implies that H (A) = 0. > A

L

Thus

is given, and let

f(L) which lies on y = sin — . We show that

bounded away from the line

form

such that

χ = 0.

S he S is

Suppose not: Then there exists

{v } of points in S which converges to a point of the

(0,a),

-1 < a < 1. For each n, choose

f (w ) = ν . Since

L

is compact,

w

e L

so that

{w ) has a limit point w, and

then it follows that

f(w) = (0,a).

connected at

More precisely, there exists a neighborhood

(0,a).

(in A) of contains f

A

fails to be locally

χ = 0.

Since

L

is locally con­

(U) contains a connected neighborhood W

of w. Since

the continuous image of a connected set is connected, on the line w

U

(0,a) such that the connected component of U which

(0,a) lies on the line

nected,

Now

χ = 0. But that implies that W

not containing any points of the sequence

the fact that

{w } has w

f(w) must lie

is a neighborhood of {w }, which contradicts

as a limit point. This contradiction

establishes (l). Let y = +1, But

B

Y be the closed region bounded by and let

B = Y - (X-A).

χ = 0,

2 χ = —,

We have as before that

TT

y = -2,

H|(Y,B) « H^(B).

is homotopically equivalent to a circle, and so H?(B) « Z.

Consequently

Ξΐ(Χ,Α) = 0, Ηρ(Υ,Β) « Z,

by the excision

and the homomorphism induced

(X,A) C (Υ,Β) is not an isomorphism.

To prove that a proper excision induces isomorphisms of singular

19½

homology groups we find. It convenient to define a new direct spectrum associated with a pair

(X,A). We define a new ordering

on S(X,A) by saying that if

and

then

if there exists a continuous

mapping

such that

the direct spectrum

The groups of

are then the same as before: as above and

then the admissible homamorphisms mapping are the homology homamorphisms induced by continuous maps

such that

claim that spectrum. contains the relation

We

I is then a direct is a directed class since the relation Properties (i) and (ii) of Definition 1.2

are immediate. Property (iii) is given by Corollary ^.3 below. We assert that our original direct spectrum cofinal in the new direct spectrum

is The first condition

of cofinality is obvious since we don't have any new groups. The second condition of cofinality is contained in the following lemma: h.2. LEMMA. If and

are continuous mappings satisfying then there exist isomorphic embeddings and

respectively, in a pair such that

4.3. COROLLARY.

and

of and a mapping and

satisfies condition (iii) of Definition

1.2 and is thus a direct spectrum.

195?

Proof of Corollary: Suppose we are given the commutative diagram below:

Here

and

are arbitrary continuous mappings. Suppose that Then by the Lemma,

a common . common

-successor

, and

-successor

-ancessor

and and

• Since

have a

and

they have a common

page 178). Consequently

have the common -successor

and

successor (and thus

have

(see have the common

-successor)

We prepare for the proof of Lemma k.2 with a few definitions and results. If

is a map of regular complexes, then for each induces a map

such that the follow-

ing diagram is carmnutative

Here k and k' are subdivision hameomorphisms.

(See VI.3 for

details.) l+.U DEFINITION. A simplicial approximation to f is a simplicial map

such that, for each

196?

g(x) lies

in the closure of the unique open simplex containing f'(x). Since g(x) and f'(x) lie in a closed simplex, there is a x

line segment joining g(x) and f'( ) Thus f h.5.

iEL

s

Sd L for each χ e Sd K.

and g are homotopic.

THEOREM. Let K he a finite regular complex, f: K

> L

a map of K to an arbitrary regular complex L. Then for some integer s > 0 there exists a simplicial approximation 3

g: Sd K

>· Sd L to f.

Proof: We proceed as in the proof of Theorem VI,3·9> except that we choose N

so that the mesh of Sd K

is less than 6/2. Let s be any g

integer larger than N. For each vertex B

of Sd K, there exists

a vertex A of Sd L such that f*(St B) C St A, since the dia­ meter of St B is less than δ. Define a function g from the vertices of Sd K to those of Sd L by choosing, for each B, a vertex A such that f (St-B) C St A, and setting g(B) = A. To show that g extends to a simplicial map of Sd K, to show that

it is sufficient

g maps the vertices of any simplex of Sd K

onto the g

a be a simplex of Sd K,

vertices of some simplex of Sd L. Let with vertices B.,Bn,...,B . Then 0' i ' ' q

erC Π St B., and so - . y J

f'(a)C

This implies that

D St(gB ). 3 3 Π St g(B.) is non-empty. Let τ be any simplex

3

°

whose interior meets Π St g(B.) in a non-empty set. For each j, 3 3 St g(B.) Π τ j= 0, and so g(B.) is a vertex of τ. Thus the vertices (g(B.)) span some face τ' of τ. It follows that g

19L,

extends to a simplicial map of

into Sd L. Let

Then

Thus the unique simplex of Sd L containing in its interior contains each

as a vertex, and so has

as a face. Therefore g(x), which of course lies in contained in the closed simplex containing

is

and so g is a

simplicial approximation to h.6. DEFINITION. Let K and L be simplicial complexes, and let be a simplicial mapping. Then the simplicial mapping cylinder

of the mapping f is the simplicial complex given as

follows. The vertices of those of L.

are the vertices of K together with

(K and L are assumed to be disjoint.) A collection of vertices spans a simplex

of

if the

span a simplex a in K and the

span

the simplex fa. We include all faces of such simplexes together with all simplexes of L. Note that K and L are embedded as subcomplexes in also that

Bote

is not homeomorphic to the mapping cylinder of

defined in V.2. ^.7. LEMMA. Let

denote the inclusions.

Then Proof: Suppose that

Then x lies in some simplex

and fx lies in the simplex

since

map. Thus the simplex

of

line segment from ix to jfx. It follows that

198?

is a simplicial contains the

Proof of Lemma ^.2: Let

be a

simplicial approximation to h. Consider the diagram below, in which k and k' are subdivision homeoanorphisms, and Thus

(1)

We define the complex

as follows. Let

be the simplicial mapping cylinder of the simplicial map is obtained from the disjoint union

by making the following identifications:

Then

is a regular complex.

Furthermore, we have embeddings

respect ively, in

(See picture.)

We want to show that It is clear that and that addition, Lemma U.7 implies that

199

i - jh'. It follows that (since (since (since (since (since We must now define the mapping (a) On

Since

we may use the

hcanotopy between f and gk'h'k to define

on

so

that

(b) On

(Sd K', Sd L ' :

Define a simplicial map

m:

by setting

for A a vertex of

and

Define f'1 on (Sd K', Sd L (c) On

1

for B a vertex of Sd K'. t o be the composition gk'm.

Set

As the reader may verify,

for as defined above is consistent with

the identifications yielding and

By construction,

The proof of Lemma h.2 is complete.

^.8. COROLLAKY.

is cofinal in

and so

is isomorphic to the limit group of the direct spectrum

200?

In the pr