HOMOTOPIES AND DEFORMATION RETRACTS THESIS

A181

HOMOTOPIES AND DEFORMATION

RETRACTS

THESIS

Presented to the Graduate Council of the

University of North Texas in Partial

Fulfillment of the Requirements

For the Degree of

MASTER OF ARTS

By

Villiam D. Stark, B.G.S.

Denton, Texas

December, 1990

Stark, William D., Homotopies and Deformation Retracts.

Master of Arts (Mathematics), December, 1990, 65 pp., 9

illustrations, references, 5 titles.

This paper introduces the background concepts necessary

to develop a detailed proof of a theorem by Ralph H. Fox

which states that two topological spaces are the same

homotopy type if and only if both are deformation retracts

of a third space, the mapping cylinder. The concepts of

homotopy and deformation are introduced in chapter 2, and

retraction and deformation retract are defined in chapter 3.

Chapter 4 develops the idea of the mapping cylinder, and the

proof is completed. Three special cases are examined in

chapter 5.

TABLE OF CONTENTS

Page

LIST OF ILLUSTRATIONS

CHAPTER

1. Introduction . . . . . . . . . . . . . . .

2. Homotopy and Deformation . . . . . . . . .

3. Retraction and Deformation Retract . .

4. The Mapping Cylinder .. . . . .

5. Applications . . . .

REFERENCES . . . .a .. ..a 0. ... 0 ...

iii

. .0 . a. 0. . . . . . . .a . .a iv

1

. 10

. 25

45

65

LIST OF ILLUSTRATIONS

Page

FIGURE

Illustration of a Homotopy ...

Homotopy for Theorem 2.6 . .

The Comb Space ...... .

Transitivity of Homotopy . .

The Mapping Cylinder

Function Diagram for Theorem 4.5

The Homotopy "r"

Special Inverse

Homotopy for Theorem 5.11 .

4

8

18

. . .22

. 26

. . .31

. . .37

. . .50

. . .61

iv

1.

2.

3.

4.

5.

6.

7.

8.

9.

CHAPTER I

INTRODUCTION

The purpose of this paper is to introduce some concepts

which can be used to describe relationships between

topological spaces, specifically homotopy, deformation, and

retraction. These concepts can also be used to characterize

an equivalence relation on the class of topological spaces.

Another topological space called a mapping cylinder is the

tool used. The idea of a mapping cylinder is described by

R. H. Fox in his paper "On Homotopy Type and Deformation

Retracts"', and this article provided the motivation for,

and much of the material in this paper. The theorems and

definitions quoted are from this paper unless otherwise

noted.

The fundamental idea of this paper., that of homotopy,

is introduced in chapter 2. The concept of homeomorphism

can be thought of as a relationship between spaces. Then

the question arises as to whether there is a more general

relationship between two spaces which are not homeomorphic.

One more general categorization of topological spaces,

homotopy type, is described in chapter 3. The purpose of

1R. H. Fox, "On Homotopy Type and Deformation Retracts,"Annals of Mathematics 44 (January 1943): 40-50.

1

2

Fox's paper is to give a proof of his theorem which states

that two spaces belong to the same homotopy type if and only

if they both can be homeomorphically imbedded in a third

space of which both are deformation retracts. The concepts

of deformation and retract are defined in chapter 3. A

generalized method of constructing this third space uses the

concept of a mapping cylinder which is introduced in chapter

4. In chapter 5 some special cases are described. The

concept of a space being inessential relative to a subset is

defined, and several theorems are proved. Special homotopy

and special homotopy inverse are defined, and a relationship

between them is described. Then E-homotopies are defined

and several results are shown.

The major result of this paper is to construct, or

reconstruct in detail, the proofs of the theorems of Fox's

paper using a more precise notation. Additional theorems

have been added to provide the background necessary to

explicate his results.

CHAPTER II

HOMOTOPY AND DEFORMATION

Throughout this paper the terms "map" or "mapping"

refer to a continuous function from one topological space to

another. If there is a question as to whether a map is

continuous, it will be referred to as a function until its

continuity is established. The notation used generally

follows that used in Willard's General Topology.2

Definition 2.1: Mappings f and g from a space A into a

space D are homotopic, written f ~ g, if and only if there

is a mapping from A x [0,1] into D such that for all a E A

6(a,0) = f(a), and 6(a,1) = g(a). The map 6 is a homotopy

between f and g.

From this definition one can see that a homotopy can be

thought of as a collection of maps 6t: A - D for every

t f [0,1]. The notation "t" will refer to a map from A

into D for a particular t e [0,1], and the notation " (a,t)"

will refer to a map from A x [0,1] into D, but it should be

clear that for every t E [0,1] 6t(a) = 6(a,t). It is often

helpful to think of the parameter t as time. Thus for the

homotopy , if 60 = f and 6, = g then the collection of

2Stephen Willard, General Topology (Reading, Mass.:Addison-Wesley Publishing Co., 1970).

3

4

t's for t c [0,1] can be thought of as a kind of

"continuous transformation" of f into g over time.

It is always important to maintain the distinction

between a function and the set which is the image of the

function; however, when a concept is illustrated

geometrically the symbol for a map and the symbol for its

image are sometimes interchanged for the sake of simplicity.

For example, figure 1 is a way to illustrate the homotopy

: A x [0,1] -+ D between the maps f : A -+ D and g : A -4 D.

t=1 g

t0 f

FIGURE 1

On the one hand the rectangle can represent the space

A x [0,1], the domain of 6. However, it can also be seen as

the collection of images of t for t E [0,1]. Thus the base

of the rectangle when t = 0 is the image of 60 = f in D, and

the top of the rectangle is the image of 6, = g. Every

point in the rectangle is the image of a point 6t(a) in D

for some t E [0,1]. The rectangle then becomes an image of

the map ( in D "stretched " over time to show the image of

6 at each time t E [0,1]. If D C A then the figure can be

5

seen both ways with the image of 6 superimposed on its

domain. For greatest clarity both the domain and the range

of-a homotopy can be shown. However, since the domain of a

homotopy is usually clear from the written description, only

its image is usually represented.

Just as 6 can be seen as a collection of maps 6t for t

e [0,1], the homotopy can also be seen as a collection of

maps 6a :[0,1] - D, and this notation will also be useful.

The proof of a simple theorem about homotopies will

illustrate some of the ways homotopies can be used to

generate a new homotopy.

Theorem 2.2: Homotopy is an equivalence relation on

the set of maps from one space to another.

Proof: Let F be the set of maps from a space A into a

space D.

Suppose f c F. f ~ f by the homotopy 6 :A x [0,1] - D

defined by (a,t) = f(a) for every t E [0,1]. Thus homotopy

is reflexive.

Suppose f ~ g by the homotopy 6 : A x {0,1] - D. Then

6(a,0) = f(a) , (a,1) = g(a), and 6(a,t) E D Va E A. Then

by substituting (1 - t) for t, 6(a,1 - 0) = 6(a,1) = g(a),

and 6(a,1 - 1) = 6(a,0) = f(a). Thus g ~ f. Therefore

homotopy is symmetric. The notation "(r)" will be used to

indicate that the homotopy, 6 is reversed by substituting

(1 - t) for t.

6

Suppose f ~ g by the homotopy , and g ~ h by the

homotopy A. Define a map : A x [0,1] - D by:

((at) = E(a,2t) for t E [0,1/2]

A(a,2t - 1) for t c [1/2,1]

Thus C(a,O) = (a,O) = f(a) , C(a,1) = A(a,1) = h(a). Since

C is defined in terms of continuous functions, to show that

C is continuous it is sufficient to show that the definition

is consistent where the different parts overlap. Thus

C(a,1/2) = 6(a,1) = g(a) = A(a,0). So C is continuous and

therefore a homotopy between f and h. Therefore homotopy is

transitive.

Therefore homotopy is an equivalence relation on the

set of maps from one space to another. 0

Definition 2.3: A path in a space A between points a0and a1 is a map p : [0,1] - A such that p(0) = a0 and

p(1) = a1 .

Lemma 2.4: Suppose ( : A x [0,1] - D is a homotopy

between 60 and 61. Then for every a E A there is a path in

D between 6(a,0) and 6(a,1).

Proof: Let a be a member of A. Define a function

Pa :[0,1] -4 D by pa(t) = 6(a,t). Thus p a(0) = (a,0) and

Pa(1) = 6(a,1). Notice that pa =a. Therefore pa is

continuous. 0

7

Definition 2.5: If A C D, then a deformation 6 is a

homotopy between the identity on A, idA, and a map

g : A -4 D. Thus 6 A x [0,1] -4 D where 6(a,O) = a and

6(a,1) = g(a) for all a E A. The set g(A) is called a

deform of A, and A is said to be deformable in D into g(A).

If A = D, one simply says that A can be deformed into g(A).

If g(A) is a singleton set then 6 is a contraction of A.

Theorem 2.6: If A can be deformed into B then a

mapping f of A into A is homotopic to the identity on A if

and only if f restricted to B, written fIB, is homotopic to

the identity on B.

Proof: Suppose A can be deformed into B. Let

f : A - A be any mapping of A into A. Since A can be

deformed into B, then B C A and there is a homotopy 6

between the identity on A and some mapping g : A -4 A where

g(A) C B. Thus 6 : A x [0,1] - A such that 6(a,0) = a and

6(a,1) = g(a) Va E A.

Suppose idB B by the homotopy n. Then

1 :B x [0,1] -1 B such that 17(b,0) = idB(b) = b and

(b,1) = fIB(b) Vb E B. It needs to be shown that

idA ~ f. Define ( :A x [0,1] -+ A by:

((a,t) = IThus the map ((a,0)

((a,1) = f(6

1

2/3

1/3

t =0

6(a,3t) for 0 < t < 1/3

r(6(a,l),3t - 1) for 1/3 < t < 2/3

f(6(a,3 - 3t)) for 2/3 t < 1

= 6(a,O) = idA(a) = a, and the map

(a,0)) = f(a).

f (A)

FIGURE 2

8

f(6(r))

f (B)

f (B)

77

6(A x {1})

g(A)

S

id

9

In order to show that C is continuous it is sufficient

to show that C is consistent at t = 1/3 and t = 2/3, since C

is defined in terms of continuous functions. The map

((a,1/3) = 6(a,3(1/3)) = 6(a,1) = g(a) E B, and g(A) = B.

Also, C(a,1/3) = r(6(a,1),3(1/3) - 1) = n(6(a,1),O) =

n(g(a),O) = idB(g(a)) = g(a). The map C(a,2/3) =

n(6(a,1),3(2/3) - 1) = n(g(a),1) = f|B(g(a)), and

n(g(A),1) = (B),1) = fIB(B) = f(B). And C(a,2/3) =

f (6(a,3-3(2/3))) = f (6(a,1)) = f (g(a)) = fjB(g(a)). Thus C

is consistent at t = 1/3 and t = 2/3 and is therefore

continuous. Therefore C is a homotopy between idA and f(A).

Conversely, suppose f ~. idA by the the homotopy (.

Thus 6 : A x [0,1] -+ A and 6(a,0) = a and 6(a,1) = f(a).

Since B C A, 61B : B x [0,1] -+ A such that (b,0) = b, and

6(b,1) = f(b) = f|B(b) Vb E B. Therefore 61B is a homotopy

between idB and fjIB'

CHAPTER III

RETRACTION AND DEFORMATION RETRACT

The concept of a retraction is related to both

set-theoretic topology and algebraic topology. In

set-theoretic topology there are several generalizations of

Tiezte's Extension Theorem which use this notion.3 In the

following material two of these extension theorems will be

used without proof, and in this paper the use of the concept

will be confined to the set-theoretic applications.

Definition 3.1: A subset B of a space A is a retract

of A if and only if there is a mapping r : A - B such that

rIB = idB. The map r is called a retraction of A into B,

and one says A is retractable into B.

Definition 3.2: A retract D of a space A is a

deformation retract if and only if a retraction r : A - D is

homotopic to idA by some homotopy,6. The retraction r is

then called a deformation retraction and the homotopy S is

called a retracting deformation.

Theorem 3.3: The subset B of a space A is a

deformation retract of A if and only if B is a retract of A,

and A is deformable into B.

3Karol Borsuk, Theory of retracts (Warszawa, Poland:PWN - Polish Scientific Publishers, 1966), 5.

10

11

Proof: Suppose B is a deformation retract of A. Then

by the definition there is a retracting deformation

6 : A x [0,1] - A such that 6(a,0) = a and 6(a,1) = r(a)

for some retraction r : A - A such that r(A) = B. Thus B is

a retract of A, and A is deformable into B.

Conversely, suppose B is a retract of A and A is

deformable into B. Since B is a retract of A, then there is

a map r : A -+ B such that rIB = idB. Then clearly

rIB ~ idB. Since A is deformable into B then by theorem

2.6, r ~ idA. Thus B is a deformation retract of A. o

As an example of a retract which is not a deform,

consider the unit circle,S1, in the plane along with the

point (0,0). Let A = S1 U {(0,0)} with the subspace

topology. The constant map k : A - A such that k(a) = (0,0)

is evidently a retraction of A into {(0,0)},but A is not

deformable into {(0,0)}. Just suppose there was such a

deformation 6 : A x [0,1] -4 A. Then 60 = idA and 61 (A) =

{(0,0)}. By lemma 2.4, for a point x of S there is a path

from 60 (x) = x to 61(x) = (0,0) in A. This is clearly a

contradiction. Therefore A is not deformable into {(0,0)}.

Definition 3.4:4 A closed subset X0 of a Hausdorff

space X is a neighborhood retract in X if and only if Xo is

a retract of some open subset of X containing Xo.

4ibid., 14.

12

Notice that every retract of X is a neighborhood

retract in X, since the space X is a neighborhood of every

subset in X. But not every neighborhood retract is a

retract. Borsuk gives the following example.5 Let Sn-1 be

na sphere in n-dimensional Euclidean space, Rn. The set,

Rn minus the origin, Rn - {O}, is a neighborhood of Sn-1

Define a map r : n - {0} _Sn-1 by r(x) = x/I|xII , where jjxjj

is the distance of x from the origin. Then r is a

retraction of Rn - {O} onto Sn-1. Then Borsuk cites a proof

of the fact that Sn-1 is not a retract of Rn given by

Hirsch. 6 Hirsch's proof relies on theorems from algebraic

topology and is beyond the scope of this paper.

Definition 3.5:7 An absolute neighborhood retract,

abbreviated ANR, is a separable, metrizable space which is a

neighborhood retract of every separable, metrizable space

which contains it as a closed subset.

Thus an ANR is a separable, metrizable space A such

that for every separable, metrizable space E which contains

A as a closed subset, there is an open neighborhood U of A

in E and a map r: U - A such that rA = idA.

5ibid., 12-14.6Morris V. Hirsch, "A Proof of the Nonretractability

of a Cell Onto Its Boundary," Proceedings of the AmericanMathematical Society 14 (1963): 364-365.

7Ralph H. Fox, "A Characterization of AbsoluteNeighborhood Retracts," Bulletin of the American MathematicalSociety 48 (1942): 271-275.

13

Theorem 3.6: If 6 is a deformation of an ANR, A, and B

is the set of fixed points of 61, then there is a homotopy C

between idA and 61 such that the points of B are fixed under

each of the maps 6t for 0 < t < 1.

Proof: Let A be an ANR, : A x [0,1] - A be a

deformation, and B be the fixed points of 6,. Thus 60 = idA

and B C 61 (A) such that Vb E B 6 1(b) = b. Define a function

7 : (A x {0} U B x [0,1] U A x {1}) x [0,1] - A by:

j((a,0),u) = a for 0 < u < 1 and a EA

)(a,t/u)for 0 < t < u <1 and a c B

t a for 0 u <t <1 and a EB

((a,1),u) = 6(a,1) for 0 < u < 1 and a e A

Here the "time" variable for 6 is t and u for 77.

To show that V is continuous it is sufficient to show

that the definition is consistent where the parts overlap.

The first line and the second line of the definition of n

overlap when t = 0 and a E B. Then i((a,0),u) = a for

0 < u < 1 and f(a,0/u) = 6(a,0) = a for 0 = t< u < 1, and

a E B. Thus when t = 0 the definition of i is consistent.

Lines two and three overlap for members of B when 0 < t = u

< 1. Then from line two j(a,t),u) = 6(a,t/u) = 6(a,1) = a,

since a E B and B is the set B of fixed points of 6,. This

14

agrees with line three. Line three and line four overlap

when t = 1 and a E B. By line four n((a,1),u) = 6(a,1)

which equals a if a e B. Thus lines three and four agree.

Therefore n is a homotopy between no and nj.

Notice that for u = 1, when t = 0, nl(a,O) = a = 6(a,O)

Va E A. And n1(a,t) = 6(a,t/1) = 6(a,t) Va E B. When t = 1

n1(a,1) = 6(a,1) for all a E A. Thus V, and 6 agree for

0 t< 1. But 6 is a map from A x [0,1] into A; so 6 is an

extension of nj from the set

{(A x {0}) U (B x [0,1]) U (A x {1})} to the set A x [0,1].

Claim: The set {(A x {0}) U (B x [0,1]) U (A x {1})}

is closed in A x [0,1]. For brevity call this set S.

Since A x {0} and A x {1} are each the product of A

with a closed set,{0} and {1} respectively, they are both

closed in A x [0,1].

Sub claim: The set of fixed points of 61 B is closed

in A.

Suppose x is a limit point of B and x 0 B. Since A is

an ANR, a separable,metrizable space, A is 2nd countable and

thus also 1st countable. Since x is a limit point of B,

there is a sequence of points of B, {bk}, which converge to

x. Since 6 is continuous, 61({bk}) = { l(bk)} is a

sequence in 61 (A) C A which converges to 6 1(x). But 6,(bk)

= bk for every k, since bk is in B. So the sequence {bk}

converges to 61(x). Therefore 61(x) = x. This implies x is

15

a fixed point of (j and x is in B. Therefore B contains all

of its limit points and is closed.

Thus B x [0,1] is closed in A x [0,1]. Therefore the

set S is the finite union of closed sets and is closed in

A x [0,1].

Borsuk's Theorem as generalized by Dowker states "Let C

be a closed subset of a space X and f and g two homotopic

mappings of C in an arbitrary absolute neighborhood retract.

Then if there is an extension F of f over X there is also an

extension G of g over X, with F and G homotopic."8

In our case C = S which is closed in A x [0,1], the

space X in our case. The maps f and g are n and no'

respectively which map S into A, an ANR. Then since there

is an extension of 77 to A x [0,1], namely 6, there is also

an extension of no to A x [0,1]. Call this extension (, and

by the theorem 6 ~ C. Thus C : A x [0,1] - A such that (IS

= no. Now when t = 0, C(a,0) = n 0(a,0) = a Va c A. So C0

= idA. When t = 1 ((a,1) = n 0(a,1) = 6(a,1). So (1 = 61.

Therefore C is a deformation of A into 61 (A). Furthermore,

((bt) = V 0(b,t) = b Vb e B for 0 ( t ( 1. So the points

of B are fixed under each of the maps, Ct for 0 < t < 1. 0

Suppose A and a subset B are ANR sets.

8Witold Hurewicz and Henry Wallman, Dimension Theory(Princeton: Princeton University Press, 1941) 86.

16

Theorem 3.7a: The set B is a deformation retract of A

if and only is there is a retracting deformation 6 of A onto

B such that the points of B are fixed under each tfor

0 < t< 1

Theorem 3.7b: The set B is a deformation retract of A

if and only if there is a deformation 6 of A onto B such

that t(B) C B for 0 < t < 1.

Proof of version a: Suppose there is a retracting

deformation ( of A onto B such that the points of B are

fixed under each 6t for 0 < t < 1. It follows from the

definition that B is a deformation retract of A.

Conversely, suppose B is a deformation retract of A by

the deformation 6 : A x [0,1] -+ A such that 60 = idA and

61(A) = B and 611B = idB. Then B is the set of fixed points

of 6 . By theorem 3.6 there is a homotopy ( such that (t(b)

= idB for 0 < t < 1 and C is a retracting deformation of A

onto B.

Proof of version b: Suppose B is a deformation retract

of A. Then by the paragraph above there is a retracting

deformation C from A onto B such that (t(B) = idB , and

therefore (B(B) C B for 0 < t < 1.

Conversely, suppose there is a deformation 6 of A onto

B such that 6t(B) C B for 0 < t 1. Then 60 = idA and

61(A) = B. So 601B = idB and 6t1B g B for every t. The map

61B is a homotopy between 601B and 61|B' Since B is an ANR

17

and contained in A, by definition it is closed in A. Since

A is an ANR it is metrizable and therefore normal. By the

Borsuk-Kuratowski Theorem which says, "If V is a closed

subset of a normal space,Z and X is an AR-set (ANR-set),

then every continuous map of V into X can be extended to Z

(to a neighborhood of V in Z)".9 Thus 6 01B can be extended

to A. Call this extension r. So r : A - B and rIB = 60=

idB. Therefore r is a retraction of A onto B. By theorem

3.3, B is a deformation retract of A. o

The necessity of the restriction that the space A be an

ANR is illustrated by the following example. Let C be a

subset of R2 , and define C as follows. Let the set

B = {(x,O)j 0 < x < 1}; the set S = {(0,y)j 0 < y 1}; and

the set Xn ={(1/n,y)j 0 < y 1, n is a natural number}.

Let C = B U Xn U S with the subspace topology. C can be

called the "comb space". Notice that C is closed and

bounded and thus compact.

9Borsuk, 5.

18

SY/

0 B 1

FIGURE 3

With a fact about ANR sets it can easily be shown that

C is not an ANR. "Each ANR(M)-space [ANR] is locally

contractible."10 A locally contractible space is a space

which is locally contractible at every point. Locally

contractible at a point x is defined to mean every

neighborhood U of x contains a neighborhood V of x which is

contractible to a point."

Claim: The comb space is not an ANR.

Proof: Just suppose that C is an ANR. The set C is a

closed subset in R2 . Consider a point (0,y) in C. Every

neighborhood of (0,y) contains a set {(1/n,y)l n > k} for

some natural number k. Let U be a neighborhood of (0,y)

which does not contain (0,0). Since C is an ANR, by the

theorem above, let V be any other neighborhood of (0,y)

1 0ibid., 87."ibid., 28.

19

contained in U which is contractible to a point. By lemma

2.4 there is therefore a path in V from every point to the

contraction point. But there cannot be a path to the

contraction point from (0,y) and (1/n,y), since (0,0) 0 V.

Therefore C is not locally contractible and thus not an ANR.

0

Claim: The comb space,C, is deformable into S so that

every point (x,y) is mapped to the point (0,y). Thus the

points in S are mapped to themselves.

Define a map 6 : C x [0,1] -+ C by:

(x,(1-4t)y) for 0 < t < 1/4

(x+(4t-1)(y-x),0) for 1/4 < t < 1/2

((3-4t)y,0) for 1/2 < t < 3/4

(0,(4t-3)y) for 3/4 < t < 1

When t = 0, 6((x,y),0) = (x,(1-0)y) = (x,y). From line

one of the definition 6((x,y),1/4) = (x,(1-1)y) = (x,0), and

from line two 6((x,y),1/4) = (x+(1-1)(y-x),0) = (x,O). From

line two 6((x,y),1/2) = (x+(2-1)(y-x),O) = (y,O), and from

line three 6((x,y),1/2) = ((3-2)y,0) = (y,0). From line

three 6((x,y),3/4) = ((3-3)y,0) = (0,0), and from line four

6((x,y),3/4) = (0,(3-3)y) = (0,0). When t = 1 6((x,y),1) =

(0,(4-3)y) = (0,y). Therefore 6 is a deformation of C into

S. Furthermore, 61I5(Oy) = (0,y); so 611S = idS.

20

Therefore S is a deformation retract of C.

Claim: There is no retracting deformation of C into

S that satisfies the condition that the points of S are

fixed under each t E [0,1].

Proof: Just suppose there is a retracting deformation

C x [0,1] - C such that at'S = ids for all t E [0,1].

Let U be a basic open neighborhood of the point (0,1) E S.

Since by lemma 2.4 (1/n,1)(t) is a path from (1/n,1) to

61(1/n,1), for every natural number n, 6(1/n,1) = (0,0) for

some t E [0,1]. Let to = inf{tj t o (1/n,1) = (0,0), and n

is a natural number}. Then inf{d(6to(1/n,1),(0,0))I n is a

natural number} = 0. Let U be a neighborhood of 6to(0,1) =

(0,1) with radius < 1/2. Let V be any neighborhood of (0,1)

such that Cto(V) U. The set V contains a set

{(1/n,1)I n > k} for some natural number k. Since

inf{d(6to(1/n,1),(0,0))} = 0, there is a natural number n

such that d(6to(1/n,1),(0,0)) < 1/2 then d(6to(1/n,1),(0,1))

> 1/2 and 6to (l/n,l) to(V). Thus 6to is not continuous.

This contradicts the supposition that 6 is a retracting

deformation. Therefore there is no retracting deformation

of C into S such that the points of S remain fixed for all

t E [0,1].

Therefore the restriction in theorem 3.7a that the

space A be an ANR is necessary and by the proof it is

sufficient.

21

If P is the set {(0,1)}, then C can be deformed into P

by adding a retracting deformation of S into P to the

retracting deformation 6 in the previous example. Then P is

a deformation retract of C. But by a similar argument used

in the first example there is no retracting deformation

: C -+ P satisfying the condition in theorem 3.7b that

t(P) C P for all t E [0,1]. Thus the restriction that the

space A be an ANR is also necessary in theorem 3.7b.

Definition 3.8: Two spaces A and B are said to belong

to the same homotopy type if and only if there are maps

f A -B and g : B - A such that fg = idA and gf = idB.

Claim: Homotopy type is an equivalence relation on the

class of topological spaces.

Proof: For any space A define f and g to be idA*

Since idA composed with idA equals idA, idAidA ~ idA. Thus

A is the same homotopy type as itself, and homotopy type is

reflexive.

Suppose the space A is the same homotopy type as the

space B. By the definition B is also the same homotopy type

as A. So homotopy type is a symmetric relation.

Suppose that the space A is the same homotopy type as

the space B, and B is the same homotopy type as the space C.

Then there are maps f : A -4 B and g : B - A such that

fg ~ idB and gf ~ idA. Also there are maps j : B -4 C and

k : C -4 B such that jk ~ idC and kj ~ idB'

22

fg

a

B

gf

OE

A

jk

7

-C

kj

6

B

FIGURE 4

Notice that the map gkjf A - A and jfgk C -4 C.

Define the map A A x [0,1] - A by:

A I(a,2t) for 0 ( t < 1/2

g(6(f(a),2t-1)) for 1/2 ( t ( 1

When t = 1/2, A (a,1/2) = 0(a,1) = gf (a) , and A (a,1/2) =

g(6(f(a),0) = g(f(a)) = gf(a). So A is well defined.

Also, A(a,0) = 8(a,0). Thus A0 =3 = idA. When t = 1

A(a,1) = g(6(f(a),1)) = g(kj(f(a))) = gkjf(a). therefore A

is a homotopy between idA and gkjf.

Define the map p :C x [0,1] - C by:

p (c,2s) for 0 < s < 1/2p(c,s) =ko

,j(a(k(c),2t-1) for 1/2 < s < 1

23

When t = 1/2 p(c,1/2) = 7(c,1) = jk(c), and p(c,1/2) =

j(a(k(c),O)) = j(k(c)) = jk(c). Thus p is well defined.

Now when t = 0 p(c,0) = 7(c,O); so p0 =70 = idC. When

t = 1 p(ci) = j(a(k(c),1) = j(fg(k(c))) = jfgk(c).

Therefore idC ~ jfgk by the homotopy p. Since jf : A -+ C

and gk : C - A, C and A are the same homotopy type.

Therefore homotopy type is a transitive relation, and

thus homotopy type is an equivalence relation on the class

of topological spaces.

Notice, in the special case when g = f- that ff'-

f- f = idB. Then ff~1(B) = B which implies that f is an

onto function. Since ff~t (a) = a for all a e A, f-is a

one-to-one function. And since f and f~ are both

continuous, f is a homeomorphism between A and B.

Thus the partition of the class of topological spaces

resulting from homeomorphism is a refinement of the

partition resulting from homotopy type.

Definition 3.9 If maps f : A - B and g : B - A are

such that gf ~ idA, then g is a left homotopy inverse of f,

and f is a right homotopy inverse of g. A two-sided

homotopy inverse of f is a map g which is both a left and

right homotopy inverse of f.

Therefore the spaces A and B are the same homotopy type

if and only if there is a map f from A into B that has a

two-sided homotopy inverse.

24

Theorem 3.10 Suppose A and B are topological spaces.

If a map f : A - B has both a right and a left homotopy

inverse, then f has a two-sided homotopy inverse.12

Proof: Let L and R be a left and a right homotopy

inverse of f, respectively. Define g = LfR. Then

fg = fLf. Since L is a left homotopy inverse of f,

Lf ~ idA. So fg fidAR = fR. Since R is a right homotopy

inverse of f, fg ~ fR ~ idB. Thus fg ~ idB, and g is a

right homotopy inverse of f.

Similarly gf = LfRf ~ LidBf = Lf ~ idA. Thus g is also

a right homotopy inverse of f, and therefore g is a

two-sided homotopy inverse of f.

12Ralph H. Fox, : "On Homotopy Type and DeformationRetracts," Annals of Mathematics 44 (January 1943):43. Theauthor credits M. M. Day with the idea for this proof.

CHAPTER IV

THE MAPPING CYLINDER

Let M and N be topological spaces and a : M - N be any

map. Without loss of generality one can assume that M f N

is empty. Let M x [0,1], be a product space with the

product topology.

Definition 4.1: The cylinder, Ca is the set:

{{(m,t)}Im E M, 0 < t < 1} U

{(a~ (a(m)) x {1}) U {a(m)} Im E M}.

Notice that Cais a partition of (M x [0,1]) U a(M)

where for every t < 1 the equivalence classes are all

singleton sets, {(m,t)}. When t = 1 for each m E M the

equivalence class is the set a~ (a(m) x {1} with the

singleton set {a(m)} attached. Let N' = {{n}jn E N - a(M)}.

Give M x [0,1] U N the disjoint union topology.

Define a map, i : ((M x [0,1]) U N) - (Ca U N') by:

{(mt)} if 0 < t < 1

i(m,t) = I (- ) i0t~(a (a(m) x {1}) U {a(m)} if t = 1

and =1n{n}1 if n E N-a(M)

(a~ (n) x {1}) U {n} if n e a(M)

Notice that if a(m) = n then i(m,1) = i(n). Thus i is

25

26

well-defined and onto the set Ca U N'; so i-'(Ca U N') is a

partition of (M x [0,1]) U N. Give Ca U N' the quotient

topology, Qi, induced by i. Then i is a quotient map, and

A C (Ca U N')is a member of Q i if and only if

i~ (A)nf (M x [0,1]) and i~ (A) f N are both open because

A x [0,1] U N has the disjoint union topology.

Definition 4.2: Let N + Ca symbolize the quotient

space,(C. U N',Qi) and call it the mapping cylinder of a.

a(m)N

0

FIGURE 5

The symbol "<m,t>" will denote the equivalence class

i(m,t) for m c M and 0 < t < 1. So the equivalence class

i(n) = <n> for n E N. Notice that if a(m) = n then <m,1> =

27

<n>. Thus N + Ca = {<m,t>I m f M,0 < t < 1} U {<n> fn E N}.

Let N denote the set {<n>In E N} =

{(a~ (a(m)) U {a(m)}j me M} U N'. Thus N Q N + Ca, and N

is the "top" of the mapping cylinder. Let Mo denote the set

{<m,0>| m E M}. So MO C N + Ca , and MO is the "base" of

the mapping cylinder. For each t e [0,1], the notation

1<M x {t}>" will denote the set {<m,t>I m e M}. Hereafter,

the symbol "t" will be reserved for the elements in the

interval [0,1] used to define a mapping cylinder and some

other letter, s or u, will be used for the elements of other

images of the unit interval. The set Ca , a subset of

N + Ca , can be described several ways: Ca = {{(mt)}|m e M,

0 t < 1} U {(a~1(a(m)) x {1}) U {a(m)} m e M}

= {<m,t>I m e M, 0 < t < 1} U {<n>| n e a(M)}

= {<mt> m e M, 0 < t < 1}.

Notice that il(M x [0,1]) U a(M) is the projection

function of the decomposition of (M x[0,1]) U a(M) since

i(a(m)) = <a(m)> and i(m,t) = <m,t> for all m e M and

0 < t < 1. However, i|M x [0,1] is not the projection

function of the decomposition of M x [0,1] since <m,1>

contains the element {a(m)} e N as well as the set

a~ (a(m)) g M x [0,1].

Lemma 4.3: Each map i1M x {t} for 0 < t < 1 is a

homeomorphism from M x {t} onto <M x {t}>.

Proof: Let t e [0,1). The image of i 1M x {t} is the

28

set <M x {t}> = {<m,t>I m E M} = {{(m,t)}j m E M}. By the

definition of i, iM x {t}(m<t) = m,t>. thus ZiM x {t} is

one-to-one and onto.

Let U be open in <M x{t}>. Then Uc, the complement of

U in <M x {t}>, is closed. Thus there is a closed set V in

Ca such that Uc = v la <Mx {t}>. Then Vc is open in C., and

Vc fM x {t}>=U. So i-1(U) = i-(Vcfa<M x {t}> =

i-(vc) fl i-(<M x {t}>). But i- (<M x {t}>) = M x {t}.

Since i is continuous i l(Vc) is open in M x [0,1]. Thus

i- - i-(vc) l (M x {t}) is open in M x {t}. Since U C

<M x {t}>, zi (U) = i-1|M x {t}(U). Since i is one-to-one,

1 <M x {t} =IM x {t}, SOiM x {t'(U) =

which is open in M x {t}. Thus flM x {t} is continuous.

Now let U be open in M x {t}. Then Uc is closed. Then

there is a closed set V in M x [0,1] such that V fl Mx {t} =

Uc, and U = Vc f M x {t}. Since iJM x [0,1] U a(M) is the

projection map for Ca 'M x [0,1] U a(M)(Vc) is open in Ca'

Since Vc C M x [0,1], iM x [0,1] U a(M)(c) =

tIM x [0 ,1](V) . so IM x [0 ,1](U) =

IM x [0. 1](Vc n M x {t}) = iM x [0 ,1](Vc) n

IM x L0,1](M x {t}) = iM x[0,1 ](Vc) a <M x {t}> which is

open in <M x {t}>. Since U C M x {t}, tiM x [0 1](U) =

IM x {t}(U) So IM x{t}(U) is open. Thus iJM x{t} is

an open map. Therefore iJM x {t} is a homeomorphism. o

The set Ca-._N={<m,t>I m e M, 0< t < 1} is also a

29

subset of N + Ca* Observe that since i(mt) = <m,t> for

every m E M and t E [0,1), then ill x [0,1) is continuous,

one-to-one, and onto. If U is an open set in M x [0,1),

then ijm x [0,1)(U) = {<m,t>I (mt) E U}, and

i| 1x [0, 1 )({<mt>I (m,t) E U}) = U. So i|m x [0,1) is an

open function. Therefore ilj x [0,1) is a homeomorphism

from M x [0,1) onto Ca-N.

Notice that Ca-N is the complement of N in N + C a* Now

i~ 1(C -N) = {(m,t)j n e M, 0 < t < 1} = M x [0,1) which is

open in M x [0,1] U N. Therefore Ca-N is open and N is

closed in N + Ca'

Define a map, h: X -+ X x {0} for any space X by

h(x) = (x,0) Vx e X where X x {0} is contained in the

product space, X x [0,1] and has the subspace topology.

Thus h is a homeomorphism between X and X x {0}. Then ,for

example, iMoh is a homeomorphism between M and Mo.

Lemma 4.4: The map 'IN is a homeomorphism from N ontoA

N.

Proof: The map 'N maps N into N by:

f<n> if n EN - a(M)

<a= (n) x {1} U {n}> if n e a(M)

Since each n is mapped onto a singleton set containing

itself unioned with a~1(n) if n e a(M), then 'IN is

30

one-to-one and onto its image.

Let U be an open subset of N. Since i is continuous,

i~ (U) is open in N U (M x [0,1]). Since N and M are

disjoint, ii(U) fl N is open in N. But i~ (U) l N =

ZIN (U); so 'N is continuous.

Now suppose U is an open set in N. Then iIN(U)

{<n>j n E U}. i~ ({n> n E U} = {a (a(m) x {1}j a(m) c U}

U {nj n E U} = (a-1 (U) x {1}) U U. Since a is continuous,

a~ (U) x [0,1] is open in M x [0,1] ; so a- (U) x {1} is open

in M x {1}. Thus iIN(U) is open in the image ofilN.

Therefore 'IN is an open function.

Therefore 'IN is a homeomorphism.10

Theorem 4.5: Let f X -4 Y and 6 X - Z be maps

defined for topological spaces X,Y, and Z. Then there is a

map g Y - Z such that 0 ~ gf if and only if Oh-2i~ |X0 can

be extended to Y + Cf.

31

Y + C

fT /

0 Oh-1 i-1

~X1z/

FIGURE 6

Proof: Suppose/0 is an extension of xh- i~1 X0 to

Y + C f. Let g= Oily. Define 6 : X x [0,1] -+ Z by 6(x,t)

=0 (L(x,t)) for 0 < t < 1. Since 6 is the composition of

continuous functions, it is continuous. 6(x,O) = 0 (L(x,0))

= * (<x,0>) = h-1 i-1|XO(<x,O>) = O(x) . and 6(x,1)=

0 (i(x,1))0 (<x,>) = 0 (<f(x)>) = gf(x). Therefore 0~

gf by the homotopy, 6.

Conversely, suppose there is a map g : Y Z such that

Y ~ gf by the homotopy 6. Then : X x [0,1] -+ Z such that

0(x,) = 0(x) and (x,) = gf(x). Define d :Y + C -+ Z by

32

0 (<x,t>) = 6t(x) for <x,t> E Cf, and 0*(<y>) = g(y) for <y>

EY.

Notice that if y = f(x), then 0((<y>) = g(y) = gf(x) =* *

6(x,l) = 0 (<x,1>). Therefore 0 is well defined.

Let V be an open subset of Z.. The set i-10*~1 (V) =

{(xt)I ((x,t) E V} U {y1 g(y) E V} = 1(V) U g'(V).

Since both 6 and g are continuous, i~ 20 (V) is open in

Y U (X x [0,1]). So by the definition of the topology on* 1 *Y + Cf, 0 ~(V) is open in Y + Cf. Therefore 0 is

continuous.

Now 0*XO(<x,O>) = 6(x,0) = 0(x) =0h' i-1 Xo(<x,0>).

Therefore 0 is an extension of h-'i'I|X from Xo to

Y + Cf .

Theorem 4.6: Let X,Y, and Z be spaces and 0 : X - Z as

in the previous theorem. Suppose g is a map from Y into Z.

Then there is a map f : X - Y such that gf ~ 0 if and only

if iO is homotopic in Z + C to a map, (i0'), from X into

YO. Note that 0' must be a map from X into Y.

Proof: Suppose i0 ~ iO' in Z + C by

SXx [0,1] - Z + C such that C(x,0) = iO(x) EcZ C Z + C

and C(x,1) = iO'(x) E YO Z + C . Define a map

w : Yo x [0,1] - Z + C by W(<y,0>,s) = <y,s> for 0 < s < 1

So w(<y,0>,0) = <y,0> or wo = idy, and w(<y,o>,1) = <y,1>.

Now each <y,1> E Z and for each y E Y, g(y) e <y,1>; so

<y,1> = <g(y)>. Thus w1 = ig. Therefore w is a homotopy

33

between idy and ig in Z + Cg. Since wo = idyo, then

woiO' =iO'. Consider w1iO' X - Z + C . Notice that

wiO"' (x) = wi(<O'(x),0>) since 6'(x) E Y. And wi(<z0'(x),0>)

= igO' (x) . Thus wiiO' = igO' . Also f or 0 ( s < 1

Ws(iO'(x)) = ws(<O'(x),0>) = <'(x),s>. Define

F : X x [0,1] -+ Z + Cg by F(x,s) = w siO'(x). Then F(x,O) =

w0iO'(x) = iO'(x) e Yo, and F(x,1) = wiO'(x) = igO'(x) E Z.

So F is a homotopy between iG' and igO' in Z + C .g

Claim: The map iG is homotopic to igO'.

Define : x [0,1] -+ Z + C by:

xs) =f((x,2s) for 0 < s ( 1/2

F(x,2s - 1) for1/2 < s < 1

Then 6(x,0) = ((x,0) = iO(x) , and (x,1) = F(x,1) =wii9'(x)

= igO'(x). Since 6(x,1/2) = ((x,1) = iO'(x) and also

6(x,1/2) = F(x,0) = woi9'(x) = iG' x), 6 is continuous and

is a homotopy between iG and ig9' in Z + C .

Claim: The map 0 is homotopic to gO'.

Define 6 : Z + C -+ Z by 6(<z>) = z and 6(<y,t>) = g(y)

Let U be open in Z. The set i-16-1 (U)

= i~'({<zz> z E U} U {<y,t>j y e g~1(U), 0 < t < 1})= {zI Z E U} U {(Y,1)I y E g1(U)} U

{(y,t) ) y ( g(U) 0 t< 1}= U U { (y, t)I Y C g~1 (U),) 0 < t <I 1} = U U (g~1 (U) x [0,21] ).

34

Since g is continuous, g~1 (U) is open, and thus 61 (U) is

the union of open sets in Z + C . Therefore 6 is

continuous.

Then the composition 66 : X x [0,1] - Z is continuous.

Also 66(x,0) = 6(iO(x)) = 6<0(x)> = 0(x), and

66(x,1) = 6(ig0'(x)) = 6(<g0'(x)>) = gO'(x). So 6 is a

homotopy between 0 and gO'. Let f = 0'. Therefore 0 ~.. gf.

Conversely, suppose there is a map f : X - Y such that

gf ~ 0 by some homotopy. Reverse that homotopy to define

6 : X x [0,1] -4 Z such that 6(x,0) = 0(x) and 6(x,1) = gf(x)

Define w : Yo x [0,1] - Z + C by w(<y,O>,s) = <y,s>. So

o = idYO and w1 = ig. Now the map, i6 : X x [0,1] -4 Z + Cg

is continuous. Also, i6(x,0) = i(0(x)) = <0(x)> E Z, and

i6(x,1) = igf(x) = <gf(x)>. Recall the map, h :Y - Y x[0,1]

such that h(y) = (y,0) is a homeomorphism. Thus igf =

wlihf. Since w Yo x [0,1] - Z + C and ih(f (x)) E Yo, then

woihf ~ wihf. But woihf = idYOihf = ihf : X - Yo. Define

A : X x [0,1] - Z + C by:

f=i6(x,2s) for 0 <s < 1/2

,x=w(ihf(x),2 - 2s) for 1/2 < s < 1

Then A(x,0) = i6(x,0) = iO(x) , and A(x,1) = w(ihf(x),0)

= ihf(x). When s = 1/2, A(x,1/2) = i6(x,1) = igf(x), and

A(x,1/2) = w(ihf(x),1) = igf(x). Therefore iO ~ ihf by the

35

homotopy A, and ihf : X -+ YO. 0

If the space Z in theorem 4.5 equals the space X, and 0

is the identity on X then the theorem becomes:

Theorem 4.5a: If f : X - Y, then there is a map

g : Y - X such that gf ~ idx if and only if idxh-'i-1 Xl can

be extended-to Y + Cf.

This gives a necessary and sufficient condition for X to be

the homeomorphic image of a retract of the mapping cylinder.

Corollary 4.7: Suppose f is a map from a space X into

a space Y. The set Xo is a retract of Y + C if and only if

the map f has a left homotopy inverse

Proof: Suppose Xo is a retract of Y + Cf. Then there

is a map r : Y + Cf -4 Xo such that rI. = id. Thus r is an

extension of idX from Xo to Y + Cf, and h1i~ I 0rIX 0 =

h1 i~ 1I|idXo = h~ = idXh' i17'XO. So h1i-Ir is

an extension of idXh'1i'I XO to Y + Cf. Then by theorem

4.5a there is a map g : Y - X such that gf ~ idX , and g is

a left homotopy inverse of f.

Conversely, suppose f has a left homotopy inverse, say

g. So g : Y -4 X such that gf ~. idX. Then by theorem 4.5a

id i~IX0can be extended to Y + Cf; call the extension r

Then r : Y + Cf -+ X where r I = idxh~Ii'IX Thus r is a

retraction and Xo is a retract of Y + Cf. 0

Lemma 4.8: Any mapping cylinder can be deformed into

its top.

36

Proof: Suppose f maps X into Y. Define a map

Y + Cf x [0,1] -+ Y + Cf by r(<x,t>,s) =

i(xt + s(i - t)) for all <x,t> E Cf and r(<y>,s) = <y> for

all <y> E Y. If f(x) = y then r(f(x)) = r(<x,1>,s) =

i(x, 1+ s(1 - 1)) = i(x,i) = <x,1> = <f(x)> = <y> =

r(<y>,s). Thus r is well defined. Also r(<x,t>,0) =

i(x,t + 0(1 - t)) = i(xt) = <x,t>, and r(<y>,0) = <y>. So

0= id Y + Cf Finally, 7(<x,t>,1) = i(x,t + 1(1 - t) -

i(x,1) = <x,1> E Y , and r(<y>,1) = <y> Thus r1(Y + Cf) = Y.

It remains to show that r is continuous.

Let U be open in Y + Cf.

Claim: Every point in r~1(U) is an interior point.

Let (<xo0 ,t0>,s0 ) E r'(U). Then r(<x 0,t0>,s0) ~~

<X0,to + s0 (1 - to> = Ox0,t'>, where t' = t0 + so(' - to)*

Case 1: Suppose t 0 < 1 and s0 < 16

Claim: The point t' E [0,1] is less than 1.

Just suppose t' =1= t0 +s 0 (1 - t0). If to # 0, then

s t0 1. But this is not the case. If t0 =0, then

1 = 0 + so(' - 0) and s0 = 1. Again, this contradicts the

hypothesis. Thus t' # 1; so t' < 1.

Y + C 2

1o . .

r- (<x,t'>)

;7x

7+C

<x00,t- - -- - - wn 4000 G

1 p2

FIGURE 7

Let V C U such that V = i(G x (p1,p2)) where G is a

basic open neighborhood of x0 in X, and (p1,p2) is a basic

open neighborhood of t' such that p2 < 1. Since jX x [0,1)

is a homeomorphism and Cf-Y is open in Y + Cf , then V is an

open neighborhood of <x0 ,t'> in Y + C,.

[o 3_1]

37

(<x 0 t >s -I

0

40~1000

de -

-4IdoI

. ~I-~---~ &

s

0

1 t 0

p10

I

eoeollo

38

Now r-1(G x {p2 }) = {(<x,t>,s)I x E G, t+s(1-t) = p2}.tQ + p

Choose t2 = -to+P . Thus t0 P 2 < 1. Since V is

open in Y + Cf, there is a p' such that t' < p' <p2 and

<xoP'>e U. So choose s2 such that so < s2 and

r(<x0 ,t2 >,s2) = <x0,p'> E V. Also r (G x {p1}) =

{(<x,t>,S) x E 0 ,t+s(1-t) = p1} so choose s I< s0 and

t < t0 such that r(<x0 ,t1 >,sl) = <x,p'> where p1 < p' < t'.

Let V = {(<x,t>,s)j x E G; s < S< <S2; t1 < t < t2}'

Then V = {<x,t> x c ox' t < t < t2} x (s,,s2) , or

i(G x (t1 ,t2)) x (s1,s2). Thus V is open in Y + Cf x [0,1],

and contains (<x0 ,t0 >,s 0).

Let (<x,t>,s) be a member of V. Then r(<x,t>,s)) =

<x,t+s(1-t)>. But x E G and t1 +s1(1-t) < t+s(1-t) <

t2 +S2(1-t2);so r(<x,t>,s) E V C U. Therefore V Q r 1 (U).

Case 2: Suppose t0 = 1. Then r(<x0 ,1>,s) = <x0 '1> for

every s E [0,1]. Then let V C Y + Cf such that V =

i(G x (p1,1]) where G is a basic open neighborhood of x0 in

X, and (p1l] is a basic open neighborhood of 1. Then V is

an open neighborhood of <xo 0 ,1> in Y + Cf. By the same

argument as above every point in r~1(U) is an interior

point.

Case 3: Suppose s0 = 1. Then t' = t + 1(1 - t) = 1

for every t E [0,1]. Then similarly every point in r-(U)

is an interior point.

39

Thus r~1(U) is open, and r is continuous.

Therefore r is a deformation of Y + Cf into Y. o

Hereafter the symbol "-r" will be used to denote this

deformation in any mapping cylinder.

When the space X is the space Z and the map 0 is the

identity on Y, theorem 4.6 becomes:

Theorem 4.6a: If g is a map from Y into Z, then there

is a map f : Z - Y such that gf ~ idz if and only if iidz is

homotopic in Z + C to a map from Z into Yo.

Using this theorem the following corollary yields a

necessary and sufficient condition for the mapping cylinder

to be deformable into its base.

Corollary 4.9: Let Y and Z be spaces and g : Y - Z be

any map. Then Z + C can be deformed into Yo if and only if

the map g has a right homotopy inverse.

Proof: Suppose Z + C can be deformed into Yo. Then

there is a deformation 6 : Z + C x [0,1] - Z + C such that

6(<y,t>,0) = <y,t>, and 6(<z>,0) = <z>. Thus 60 = idz + C g

Also 6(Z + C x {1}) = YO. For every s E [0,1] , the map

6s ilZ is the composition of continuous maps, and 6s maps

ij(Z), a subset of Z + C , into Z + C . Thus

6| : ilz(Z) x [0,1] -4Z + C such that 61i(ilz(Z) x {0}) =

61j(Z x {0}) = id% , and 61i(ilZ(Z) x {1}) = 61j(Z x {1}) Q

Yo. Thus 6bj is a homotopy between the map idz = iIgidZ =

40

iidz and the map 611ji1Z which maps Z into Yo. Therefore by

theorem 4.6a, there is a map f : Z - Y such that gf ~ idz;

so g has a right homotopy inverse.

Conversely, suppose g has a right homotopy inverse,

say f. Then by theorem 4.6a, iidz is homotopic to a map,

say iG', which maps Z into Yo. Call this homotopy (. Thus

: i(Z) x [0,1] -. Z + C such that ((i(Z) x {0}) = iidz(Z)Ag

Z , and ((i(Z) x {1} = iO'(Z) = Yo. By lemma 4.8, Z + Cg

can be deformed into Z by the deformation r where

70 =idZ + C and r(Z + C ) = Z. Define

6: Z + C x [0,1] -. Z + C by:

6(cyt>,s)=Ir(<y,t>,2s) for 0 ( s < 1/2

((r1 (<y,t>,2s - 1)) for 1/2 < s I 1

and

r(<z>,2s) for 0 < s < 1/2

((r1 (<z>),2s - 1) for 1/2 < s ( 1

Notice that 6(<y,t>,1/2) = r(<y,t>,1) = <y,1>, and

6(<y,t>,1/2) = ((r1(<y,t>),2(1/2) -1) = ((r (<y,t>),0) =

((<y,1>,0) = <y,1>. Also 6(<z>,1/2) = r(<z>,1) = <z>, and

6(<z>,1/2) = ((rl<z>,0) = ((<z>,0) = <z>. Clearly, when

g(y) = z then 6(<z>,s) = 6(<g(y)>,s) = 6(y,1>,0. So 6 is

well defined. Then 6(<y,t>,0) = r(<y,t>,0) = r0(<y,t>) =

<y,t>, and 6(<z>,0) =r(<z>,0) = <z>. In other words, 60

41

idz + C . Also 6(<y,t>,1) = ((r (<y,t>),2(1) - 1) =

((<y,1>,1) by the definition of r. So 6(<y,t>,1) =

((<Y,1>,1) = ((<g(y)>,1) =((t(g(y),l) = to'(g(y)) E Yo.

Finally, 6(<z>,1) = ((7-1(<z>),2(1) - 1) = ((<z>,1) again, by

the definition of r. So 6(<z>,1) = ((<z>,1) = ((i(z),1) =

i0'(z) E Yo. Thus 61 (Z + C ) C YO.

Therefore 6 is a deformation of Z + C into Yo. 0

Theorem 4.10: There is a map f : X - Y such that X0 is

a retract of Y + Cf if and only if there is a map g : Y -4 X

such that X + C can be deformed into YO.

Proof: Suppose there is a map f : X - Y such that X0

is a retract of Y + Cf. By corollary 4.7, f has a left

homotopy inverse, say g. Then g : Y -a X such that gf ~ idX.

Thus g has a right homotopy inverse. Then by corollary 4.9

X + C can be deformed into Yo.

Conversely, suppose there is a map g : Y -i X such that

X + C can be deformed into Yo. By corollary 4.9, g has a

right homotopy inverse, say f. Then f : X - Y such that

gf ~ idx. Thus f has a left homotopy inverse. Then by

corollary 4.7, Xo is a retract of Y + Cf. 0

Theorem 4.11: Let X, Y, and Z be topological spaces

and 0 be a map from X into Z. There is a map f : X -4 Y such

that the map Oh~i|j1 which maps Xo into Z can be extended

to Y + Cf, the mapping cylinder of f, if and only if there

42

is a map g : Y -. Z such that iG is homotopic in Z + Cg, the

mapping cylinder of g, to a map from X into Yo.

Recall that X0 is the base of the mapping cylinder of f

and Yo is the base of the mapping cylinder of g. The map i

is the projection map for a mapping cylinder. Which

cylinder should be clear from the context. The map h is the

homeomorphism from any space M to M x {Q}. Its particular

domain and range should also be clear from the context.

Proof: Suppose there is a map f : X - Y such that

9h' i~ IXO can be extended to Y + Cf. By theorem 4.5 there

is a map g : Y -+ Z such that gf ~ 0. Then by theorem 4.6,

iG is homotopic in Z + C to a map from X to Yo.

Conversely, suppose there is a map g : Y -4 Z such that

iO is homotopic to a map from X to Yo. By theorem 4.6 there

is a map f : X -+ Y such that fg ~ 0. Then by theorem 4.5

0h' i_1'XO can be extended to Y + Cf.

Theorem 4.12: Suppose X and Y are topological spaces,

and f is a map from X into Y. Then X0 is a deformation

retract of Y + Cf if and only if f has a two-sided homotopy

inverse.

Proof: Suppose X0 is a deformation retract of Y + Cf.

By theorem 3.3, X0 is a retract of Y + Cf, and Y + Cf can be

deformed into X0. By corollary 4.7, since X0 is a retract

of Y + Cf, f has a left homotopy inverse. By corollary 4.9,

since Y + Cf can be deformed into Xo, f has a right homotopy

43

inverse. By theorem 3.10, since f has both left and right

homotopy inverses, f has a two-sided inverse.

Suppose, conversely, that f has a two-sided homotopy

inverse. Since f a left homotopy inverse, by corollary 4.7

Xo is a retract of Y + Cf. And, since f has a right

homotopy inverse, by corollary 4.9, Y + Cf can be deformed

into Xo. Then by theorem 3.3, Xo is a deformation retract

of Y + Cf

Theorem 4.13: Two topological spaces belong to the

same homotopy type if and only if they both can be embedded

in a third space, V, in such a way that their images are

both deformation retracts of V.

Proof: Let X and Y be two topological spaces. Suppose

X and Y belong to the same homotopy type. As a consequence

of the definition of homotopy type, there is a map f : X - Y

such that f has a two-sided homotopy inverse. Then by

corollaries 4.7 and 4.9, Xo is a deformation retract of

Y + Cf. Notice that Y is a retract of Y + Cf by the map

r :Y + Cf -+ Y where r(<x,t>) = <x,1> and r(<y>) = <y>. By

lemma 4.8, Y + Cf can be deformed into Y. Thus by

theorem 3.3, Y is a deformation retract of Y + Cf.

Therefore X and Y are homeomorphically embedded in Y + Cf

and both are deformation retracts of Y + Cf.

Conversely, suppose X and Y are embedded in a space V

such that the homeomorphic images of X and Y are both

44

deformation retracts of V. Then there is a deformation

6 : V x [0,1] -+ V such that 6(w,0) = w for all w E V, and

6(V x {1}) = Xo, and 611X" = idX. Notice also that

61|y : YO - Xo. There is also a deformation

A : V x [0,1] -+ V such that A(w,0) = w for all w E V, and

A (V x {1}) = Yo, and A 1 yO = idyO. Similarly,

A11O : XO YO

Define the homotopy, : V x [0,1] -4 V by

(wt) = A(6(w,t),t). Then 6(w,0) = A(6(w,0),0) =

A(w,0) = w. That is, 60 = idw. Also, 6(w,1) = A(6(w,1),1)

= A(61(w),1) = A1 61(w). In other words, 61 = A1 61 =

A IX01. Thus idV A i I X061 by the homotopy (. Therefore

idy0 = idiy0 = 1 1x061yo.

Similarly, define the homotopy ( V x [0,1] - V by

((wt) = 6(A(w,t),t). Then ((w,0) = 6(A(w,0),0) =

6(w,0) = w; so (0 = id- . Also ((w,1) = 6(A(w,1),1) =

6(Al(w),1) = 61 A,(w); so (1 = 611y A1. Thus idy = 61Iy0Ai

and idxo =611YOA 11xO. Therefore the map, 6iiy has a

two-sided homotopy inverse, namely, AiXo. By the

definition Xo and Yo are the same homotopy type, and thus

their homeomorphic images, X and Y, are the same homotopy

type.

CHAPTER V

APPLICATIONS

Definition 5.1: A space X is said to be inessential

relative to a subset B if and only if there is a deformation

of X into a proper subset D such that the points of B remain

fixed during the deformation.

Theorem 5.2: Let f : X -+ Y such that Y + Cf is an ANR,

and suppose f has a right homotopy inverse. If V is a

proper subset of Y + Cf containing X0 and k : Y -+ V such

that kfh- 1 X1 x ~ idyIl 0 , then Y + Cf is inessential

relative to X0.

Proof: Letf : X -+ Ybe a map such that Y + Cf is an

ANR and f has a right homotopy inverse.

Suppose V is a proper subset of Y +Cf which contains

X0 and k : Y -+ V is a map such that kfh'1i~I ~id1X'

Notice that by identifying V with Z in theorem 4.5 and

o with idX ih : X -+ V then k can play the role of g. Then

theorem 4.5 says: Given spaces X, Y, and V; the map

f: X -+ Y; and the map idX ih : X -+ V; then there is a map

k : Y -+ V such that idxoih ~ kf if and only if

(idxih)h- i- |XO = idX 0 can be extended to Y + Cf.

Since ijX0 and h are homeomorphisms, kfhFii 1

idVIX0 implies that kf ~ idVlXoih. Thus by the above

45

46

version of theorem 4.5, idyI = idX: Xo -4 Xo C V can be

extended to Y + Cf. Let A : Y + Cf -+ V be an extension; so

A| = idX. Since f has a right homotopy inverse, by

corollary 4.9, Y + Cf can be deformed into Xo. By theorem

2.4, A is homotopic to idY + Cf and there is a deformation

r such that no = idY + Cf and = A. Since Y + Cf is an

ANR, by theorem 3.6 there is a deformation 6 such that the

fixed points of A, namely X0, remain fixed for every 6 t

Therefore Y + Cf is inessential relative to X0. 0

Theorem 5.3: If f maps the ANR X into Y = {y} and X0

can be contracted in a proper subset of Y + Cf, then Y + Cf

is inessential relative to Xo.

Proof: Since Xo can be contracted in a proper subset

of Y + Cf , let V C Y + Cf be that set. Thus there is a

deformation p : V x [0,1] -+ V, and a point xc in Xo such

that p0 = idV and p,(Xo) = {xc} gXo. Clearly pIj0 = idXO.

Let k : {y} -4X be a map defined by k(y) = xc. Since fk

{y} -4 {y}, fk = idy, and fk ~ idy. Thus f has a right

homotopy inverse. Now kfh'i1IXO is a map from X0 into

{xc} so kfh~'i~ XO p1j XOSince pjXOI ~ 0|XO, then

kfhf'i-|X0 ~ idXO.

It only remains to show that Y + Cf is an ANR in order

to conclude from theorem 5.2 that Y + Cf is inessential

relative to Xo.

47

Notice that since Y = {y}, Y + Cf = Cf =

{<x,t>| x E X, 0 < t < 1}

Claim: The set Y + C, is an ANR.

Let E be a separable, metrizable space which contains

Y + Cf as a closed subset. Define a function

PI: Y + Cf -+ [0,1] by P1(<x,t>) = t. Let V be an open

subset of [0,1]. Then P i(V) = {<x,t>I x E X, t E V}. Now

i~ (<x,t>I x e X, t e V}) = X x V possibly unioned with Y if

1 E V. But X x V is a subbase element of X x [0,1]; so

i~ (<x,t>I x c X, t e V} is open in X x [0,1] U Y. Thus

P~ (V) is open in Y + Cf. Therefore PI is continuous.

Since E is metrizable and therefore normal and Y + Cf is

closed in E, then by Tietze's Extension Theorem13 P can be

extended to all of E. Call this extension Pj. Thus

Pj : E -+ [0,1] such that Pj|y + Cf = I

Define a function PX : Y +Cf -+ X by PX(<x,t>) = x.

Let U be an open subset of X. The set Pj1(U) =

{<x,t>l x E U, 0 < t < 1}. The set i~1P 1 (U) =

{(xt)j x E U, 0 < t < 1} U Y = ({xj x e U} x [0,1]) U Y.

Thus i~'(Px'(U)) is open in Y + Cf. Therefore PX is

continuous.

Since X is an ANR and Y + Cf is closed in E, then by

'3Stephen Willard, General Topology (Reading, Mass.:Addison-Wesley Publishing Co.,1970) 103.

48

the Borsuk-Kurotowski Theorem'4 Px can be extended to an open

neighborhood,W, of Y + C, in E. Call the extension Pj.

Thus P : V - Y + Cf such that PjIY + Cf = X

Define a function r :V - Y + Cf by

r(e) = <P'(e),Pf(e)>. Let U be an open subset of Y + Cf.

Then r 1(U) = {ei P(e) = x, Pf(e) = t for some <x,t> e U} =

{ei Pj(e) = x, O <t <l1 for some <x,t> e U}fl

{eij Pj(e) = t, x E X for some <x,t> E U} =

Pf'(U) fl Pj(U). Since both P' and Pf are continuous,

r~ (U) is the intersection of open sets and r is continuous.

Notice that if e E Y + Cf, then e is some <x,t>, and r(e) =

r(<x,t>) = <P(x,t>),P'(<x,t>)> = <x,t>. Thus ry + C

idY + Cf. So r is a retraction of V onto Y + Cf, and Y + Cf

is an ANR.

Therefore by theorem 5.2, Y + Cf is inessential relative to

X0. 0

Fox describes an interesting subclass of retracting

deformations, and shows a correspondence with a subclass of

two-sided inverses.15

14Ralph H. Fox, "A Characterization of Absolute Retracts,"Bulletin of the American Mathematical Society 48 (1942) 273.

15Ralph H. Fox, "On Homotopy type and DeformationRetracts," Annals of Mathematics 2, 44 (1943) : 47-48.

49

Definition 5.4: Let Y + Cf be the mapping cylinder of

f X - Y and p be the retraction of Y + Cf into Y defined

by p(<x,t>) = <f(x)> = <x,1> and p(<y>) = <y>. Then a

retracting deformation 6 of Y + Cf into Xo is special if and

only if in addition to the usual conditions:

6 : Y + Cf x [0,1] - Y + Cf such that 60 = idY + C f

61 (Y + Cf) = X0, and 61XO = idXO; 6 also satisfies the

following two conditions:

i) For every s E [0,1] 6s'X0 = idXO.

ii) For every x E X whenever t = s

p(6(<f(x)>,s)) = p(6(<xt>),1)

Here s is the deformation parameter, and t is the mapping

cylinder parameter.

Definition 5.5: A two-sided inverse, g, of the map

f : X -4 Y is special if and only if there are homotopies

F and G such that in addition to the usual properties,

F : X x [0,1] - X such that F0 (x) = x and F,(x) = gf(x),

G : Y x [0,1] -4 Y such that GO(y) = y and G1 (y) = fg(y), the

condition f(F(x,s)) = G(f(x),s) is satisfied.

50

gf

F

idXfF Gf

Y

fg

G

f()idy

FIGURE 8

Theorem 5.6: The map f: X -+ Y has a special two-sided

inverse if and only if there is a special retracting

deformation of Y + Cf into X0 .

Proof: Suppose there is a special retracting

deformation 6 of Y + Cf into X0. Let g be a map such that

g(y) = h 1i~ X01(<y>). Notice that i~ -= x {o}'

and i_ j = ilj. Since i_1 X and h are homeomorphisms,

ijX x {0 }hg(y) =1<y>).

For each s E [0,1] define the map F : X x [0.1] -+ X by

F(x,s) = h-1C1X61(i(x,s)), and define the maps

G : Y x [0,1] -+ Y by G(y,u) = i'Ily6u(i(y)). Thus F(x,0) =

h~ i 1 IX061(<x,O>) = h'1i~xo(<x,0>), since 6s(<x,0>) =

<x,0> for all s E [0,1]. So F(x,0) = h-1i-I Xo(<x,0>) =

h~1 (X,0)= x. In other words, F0 = idx. And F(x,1) =

51

h~ i X61 (<x,1>) = h-1 i-1 IX61 (cf x)) = g(f(X))' So

F1 = gf. Also G0(y) = 0 p6 (<y>,0) = ijy1p(<y>) =

iJy (<y>) = y. So G = id . And G(y,1) = i l~ p6(<y>,1) =

iIl'p(i X x {0}hg(y)) since 6(<y>,1) = jX x {01 hg(y). So

G(y,1) = il|~1p(iX. x {0}(g(y),0)) = il~Iip(g(y),O) =

ijyj<f(g(y)> = f(g(y)). So Gi = fg. Then idx ~ gf by the

homotopy F, and idy ~ fg by the homotopy G. Therefore g is

a two-sided homotopy inverse of f. In additionhowever,

f(F(x,s)) = f(h-1 z I, 6 (<x,s>). Now f(h- i-1lX b1(<x, s>)

= f(h~ &'I(<x',s>) for some x' c X. The map

i- Y1p6(<xs>) = il-1p(<x',0>), for the same x' E X, so

i|~Y p6(<x,s>) = i|~1 (<f(x')>) = f(x'). Thus f(F(x,s)) =

il1 pol(<x,s>) for every s E [0,1]. Since 6 is a special

retracting deformation, p61(<x,s>) = p6s(<f(x)>). Thus

f(F(x,s)) = i l~ pI(<x,s>) = iIl~- p6s(<f(x)>) = G(f(x),s).Therefore g is a special two-sided homotopy inverse of f.

Conversely, suppose g is a special two-sided homotopy

inverse of f, and F and G are the homotopies satisfying the

condition f(F(x,s)) = G(f(x),s). Recall that Cf C Y + Cfand Cf = {<x,t>I x E X , 0 < t < 1} where f(x) E c<x,1> for

every x e X. Define a function : Cf x [0,1] -+ Cf by:

ihF(x,t(2u+1)); 0 < t < 2u/2u+1, 0 ,u)= <F(x,2u),t(2u+1)-2u)>; 2u/2u+1 (t< 1, 0 u 1/2

ihF(x,2t); 0 < t < 1/2, 1/2 ( u < 1

.g(f(x)), 2t-1)(2-2u)>; 1/2 < t< 1, 1/2 < u < 1

52

Line one and line two of the definition of 6 overlap

when t = 2u/2u+1 and 0 ,u) =

ihF(x,(2u/2u+1)(2u+1)) = ihF(x,2u) = <F(x,2u),0>. Also

f(<x,2u/2u+1>,u) = <F(x,2u),(2u/2u+1)(2u+1)-2u> =

<F(x,2u),0>

Line three and line four of the definition of 6 overlap

when t = 1/2 and 1/2 ,u) =

ihF(x,2(1/2)) = ihF(x,1)) = <g(f(x)),0>. And 6(<x,1/2>,u) =

<g(f(x)),(2(1/2)-1)(2-2u)> = <g(f(x)),0>.

When line two and three overlap u = 1/2. The condition

2u/2u+1 < t < 1 becomes 1/2 < t < 1. Thus when line two and

line three overlap t = 1/2 also. Then 6(<x,1/2>,1/2) =

ihF(x,2(1/2)) = <F(x,1),0> = <g(f(x)),0>. And

6(<x,1/2>,1/2) = <F(x,1),(2(1/2)-1)(2-2(1/2))> =

<g(f(x)),0>. Since <x,1> = <f(x)>, when 0 ,u) = 6(<x,1>,u) = <F(x,2u),(1)(2u+1) -2u)> =

<F(x,2u),1> = <f(F(x,2u)> = <G(f(x),2u> E Y. When

1/2 ,u) = 6(<x,1>,u) = <g(f(x)),(1)(2-2u)>.

Therefore 6 is well defined on Cf and is continuous. Thus (

is a deformation of Cf onto X0 .

For every <x,0> E Xo, (<x,O>,u) = ihF(x,2u) = <x,0>

for every u e [0,1]. Thus 6 is a retracting deformation

that satisfies condition i.

Claim: Condition ii is also satisfied by 6'.

Condition ii requires that p(M(<f(x)>,u)) =

53

p( (<x,u>,1)). If 0 ,u) = p(6u(cx,1>)

<F(x,2u),1> = <fF(x,2u)>, and p6(<x,u>,1) = pihF(x,2u) =

p<F(x,2u),0> = <fF(x,2u)>. If 1/2 ,u) =

p(<g(f(x)),2-2u>) = <fgf(x)>, and p6(<x,u>,1) = p<g(f(x),0>

= <fgf(x)>. Thus whenever u = t p6(<f(x)>,u) = p (<x,t>,1).

Therefore 6 is a special retracting deformation of Cf into

Xo.

Since G and g are both defined for all of Y, the

definition for 6 can be extended to Y + Cf by:

S((<x,1>,u) for <y> e Cf

'(<y>,u) = <G(y,2u)> for <y> E Y - Cf, 0 for <y> EY - Cf, 1/2 < u ( 1

Clearly 6' agrees with 6 if y = f(x). therefore 6' is

a retracting deformation Y + Cf onto Xo. 0

Definition 5.7: A homotopy 6 : X x [0,1] -+ M where M

is a metric space with metric d is called a c-homotopy if

and only if d(6(x,t 1), (xt2)) < E for every t1 and t2 in

[0,1] and every x in X. In particular,

d(6(x,0),6(x,1)) < E. If the homotopy is between maps f and

g, write f ~f g . If 6 is a deformation, call it a

e-deformation.

Definition 5.8: Let X and Y be metric spaces and

f : X -+Yand g : Y -+ X be maps such that gf ~ idX. Then g

54

is a left e-homotopy inverse of f and f is a right

e-homotopy inverse of g.

Theorem 5.9: Suppose X and Y are metric spaces. Then

a map f X - Y has a left E-homotopy inverse for every

E > 0 if and only if f is a homeomorphism, and for every

c > 0, idf(X) is E-homotopic in f(X) to a map extendable to

Y.

Proof: Let E > 0 be arbitrary. Suppose gE is a left

E-homotopy inverse of f. Then idXL~ gEf by a E-homotopy

f : X x [0,1] -4 X such that (6(x,O) = x , 6E(x,l) = gEf(x),

and for every t1 ,t2 c [0,1] and x e X d(6(x,t1 ),6(x,t2 )) < 6

In particular, for every x E X, d(6(x,0),6(x,1)) =

d(x,gEf(x)) < e.

Claim: The map f is one-to-one.

Let f(x1) = f(x2). By the hypothesis d(x,gEf(xl)) < E

and d(x2 )g f(x2 )) < E for every E > 0. Since f(xj) = f(x2)'

then d(x2 )gEf(x1)) < E. By the triangle inequality

d(xl,x2 ) < d(xl,g'f(x,)) + d(x2,g1f(xl)) < 2E for every

E > 0.

Just suppose x1 # x2 . Let d(x1 ,x2) =r > 0. Choose

c < r/2. Then 2E< r = d(x ,x2 ) < E. But this is a

contradiction. Therefore x 1 = x2 and f is one-to-one.

Clearly f is onto its image.

Claim: The map f is an open map.

Let U be an open subset in X and y0 be a member of

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f(U). Let {yk- 4y0 be a sequence in f(X) which converges

to y0. To prove that f is an open map it is sufficient to

show that {yk} is eventually in f(U). Since f is

one-to-one, f~1 (yk) is a single member of X for every k, and

f 1 (y0) EU. Let E > 0 be chosen. Since f~1 (yk) E X, and

gE is a left e-homotopy inverse of f,

d(idX(fl(yk)),gcf(fl (yk))) = d(f- (yk),gE(yk)) ' E.

Similarly, d(f~ (y0 )gE(y 0)) < c. Since g6 is continuous,

g'(yk g(Yk)} is a sequence in X which converges to

g(y ). Thus there is a natural number NE such that Vn > NE

d(g'(yn),gE(y 0 )) < E. Therefore if n > NE

d(f~ (y0),f n(yn)) d(f-(yO),g'E(y0)) + d(g'(y0)g(Yn))

+ d(g(y nI n)) < 3E. Therefore the sequence {f- (yk)

converges to f-1 (y0) in X. Since f~ (y0) is in U which is

open in X, there is a natural number M such that Vm > M

f 1 (ym) is in U. So ff-(ym) = ym is in f(U). Therefore

{yk} is eventually in f(U), and f is an open map.

Therefore f is a homeomorphism.

Since gcf ~ idx and f is a homeomorphism, then

fgcff~l ~ fidXf'. But fidXf = ff-1 = idf(x) , and

fgEff1 = fgidf(X) = fg'f(). Thus fg|f() ~ idf(x) for

every E > 0. Clearly fgc|f() 9is a map that can be

continuously be extended to Y by fg*.

Conversely, suppose the map f : X - Y is a

homeomorphism and for every E > 0, idf(x )is E-homotopic to

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a map extendable to Y. Since f is a homeomorphism, there is

no loss of generality to assume that X = f(X) or that X C Y

and that f = idX. For each E>0 let ge : Y -4X be the

extension of the map ge|f() which is e-homotopic to idX.

Thus glf(X) : f(X) -+ Y. But since f(X) = X,

gE f(X) X -+ X, and ge f()!~f idX. So for every c>0

there is a map (c : X x [0,1] -+ X such that (c(x,O) = x,

(x,1)=gc|f X (x), and for every x E X

d( '(x's1),66(xcs2)) < E for every s 1s2 in [0,1]. In

particular, d(x,gElf(X)(x)) < c. But since f = idX , for

all x c X f(x) = x. Thus (6(x,i) = gcf(x), and

d((x,sl),(xs2)) = d((f(x),s1),(f(x),s2 )) < E for all x E X.

Therefore idX ~E g!f, and gE is a e-homotopy inverse of f

for every E > 0.

Theorem 5.10: Suppose Y is a metric space and X is an

ANR. Then a map f X - Y has a left E-homotopy inverse for

every E > 0 if and only if f is a homeomorphism from X onto

f(X), and f(X) is a retract of Y.

Proof: Let X be an ANR and Y be a metric space.

Suppose a map f : X -+ Y has a e-homotopy inverse for every

E > 0. Since an ANR is a separable metric space, by

theorem 5.9, f is a homeomorphism and idf(x) is homotopic to

a map extendable to Y. Let 6 : Y -4 f(X) be an extension so

that idf(x ~I f(X). The set f(X) is closed in Y by the

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same argument used in theorem 5.9 to show that f is an open

map. Also since f is a homeomorphism and X is an ANR, then

f(X) is an ANR. Then by the Borsuk-Dowker Theorem16 idf(x)

is also extendable to Y. Call this extension r; so r: Y

f(X) such that rlf(X) = idf(x). Therefore r is a retraction

and f(X) is a retract of Y.

Conversely, suppose the map f X - Y is a

homeomorphism onto f(X), and f(X) is a retract of Y. Then

there is a map r : Y - f(X) such that rlf(X) = idf(x). Then

clearly rlf(X) ~ idf(x )and r is an extension of rlfg) to

Y. Therefore by theorem 5.9, f has a left e-homotopy

inverse for every e > 0.

Theorem 5.11: Suppose that X and Y are metric spaces

and f X - Y such that Y + Cf is metrizable. The map f has

a right e-homotopy inverse for every E > 0 if and only if Y

is r7-deformable'into Cf-Y for every r > 0.

The term "ry-deformable" means that the deformation is a

it-homotopy. Recall that C, = {<x,t>I x e X, 0 ( t < 1}, and

<x,1> = <f(x)> for every x e X. Thus C-f=-Y

{<x,t>| x f X, 0 < t <1}.

Proof: Suppose f has a right e-homotopy inverse for

every E > 0, say gf. So for every E > 0, idy ~l fgE by a

homotopy ff : Y x [0,1] -4 Y where (f(y,0) = y, (f(y,1) =

'6Witold Hurewicz and Henry Wallman, Dimension Theory(Princeton: Princeton University Press, 1941), 86.

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fgE(y), and dy(((y,s1 ),(((y,s 2 )) < c for every si, s2 in

[0,1]. For each e > 0 define 66E: Y x [0,1] -* Y + Cf by:

(<y>u)=ti66(y,2u) 0 ,0) = ij((y,0) = i(y) = <y>; so 6= id.

And 66(<y>,1) = i(g6(y),1-E) = <g(y),1-e> E Cf- Y, since

g9(y) e X and E > 0. When u = 1/2, 6(<y>,1/2) = i(6(y,1) =

i(fg'(y)) = <fg'(y)>, and 66(<y>,1/2) = i(gE(y),1-E(0)) =

i(g'(y),1) = <g'(y),1> = <fgE(y)>. Thus 66 is well defined

and continuous for every E > 0.

Since Y + Cf is metrizable, call its metric dc. Let dP

be the metric in Y + Cf x [0,1] defined by

dp ((<qj>,sj),(2>,s2)) = d c(<ci>,<c2 >)2 + di(sl,s2

where <q> represents any element of Y + Cf, and d is the

usual distance in the unit interval. Similarly, let dx be

the metric in X x [0,1], and dY be the metric in Y. Let

17 > 0 be chosen.

Case 1: Suppose u1 and u2 are in [0,1/2]. Now for

every E > 0, dp (66(<y>,u1 ),6(6(<y>,u2 )) =

d( (i((y,2uj),i(66(y,2u2)). Since i is continuous, if

dy((6(y,2u1 ),(6(y,2u 2 )) ( E, then

dp (i(6(y,2u1 ),i((6(y,2u 2 )) < V for every q > 0. So for the

chosen V choose an e > 0 such that

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d P(561(<y>,u 1),5 1(<y>,U 2)) *

Case 2: Suppose u1 and u2 are in [1/2,1]. Then

d, (6 5 (<y>,u 1 ), 6 '(<y>,u2 )) =

dP(i(gE(y),1-c(2u1 -1)),i(g'(y),1-c(2u2-1)) for every E > 0.

Notice that 1/2 0. Again since i is continuous,

choose c2 > 0 such that 2E2Iu2-u1 I< 2E211-01 < n. Then

d (62(<y>,u 1)),6 f2 (<y>,u2)) I-

Case 3: Suppose u1 E [0,1/2] and u2 E [1/2,1]. Since,

when u = 1/2, i(y,) = i(g(y)), then

d, (6(<y>,u1),6'(<y>,u 2)) =

d P( i f(y,2u,) ,i(g6(y) ,1-e(2u2-1))) i d P( i 6(y,2u1) , i 6(y ,1))

+ d, (i(gE(y),1-e(2u2-1)),i(gE(y),1)) for every E > 0. Now

from case 1 and case 2, d P(i6 1 (y,2u1 ),i6 1 (y,1)) +

d (i(g2(y),1-E(2u2-1)),i( 2(y),1)) < 2. So choose c3 so

that dP(6(<y>,u 1 ),6E(<y>,u2)) < 2n/2.

Choose E < min {EE 2E 3 }. Then

d, (6E(<y>,u1 ),6'(<y>,u2 )) < n. Therefore for every 77 > 0, Y

is q-deformable into Cf-Y by 6f.

Conversely, suppose that Y is n -eformable into Cf-Y

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for every n > 0. Then for every E > 0 there is a map

61 :Y x [0,1] -+ Y + Cf such that 6'(<y>,0) = <y>,

61(<y>,1) E Cf-Y, and d P(6c(<y>,u1) ,67(<y>,u2)) < < In

particular, dp(<y>,6n(<y>,1)) < 1. Let E > 0 be chosen.

Define v : Cf-Y -4 X by v(<x,t>) = x. Suppose U is an

open set in X. Then V-1(U) = i{(x,t)I x E U, t E [0,1)} =

U x [0,1) which is open in X x [0,1]. So v-1(U) is open in

Cf-Y and v is continuous.

Define g Y -+ X by g'(y) = v(6'(i(y),1)).

Notice that since 6c(<y>) e Cf-Y, then 6c(<y>) =<x,t>

for some <x,t> E Cf-Y. Also by the definition of v, every

<x,t> E Cf-Y is equal to <v(<x,t>),t>.

Define A : Cf-Y x [0,1] -+ Y + Cf by A(<x,t>,u) =

<x,t+u(1-t)>. Notice that A is defined similarly to -r in

lemma 4.8, = riC- y; so A is continuous.

Define ( :Y x [0,1] -+ Y by:

v = (<y>,2v) 0 < V 1/2

SA (6c(<y>) ,2v-1) 1/2 < v < 1

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{<f V6 (<y>)>|<y> E} = {<fgE(i(x))> y E Y}

{<v(<x,t>),1>I <x,t> = 66(<y>), <y> e }

A

{<w(<x,t>),t>l <x,t> = 6(<y>), <y> E Y}

66(Y)

6

id

FIGURE 9

When v = 0, 6E(<y>,o) = 6E(<y>,o) = <y>; so = ide.

When v = 1/2, 66(<y>,1/2 ) = 6 (<y>,1) = 6(<y>), and

6 (<y>,1/ 2 ) = A(66(<y>),0) = <x,t+0> where <x,t> = 6(<y>).

When v = 1, ( (<y>,1) = A(66(<y>),2-1). Now 6(<y>) is some

<xt> E Cf-Y; so ((<y>,1) = A(<x,t>,1) = <x,t+1(1-t)> =

<x,i>. Notice that v(61(i(y)) = v(61(<y>) = v(<x,t>) = x.

Thus ((<y>,1) = <x,1> = <v(6(i(y)),1> = <fv6(i(y))> =

<fg6(y)> = ilyfgf i | , since ily is a homeomorphism. Thus

(E is a homotopy between id^ and ( (() = ilyfgi~11^. Then

also idy ~ fgE. Since E was arbitrarily chosen, a homotopy

C can be constructed for every e > 0. It remains to show

that 6E is an e-homotopy.

62

Case 1: Suppose vi and v2 are in [0,1/2]. Then

de 1 2)) = dc(6E((y> v)1(6E(<y>,'v2)) 'Case 2: Suppose v1 and v2 are in [1/2,1]. Then

de1 2)) = d1c(A61(y)v1),A6(y)v2))

Since A is continuous, for every 6 > 0 there is an 17 > 0

such that if dc(A67(1y>),vi),A(<y>),v<2)) <6, then

d1 112)d c~I(<>2v1 (6(<y>v2) < 77' So choose n such that

dc(A6q(<y>)v)'A6(<y>),v2)) ( 6.

Case 3: Suppose v. E [0,1/2] and v2 E [1/2,1]. Recall

that when v = 1/2, (e(<Y>,1/2) = 66(<y>,1) = A6(<y>),0).

Then d c(66(<Y>'vi) (<y>v2 )) dc(6(y>jvi),(<y>,1/2))

+ dc(66(<y>,1/2) ,((<y>,v2)) = d(6 6('y>,v 1),(6 C(<y>,l)) +

dc(Ab6(<y>),0),2A6(<y>),vj)). From case 2 there is an 17 > 0

such that d ( 2)) < e Choose n' such

that dc(6n (y>jv 1 ) ,(6 (<y>,1)) < e/2 and

dc(A6 (<y>),0),A67(<y>),vl)) < c/2. Then

dc(A6 (<Y>),v),A61( (<y>),v2)) C<. Choose 7 =

min {e,j,n'}. Then (7 is an e-homotopy.

Since i' I is a homeomorphism, by a similar argument

there is a 7' > 0 such that iY 7 / is an e-homotopy

between idy and fgC for any c > 0. 0

Theorem 5.12: Suppose X is a compact metric space; Y

is a metric space; and f : X -+ Y is a map such that Y + Cfis

metrizable. Then f has a two-sided e-homotopy inverse for

every e > 0 if and only if f is a homeomorphism of X into Y.

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Proof: Suppose f has a two-sided e-homotopy inverse

for every e > 0.

Claim: The closure of f(X) is equal to Y.

Clearly T(XT C Y. Let y E Y. Let N be a basic open

neighborhood of y in Y. Then iN = ily(N) is an open

neighborhood of <y> in Y containing some basic open set G of

some radius, say p. By the previous theorem, Y is

E-deformable into Cf-Y; so for E = p there is a

6" : Y x [0,1] -+ Y + Cf such that 6(<y>,0) = <y>,

6P(<y>,1)fE Cf-Y, and dc(6P(<y>,s1 ),6P(<y>,s2)) < c. Since

6p(<y>) E Cf-Y, then 6P(<y>) is some <x,t> where t < 1. So

dc(<y>,<x,t>) < E. Since 6 is a homotopy, 6P : [0,1]-+(y>

Y + Cf is a path in Y + Cf from <y> to <x,t>. Furthermore,

the distance between any two points in this path is less

than p. Since 6P is continuous, there is some s e [0,1]

and some x' E X such that 6 y>(s) = <x',1> = <f(x')>, and

dc(<y>,<x',1>) < p. So <f(x')> E G. Now

f(x') E il~ (<f(x')>) g N; so f(x') E N. Therefore y is a

closure point of f(X), and I(X) = Y.

Since X is compact and f is continuous , f(X) is

compact. Since Y is a metric space, f(X) = TfXJ = Y. Then

idf(x) = idy; so idf(x) ~ idy for every E > 0. Then by

theorem 5.9, f is a homeomorphism from X onto Y.

Conversely, suppose f is a homeomorphism from X onto Y.

Then every y = f(x) for some x e X. Thus every <y> E Y can

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be written as <f(x)> or <x,1>.

Let e > 0 be chosen. Define 6 Y: x [0,1] -+ Y + Cf by

6E(<ff(x)>,s) = <x,1-sE>. Thus 66(<ff(x)>,0) = <x,1-0> =

<x,1> = <f(x)>; so 6 = id'. When s = 1, 6E(<f(x)>,1) =

<x,1-E> E Cf-Y. Also dc(6E(<f(x)>,s1),6E(<f(x)>,s2)) =

dc6E(<x,1>,sl),6E(<x,1>,s2 )) dc(<x,1>,<x,1-e>) = 7 Thus

for every n > 0 Y is r-deformable into Cf-Y. Then by

theorem 5.11 f has a right E-homotopy inverse for every E >

0.

Since f is a homeomorphism from X onto Y, f(X) = Y and

idf(x) = idy; so by theorem 5.9, f has a left e-homotopy

inverse for every E > 0.

Let E > 0 be chosen, and let 16 be a left e-homotopy

inverse and rf be a right e-homotopy inverse of f. Define a

map ge = lfrE. Now fgE = flEfrE. Since l f ~, id x then

for some a > 0, flffrE is a-homotopic to fidXrE = frE.

Since fre ~, f idy , then flEfrE is a'-homotopic to idY for

some a' < 0.

Similarly, gf = lfrEf ~ lEidX =f'Ef idy. So

for p = min {7',a'} fgE ~ idX and g'f ~ idy. Therefore

gf is a two-sided p-homotopy inverse of f.

Then for every p > 0 there is some E > 0 such that ge

is a two-sided p-homotopy inverse of f. 0

REFERENCES

Borsuk, Karol. Theory of Retracts. Varsawa, Poland: PWN-Polish Scientific Publishers, 1967.

Fox, Ralph H. "A Characterization of Absolute NeighborhoodRetracts." Bulletin of the American MathematicalSociety. 48 (1942): 271-275.

. "On Homotopy Type and Deformation Retracts."Annals of Mathematics. (2) 44 (1943): 40-50.

Hurewicz, Withold and Henry Wallman. Dimension Theory.Princeton: Princeton University Press, 1941.

Willard, Stephen. General Topology. Reading, Massachusetts:Addison-Wesley Publishing Co., 1970.

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