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 . .0 . a. 0...... a . .a iv CHAPTER 1. Introduction ...... 1 2. Homotopy and Deformation ......
3. Retraction and Deformation Retract . . . 10 4. The Mapping Cylinder ...... 25 5. Applications . . . . 45
REFERENCES . . . .a .. ..a0. ... 0 ... 65
iii LIST OF ILLUSTRATIONS
Page FIGURE
1. Illustration of a Homotopy ... 4
2. Homotopy for Theorem 2.6 . . 8
3. The Comb Space ...... 18
4. Transitivity of Homotopy . . . . .22
5. The Mapping Cylinder . 26
6. Function Diagram for Theorem 4.5 . . .31
7. The Homotopy "r" . . .37
8. Special Inverse . . .50
9. Homotopy for Theorem 5.11 . . . .61
iv 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
2 Stephen Willard, General Topology (Reading, Mass.: Addison-Wesley Publishing Co., 1970).
3 4
of t's for t c [0,1] can be thought of as a kind "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:
t E [0,1/2] ((at) = E(a,2t) for 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 a0 and 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 -4A 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: 8
6(a,3t) for 0 < t < 1/3
((a,t) = r(6(a,l),3t - 1) for 1/3 < t < 2/3
I f(6(a,3 - 3t)) for 2/3 t < 1
Thus the map ((a,0) = 6(a,O) = idA(a) = a, and the map
((a,1) = f(6 (a,0)) = f(a).
1 f (A)
f(6(r))
2/3 f (B) f (B)
77
1/3 6(A x {1}) g(A)
S
t =0 id
FIGURE 2 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 B. ((a,1/3) = 6(a,3(1/3)) = 6(a,1) = g(a) E B, and g(A) = 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) = and n(6(a,1),3(2/3) - 1) = n(g(a),1) = f|B(g(a)), = n(g(A),1) = (B),1) = fIB(B) = f(B). And C(a,2/3) Thus C f (6(a,3-3(2/3))) = f (6(a,1)) = f (g(a)) = fjB(g(a)). 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 (. f(a). Thus 6 : A x [0,1] -+ A and 6(a,0) = a and 6(a,1) =
Since B C A, 61B : B x [0,1] -+ A such that (b,0) = b, and 1B is a homotopy 6(b,1) = f(b) = f|B(b) Vb E B. Therefore 6 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.
3 Karol 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 n a 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 American Mathematical Society 14 (1963): 364-365. 7Ralph H. Fox, "A Characterization of Absolute Neighborhood Retracts," Bulletin of the American Mathematical Society 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 ((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) When t = 1 Va E A. And n 1(a,t) = 6(a,t/1) = 6(a,t) Va E B. E A. Thus V, and 6 agree for n1(a,1) = 6(a,1) for all a is an 0 t< 1. But 6 is a map from A x [0,1] into A; so 6 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 6 1 61 (A) = B and 1 B = 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 6 61 (A) = B. So 01B = idB and 6 t1B g B for every t. The map 6 1B is a homotopy between 601B and 6 1|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 6 S. Furthermore, 1 I5 (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 gf jk kj a OE 7 6 B A -C 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 < 1/2 p(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 Deformation Retracts," Annals of Mathematics 44 (January 1943):43. The author 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 if n E N-a(M) and =1n{n}1 (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 Qi 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 " Let N denote the set { {(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 { 1 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} = { 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)) = 0 < t < 1. However, i|M x [0,1] is not the projection function of the decomposition of M x [0,1] since Lemma 4.3: Each map i1M x {t} for 0 < t < 1 is a homeomorphism from M x {t} onto Proof: Let t e [0,1). The image of i 1M x {t} is the 28 set Let U be open in Ca such that Uc = v la Vc fM x {t}>=U. So i-1(U) = i-(Vcfa 1 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 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={ subset of N + Ca* Observe that since i(mt) = i| 1x [0, 1 )({ 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 onto A N. Proof: The map 'N maps N into N by: f 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) { 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 -4Y 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 / ~X1 z/ 0 Oh-1 i-1 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)) = * ( 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 ( EY. Notice that if y = f(x), then 0(( {(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( 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( Now each between idy and ig in Z + Cg. Since wo = idyo, then woiO' =iO'. Consider w1iO' X - Z + C . Notice that wiO"' (x) = wi( F : X x [0,1] -+ Z + Cg by F(x,s) = wsiO'(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: 0 < s ( 1/2 xs) =f((x,2s) for 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( Let U be open in Z. The set i-16-1 (U) = i~'({ {(y,t) ) y ( g(U) 0 t< 1} = U U { (y, t)I Y C g~1 (U),) 0 < t 1 Since g is continuous, g~1 (U) is open, and thus 6 (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 a 66(x,1) = 6(ig0'(x)) = 6( Conversely, suppose there is a map f : X - Y such that homotopy to define gf ~ 0 by some homotopy. Reverse that 6(x,1) = gf(x) 6 : X x [0,1] -4 Z such that 6(x,0) = 0(x) and So Define w : Yo x [0,1] - Z + C by w( - Y x[0,1] i6(x,1) = igf(x) = 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 Define woihf ~ wihf. But woihf = idYOihf = ihf : X - Yo. A : X x [0,1] - Z + C by: f=i6(x,2s) for 0 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 by the A(x,1/2) = w(ihf(x),1) = igf(x). Therefore iO ~ ihf 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( 0= id Y + Cf Finally, 7( Let U be open in Y + Cf. Claim: Every point in r~1(U) is an interior point. Let ( Case 1: Suppose t0 < 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. 37 Y + C 2 [o 3_1] I 1o . . t >s -I ( 0 s 40~ 1000 de - -4I doI . ~I-~---~ & eoeollo 0 ;7x 1 t 0 7+C 0 1 p2 p1 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 Now r-1(G x {p2 }) = {( open in Y + Cf, there is a p' such that t' < p' r( {( t < t0 such that r( Let V = {( i(G x (t1 ,t2 )) x (s1,s2). Thus V is open in Y + Cf x [0,1], and contains ( Let ( t2 +S2(1-t2);so r( Case 2: Suppose t0 = 1. Then r( 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 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( 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 + C g 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( ((r1 ( ((r1 ( Notice that 6( 6( (( 6( idz + C . Also 6( (( (( Finally, 6( 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( 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 = 6 A IX01. Thus idV Ai I X0 1 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 Iy Ai 6(Al(w),1) = 61 A,(w); so (1 = 611y A1. Thus idy = 1 0 6 a and idxo =61 1YOA 11xO. Therefore the map, iiy has 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 1 that kfh- 1 Xx ~ 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 = idX0 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 6t 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 = { 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( 1 E V. But X x V is a subbase element of X x [0,1]; so i~ ( 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( Let U be an open subset of X. The set Pj1(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) = Then r 1(U) = {ei P(e) = x, Pf(e) = t for some {eij Pj(e) = t, x E X for some 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 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 Deformation Retracts," 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( 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: 6 i) For every s E [0,1] s'X0 = idXO. ii) For every x E X whenever t = s p(6( 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 fg F G fF Gf f() idX idy Y 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 1 = x {o}' g(y) = h i~ X01( and i_ j = ilj. Since i_1 X and h are homeomorphisms, ijX x {0 }hg(y) =1 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( h~1 (X,0)= x. In other words, F0 = idx. And F(x,1) = 51 1 1 h~ i X61 ( 6 F1 = gf. Also G0(y) = 0 p ( iJy ( iIl'p(i X x {0}hg(y)) since 6( G(y,1) = il|~1p(iX.x {0}(g(y),0)) = il~Iip(g(y),O) = ijyj i-Y1 p6( i|~Y p6( il1 pol( 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 + Cf and Cf = { ihF(x,t(2u+1)); 0 < t < 2u/2u+1, 0 < u ( 1/2 ( .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 < 1/2. Then 1( Line three and line four of the definition of 6 overlap when t = 1/2 and 1/2 < u < 1. Then 6( 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( 6( 6( Therefore 6 is well defined on Cf and is continuous. Thus ( is a deformation of Cf onto X0 . For every Claim: Condition ii is also satisfied by 6'. Condition ii requires that p(M( p( ( p = 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(( 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 t ,t c [0,1] 1 2 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,x 2 ) < 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 x1 = 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 55 f(U). Let {yk- 4y 0 be a sequence in f(X) which converges to to y0 . To prove that f is an open map it is sufficient 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 1 1 E X, and f (y0) EU. Let E > 0 be chosen. Since f~ (yk) 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, to g'(yk g(Yk)} is a sequence in X which converges that Vn > NE g(y ). Thus there is a natural number NE such d(g'(yn),gE(y 0 )) < E. Therefore if n > NE d(f~ (y0),f n(yn)) d(f-(yO),g'E(y 0)) + d(g'(y 0)g(Yn)) + d(g(y nI n)) < 3E. Therefore the sequence {f- (yk) 1 converges to f- (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 open map. {yk} is eventually in f(U), and f is an 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 56 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 -4Xbe 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),s 2 )) < 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 57 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, = { { 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] -4Y where (f(y,0)= y, (f(y,1)= '6Witold Hurewicz and Henry Wallman, Dimension Theory (Princeton: Princeton University Press, 1941), 86. 58 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: 0 < u ( 1/2 ( Thus 6E( And 66( Since Y + Cf is metrizable, call its metric dc. Let dP be the metric in Y + Cf x [0,1] defined by 2 dp (( 17 > 0 be chosen. Case 1: Suppose u1 and u2 are in [0,1/2]. Now for every E > 0, dp (66( d P(561( Case 2: Suppose u1 and u2 are in [1/2,1]. Then d, (6 5 ( 1-E < 1-E(2u-1) 1. Thus dX ((g'(y) ,1-e(2u,-1)) ,(g6(y) ,1-e(2u2l) 2 2 0 + (d1 (1-e(2u1 -1)),(1-E(2u2 -1))) = (2e(u2 -u)) = 2E Iu2 -u11 for every E > 0. Again since i is continuous, choose c2 > 0 such that 2E2 Iu2-u1 I< 2E 2 11-01 < n. Then d (62( 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( 1 1 from case 1 and case 2, dP(i6 (y,2u1 ),i6 (y,1)) + d (i(g2(y),1-E(2u2 -1)),i( 2(y),1)) < 2. So choose c3 so that dP(6( Choose E < min {EE 2E 3 }. Then d, (6E( is q-deformable into Cf-Y by 6f. Conversely, suppose that Y is n -eformable into Cf-Y 60 for every n > 0. Then for every E > 0 there is a map 61 :Y x [0,1] -+ Y + Cf such that 6'( 61( Define v : Cf-Y -4 X by v( Define g Y -+ X by g'(y) = v(6'(i(y),1)). Notice that since 6c( Define A : Cf-Y x [0,1] -+ Y + Cf by A( lemma 4.8, = riC- y; so A is continuous. Define ( :Y x [0,1] -+ Y by: v =( { { A { 6 id FIGURE 9 When v = 0, 6E( When v = 1/2, 66( 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( 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( Case 3: Suppose v. E [0,1/2] and v2 E [1/2,1]. Recall that when v = 1/2, (e( Then dc(66( + dc(66( 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. 63 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 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( 6P( 6p( dc( than p. Since 6P is continuous, there is some s e [0,1] and some x' E X such that 6 y>(s) = dc( f(x') E il~ ( 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 be written as Let e > 0 be chosen. Define 6 Y: x [0,1] -+ Y + Cf by 6E( 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 Neighborhood Retracts." Bulletin of the American Mathematical Society. 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. 65 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