The Fundamental Group and Covering Spaces

Chapter 8 The Fundamental Group and Covering Spaces In the first seven chapters we have dealt with point-set topology. This chapter provides an introduction to algebraic topology. Algebraic topology may be regarded as the study of topological spaces and continuous functions by means of algebraic objects such as groups and homomorphisms. Typically, we start with a topological space (X, ) and associate with it, in some specific manner, a group G X (see Appendix I). This is done in such a way that if (X, ) and (Y, ) are homeomorphic, then G X and G Y are isomorphic. Also, a continuous function f : X Y induces a homomorphism f * : G X G Y, and, if f is a homeomorphism, then f * is an isomorphism. Then the structure of G X provides information about the structure of (X, ). One of the purposes of doing this is to find a topological property that distinguishes between two nonhomeomorphic spaces. The group that we study is called the fundamental group. 8.1 Homotopy of Paths In Section 3.2 we defined paths and the path product of two paths. This section studies an equivalence relation defined on a certain collection of paths in a topological space. Review Theorems 2.15 and 2.27, because these results about continuous functions will be used repeatedly in this chapter and the next. The closed unit interval [0, 1] will be denoted by I. Definition. Let (X, ) and (Y, ) be topological spaces and let f, g: X Y be continuous functions. Then f is homotopic to g, denoted by f g, if there is a continuous function H: X I Y such that H(x, 0) f(x) and H(x, 1) g(x) for all x X. The function H is called a homotopy between f and g. 265

266 Chapter 8 The Fundamental Group and Covering Spaces In the preceding definition, we are assuming that I has the subspace topology determined by the usual topology on. Thus X I has the product topology determined by and. We think of a homotopy as a continuous one-parameter family of continuous functions from X into Y. If t denotes the parameter, then the homotopy represents a continuous deformation of the function f to the function g as t goes from 0 to 1. The question of whether f is homotopic to g is a question of whether there is a continuous extension of a given function. We think of f as being a function from X {0} into Y and g as being a function from X {1} into Y, so we have a continuous function from X {0, 1} into Y and we want to extend it to a continuous function from X I into Y. EXAMPLE 1. Let X and Y be the subspaces of defined by X {(x, y) : x 2 y 2 1} and Y {(x, y) : (x 1) 2 y 2 1 or (x 1) 2 y 2 1}. Define f, g: X Y by f(x, y) (x 1, y) and g(x, y) (x 1, y). (See Figure 8.1). X f g Y Figure 8.1 Then f is not homotopic to g. EXAMPLE 2. Let X be the subspace of defined by X {(x, y) : x 2 y 2 1} and let Y X X. Define f, g: X Y by f(x, y) ((1, 0), (x, y)) and g(x, y) ((0,1), (x, y)). Then the function H: X I Y defined by H((x, y), t) (( 1 t 2, t), (x, y)) is a homotopy between f and g so f g. Notice that X is a circle and Y is a torus, f wraps the circle X around {(1, 0)} X, and g wraps the circle X around {(0, 1)} X (see Figure 8.2). Let A denote the arc from ((1, 0), (1, 0)) to ((0, 1), (1, 0)) (see Figure 8.2). Then H maps X I onto A X.

8.1 Homotopy of Paths 267 (0, 1) g A ((0, 1), (1, 0)) (1, 0) H (( 3/2, 1/2), (1, 0)) (X 3 {1/2}) f ((1, 0), (1, 0)) Figure 8.2 THEOREM 8.1. Let (X, ) and (Y, ) be topological spaces. Then is an equivalence relation on C(X, Y) (the collection of continuous functions that map X into Y). Proof. Let f C(X, Y). Define H: X I Y by H(x, t) f(x) for each (x, t) X I. Then H is continuous and H(x, 0) H(x, 1) f(x), so f f. Suppose f, g C(X, Y) and f g. Then there is a continuous function H: X I Y such that H(x, 0) f(x) and H(x, 1) g(x) for all x X. Define K: X I Y by K(x, t) H(x, 1 t). Then K is continuous, and K(x, 0) H(x, 1) g(x) and K(x, 1) H(x, 0) f(x) for all x X. Therefore g f. Suppose f, g, h C(X, Y), f g, and g h. Then there are continuous functions F, G: X I Y such that F(x, 0) f(x), F(x, 1) g(x), G(x, 0) g(x), and G(x, 1) h(x) for all x X. Define H: X I Y by H(x, t) { F(x, 2t), for all x X and 0 t 1 2 G(x, 2t 1), for all x X and 1 2 t 1. Then H is continuous, and H(x, 0) F(x, 0) f(x) and H(x, 1) G(x, 1) h(x) for all x X. Therefore f h. If and are paths in a topological space, we define a relation between them that is stronger than homotopy. Definition. Let and be paths in a topological space (X, ). Then is path homotopic to, denoted by p, if and have the same initial point x 0 and the same terminal point x 1 and there is a continuous function H: I I X such that H(x, 0) (x) and H(x, 1) (x) for all x X, and H(0, t) x 0 and H(1, t) x 1 for all t I. The function H is called a path homotopy between f and g.

x 0 x 1 268 Chapter 8 The Fundamental Group and Covering Spaces The question of the existence of a path homotopy between two paths is again a question of obtaining a continuous extension of a given function. This time we have a continuous function on (I {0, 1}) ({0, 1} I) and we want to extend it to a continuous function on I I (see Figure 8.3). I 3 I X Figure 8.3 EXAMPLE 3. Let X {(x, y) 2 : 1 x 2 y 2 4} (see Figure 8.4), let : I X be the path that maps I in a linear fashion onto the arc A from ( 1, 0) to (1, 0), and let : I X be the path that maps I in a linear fashion onto the arc B from ( 1, 0) to (1, 0). Then is homotopic to (by a homotopy that transforms into in the manner illustrated in Figure 8.5), but is not path homotopic to because we cannot get from to and keep the end points fixed without crossing the hole in the space. If (X, ) is a topological space and x 0 X, let (X, x 0 ) { : I X: is continuous and (0) (1) x 0 }. Observe that if, (X, x 0 ), then the product (defined in Section 3.2) * (X, x 0 ). It is, however, easy to see that * is not associative on (X, x 0 ). Each member of (X, x 0 ) is called a loop in X at x 0. The proof of the following theorem is similar to the proof of Theorem 8.1 and hence it is left as Exercise 1. A (21, 0) (1, 0) B Figure 8.4

8.1 Homotopy of Paths 269 (I) (I) (I) (I) (I) (I) (I) (I) (I) (I) (I) (I) Figure 8.5 THEOREM 8.2. Let (X, ) be a topological space, and let x 0 X. Then p is an equivalence relation on (X, x 0 ). Let (X, ) be a topological space and let x 0 X. If (X, x 0 ), we let [ ] denote the path-homotopy equivalence class that contains. Then we let 1 (X, x 0 ) denote the set of all path-homotopy equivalence classes on (X, x 0 ), and we define an operation on 1 (X, x 0 ) by [ ] [ ] [ * ]. The following theorem tells us that is well-defined. THEOREM 8.3. Let (X, ) be a topological space and let x 0 X. Furthermore, let 1, 2, 1, 2 (X, x 0 ) and suppose 1 p 2 and 1 p 2. Then 1 * 1 p 2 * 2. Proof. Since 1 p 2 and 1 p 2, there exist continuous functions F, G: I I X such that F(x, 0) 1 (x), F(x, 1) 2 (x), G(x, 0) 1 (x), and G(x, 1) 2 (x) for all x I, and F(0, t) F(1, t) G(0, t) G(1, t) x 0 for all t I. Define H: I I X by Then H is continuous, H(x, t) { F(2x, t), if 0 x 1 2 and 0 t 1 G(2x 1, t), if 1 2 x 1 and 0 t 1. H(x, 0) { F(2x, 0), if 0 x 1 2 ˇG(2x 1, 0) if 1 2 x 1 { 1(2x), if 0 x 1 2 1 (2x 1), if 1 2 x 1 ( 1 * 1 )(x)

270 Chapter 8 The Fundamental Group and Covering Spaces and H(x, 1) { F(2x, 1), if 0 x 1 2 ˇG(2x 1, 1) if 1 2 x 1 { 2(2x), 0 x 1 2 1 2 (2x 1), 2 x 1 ( 2 * 2 )(x) for all x X, and H(0, t) F(0, t) x 0 and H(1, t) G(1, t) x 0 for all t I. Therefore 1 * 1 p 2 * 2. The following three theorems show that if (X, ) is a topological space and x 0 X, then ( 1 (X, x 0 ), ) is a group. THEOREM 8.4. Let (X, ) be a topological space, let x 0 X, and let [ ], [ ], [ ] 1 (X, x 0 ). Then ([ ] [ ]) [ ] [ ] ([ ] [ ]). Proof. We show that x I, ( * ) * p * ( * ). First we observe that for each and [( * ) * ](x) { ( * )(2x), if 0 x 1 2 (2x 1), if 1 2 x 1 1 4 { (4x), if 0 x (4x 1), if 1 4 x 1 2 (2x 1), if 1 2 x 1 { (2x), if 0 x 1 2 [ * ( * )](x) (4x 2), if 1 2 x 3 4 (4x 3), if 3 4 x 1. Next we draw a picture to illustrate how we arrive at the definition of the desired path homotopy H. (0, 1) 1 3 ( 2, 1) ( 4, 1) (1, 1) L t (0, 0) 1 1 (, 0) (1, 0) 4 ( 2, 0)

8.1 Homotopy of Paths 271 If t I, then the intersection of the horizontal line L and the line segment joining ( 1 4, 0) and ( 1 2, 1) has coordinates ((1 t) 4, t), and the intersection of L and the line segment joining ( 1 2, 0) and ( 3 4, 1) has coordinates ((2 t) 4, t). Therefore for each t I, we want H to be defined in terms of if x (1 t) 4, in terms of if (1 t)/4 x (2 t) 4, and in terms of if x (2 t) 4. Also for each t I, we want H(0, t) H((1 t) 4, t) H((2 t) 4, t) H(1, t) x 0. Using elementary analytic geometry, we arrive at the definition of H. Define H: I I X by H(x, t) { ( 4x 1 t) ( 1 t, if 0 t 1 and 0 x 4 (4x 1 t), if 0 t 1 and 1 t x 2 t 4 4 4x 2 t 2 t 2 t ), if 0 t 1 and x 1. 4 Thus H is continuous, and it is a routine check to show that H(x, 0) [( * ) * ](x) and H(x, 1) [ * ( * )](x) for all x I, and H(0, t) H(1, t) x 0 for all t I. We have proved that is associative on 1 (X, x 0 ). Now we prove that ( 1 (X, x 0 ), ) has an identity element. THEOREM 8.5. Let (X, ) be a topological space, let x 0 X, and let e: I X be the path defined by e(x) x 0 for each x I. Then [ ] [e] [e] [ ] [ ] for each [ ] 1 (X, x 0 ). Proof. We prove that if [ ] 1 (X, x 0 ) then * e p and e * p. Let [ ] 1 (X, x 0 ). We draw a picture to illustrate how we arrive at the definition of the path homotopy H to show that * e p. (0, 1) (1, 1) L t e (0, 0) 1 ( 2, 0) (1, 0)

272 Chapter 8 The Fundamental Group and Covering Spaces For each t I, the intersection of the line L with the line segment joining ( 1 2, 0) and (1, 1) has coordinates ((1 t) 2, t). Therefore for each t I, we want the homotopy defined in terms of if x (1 t) 2. Using analytic geometry, we arrive at the formula { ( 2x 1 t, if 0 x 1 t) 2 H(x, t) x 0, if 1 t x 1. 2 It is a routine check to see that H(x, 0) ( * e)(x) and H(x, 1) (x) for each x I and H(0, t) H(1, t) x 0 for each t I. The proof that e * p is left as Exercise 2. We have proved that [e] is the identity element of 1 (X, x 0 ). We complete the proof that 1 (X, x 0 ), ) is a group by showing that each member of 1 (X, x 0 ) has an inverse in 1 (X, x 0 ). THEOREM 8.6. Let (X, ) be a topological space, let x 0 X, and let [ ] 1 (X, x 0 ). Then there exists [ ] 1 (X, x 0 ) such that [ ] [ ] [ ] [ ] [e]. Proof. We prove that there is an (X, x 0 ) such that * p e and * p e. Define : I X by (x) (1 x) for each x X. We draw a picture to illustrate how we arrive at the definition of the path homotopy H to show that * p e. (0, 1) (1, 1) S 1 L S 2 t (0, 0) 1 (, 0) (1, 0) 2 For each t I, we map the part of L that lies between S 1 and S 2 into the point f(1 t), we map the segment [(0, t), ((1 t) 2, t)] in the same manner that maps the interval [0, 1 t], and we map the segment [((1 t) 2, t), (1, t)] in the same

8.1 Homotopy of Paths 273 manner that maps the interval [t, 1]. Observing that 2((1 t) 2) 1 t and 2((1 t) 2) 1 t, we arrive at the formula { (2x), H(x, t) (1 t), (2x 1), It is easy to see that H(x, 0) ( * )(x) and H(x, 1) e(x) for all x I and H(0, t) H(1, t) x 0 for each t I. The proof that * p e is left as Exercise 3. We have proved that ( 1 (X, x 0 ), ) is a group. Definition. Let (X, ) be a topological space and let x 0 X. Then ( 1 (X, x 0 ), ) is called the fundamental group of X at x 0. The fundamental group was introduced by the French mathematician Henri Poincaré (1854 1912) around 1900. During the years 1895 1901, he published a series of papers in which he introduced algebraic topology. It is not true that algebraic topology developed as an outgrowth of general topology. In fact, Poincaré s work on algebraic topology preceded Hausdorff s definition of a topological space. Neither was Poincaré s work influenced by Cantor s theory of sets. Poincaré was surpassed only by Leonard Euler (1707 1783) as the most prolific writer of mathematics. Poincaré published 30 books and over 500 papers on mathematics, and he is regarded as the founder of combinatorial topology. Throughout the remainder of this text, we denote ( 1 (X, x 0 ), ) by 1 (X, x 0 ). The natural question to ask is whether the fundamental group of a topological space depends upon the base point x 0. This and other questions will be answered in the remaining sections of this chapter. We conclude this section with the definition of a contractible space and two theorems whose proofs are left as exercises. Definition. A topological space (X, ) is contractible to a point x 0 X if there is a continuous function H: X I X such that H(x, 0) x and H(x, 1) x 0 for each x X and H(x 0, t) x 0 for each t I. Then (X, ) is contractible if there exists x 0 X such that (X, ) is contractible to. THEOREM 8.7. Let (X, ) be a topological space and let (Y, ) be a contractible space. If f, g: X Y are continuous functions, then f is homotopic to g. THEOREM 8.8. Let (X, ) be a contractible space and let (Y, ) be a pathwise connected space. If f, g: X Y are continuous functions, then f is homotopic to g. x 0 if 0 x if 1 t 2 if 1 t 2 1 t 2 x 1 t 2 x 1.

274 Chapter 8 The Fundamental Group and Covering Spaces EXERCISES 8.1 ˇ 1. Let (X, ) be a topological space and let x 0 X. Prove that p is an equivalence relation on (X, x 0 ). 2. Let (X, ) be a topological space, let x 0 X, and define e: I X by e(x) x 0 for each x X. Show that if (X, x 0 ), then e * p. 3. Let (X, ) be a topological space, let x 0 X, and define e: I X by e(x) x 0 for each x X. Let (X, x 0 ), and define : I X by (x) (1 x) for each x X. Prove that * p e. 4. Let (X, ), (Y, ) and (Z, ) be a topological spaces, let f 1, f 2 : X Y be continuous functions such that f 1 f 2, and let g 1, g 2 : Y Z be continuous functions such that g 1 g 2. Prove that g 1 f 1 g 2 f 2. 5. Let n. subset of n is convex if for each x, y and t I, (1 t)x ty X. Let X be a convex subset of n and let and be paths in X such that (0) (0) and (1) (1). Prove that is path homotopic to. 6. Let (X, ) be a topological space and let f, g: X I be continuous functions. Prove that f is homotopic to g. 7. Let (X, ) be a pathwise connected space and let, : I X be continuous functions. Prove that is homotopic to. 8. (a) Prove that (with the usual topology) is contractible. (b) Prove that every contractible space is pathwise connected. 9. Prove Theorem 8.7. 10. Prove Theorem 8.8. 11. Let X be a convex subset of n and let x 0 X. Prove that X is contractible to. x 0 12. Let (G,, ) be a topological group (see Appendix I), and let e denote the identity of (G, ). For, (G, e), define by ( )(x) (x) (x). (a) Prove that is an operation on (G, e) and that ( (G, e), ) is a group. (b) Prove that the operation on (G, e) induces an operation on 1 (G, e) and that ( 1 (G, e), ) is a group. (c) Prove that the two operations and on 1 (G, e) are the same. (d) Prove that 1 (G, e) is abelian.

8.2 The Fundamental Group 275 8.2 The Fundamental Group In this section we establish some properties of the fundamental group. The first result is that if (X, ) is a pathwise connected space and x 0, x 1 X, then 1 (X, x 0 ) is isomorphic to 1 (X, x 1 ). Before we begin the proof, we draw a picture (Figure 8.6) to illustrate the idea. It is useful to look at this picture as you read the proof. X x 1 x 0 Figure 8.6 THEOREM 8.9. Let (X, ) be a pathwise connected space, and let x 0, x 1 X. Then 1 (X, x 0 ) is isomorphic to 1 (X, x 1 ). Proof. Let : I X be a continuous function such that (0) x 0 and (1) x 1, and define : I X by (x) (1 x) for each x X. Then define : 1 (X, x 0 ) 1 (X, x 1 ) by ([ ]) [( * ) * ] for each [ ] 1 (X, x 0 ). We want to prove that is an isomorphism. First observe that if (X, x 0 ), then ( * ) * (X, x 1 ). In order to show that is welldefined, it is sufficient to show that if and are members of (X, x 0 ) and P, then ( * ) * P ( * ) *. The proof of this fact is left as Exercise 1(a). Now let [ ], [ ] 1 (X, x 0 ). Then ([ ] [ ]) ([ * ]) [( * ( * )) * ], and ([ ]) ([ ]) [( * ) * ] [( * ) * ] [( * ) * ) * (( * ) * )]. Thus in order to show that is a homomorphism, it is sufficient to show that ( * ( * )) * p (( * ) * ) * (( * ) * ). The proof of this fact is left as Exercise 1(b). Now we show that is one-to-one. Suppose [ ], [ ] 1 (X, x 0 ) and ([ ]) ([ ]). Then [( * ) * ] [( * ) * ], and so ( * ) * p ( * ) *. In order to show that [ ] [ ], it is sufficient to show that p, and we leave the proof of this fact as Exercise 1(c). Now we show that maps 1 (X, x 0 ) onto 1 (X, x 1 ). Let [ ] 1 (X, x 1 ). Then (X, x 1 ) and so * ( * ) (X, x 0 ). Then ([ * ( * ]) [( * ( * ( * )) * ]. Thus it is sufficient to show that ( * ( * ( * ))) * p, and the proof of this fact is left as Exercise 1(d).

276 Chapter 8 The Fundamental Group and Covering Spaces EXAMPLE 4. Let X be a convex subset (see Exercise 5 of Section 8.1) of n. Then X is pathwise connected because if x, y, the function : I X defined by (t) (1 t)x ty is a path from x to y. Thus up to isomorphism the fundamental group of X is independent of the base point. Let x 0 X and let (X, x 0 ). Define H: I I X by H(x, t) tx 0 (1 t) (x). Then H(x, 0) (x) and H(x, 1) x 0 for each x X. Also H(0, t) tx 0 (1 t) (0) x 0 and H(1, t) tx 0 (1 t) (1) x 0. Therefore is path homotopic to the path e: I X defined by e(x) x 0 for each x X. Hence the fundamental group of X at x 0 is the trivial group (that is, the group whose only member is the identity element), and, by Theorem 8.9, if x is any member of X, then 1 (X, x) is the trivial group. Let (X, ) be a topological space, let x 0 X, and let C be the path component of X that contains x 0. Since the continuous image of a pathwise connected space is pathwise connected (see Exercise 7 of Section 3.2), we have the following results. If (X, x 0 ), then is also a member of (C, x 0 ). Also if H: I I X is a continuous function such that H(x, t) x 0 for some (x, t) I I, then H(I I) C. Thus 1 (X, x 0 ) 1 (C, x 0 ). Therefore 1 (X, x 0 ) depends only on the path component C of X that contains x 0, and it does not give us any information about X C. Let (X, ) be a pathwise connected space, and let x 0, x 1 X. While Theorem 8.9 guarantees that 1 (X, x 0 ) is isomorphic to 1 (X, x 1 ), different paths from x 0 to x 1 may give rise to different isomorphisms. If (X, ) is a pathwise connected space, it is common practice to speak of the fundamental group of X without reference to the basepoint. We show that the fundamental group is a topological property by introducing a homomorphism induced by a continuous function. Definition. A topological pair (X, A) is an ordered pair whose first term is a topological space (X, ) and whose second term is a subspace A of X. We do not distinguish between the topological pair (X, ) and the topological space (X, ). If x 0 X, we let (X, x 0 ) denote the topological pair (X, {x 0 }). Definition. If (X, A) and (Y, B) are topological pairs, a map f : (X, A) (Y, B) is a continuous function f : X Y such that f(a) B. THEOREM 8.10. Let (X, ) and (Y, ) be topological spaces, let x 0 X and y 0 Y, and let h: (X, x 0 ) (Y, y 0 ) be a map. Then h induces a homomorphism h * : 1 (X, x 0 ) 1 (Y, y 0 ). Proof. Let [ ] 1 (X, x 0 ). Then : I X is a continuous function such that (0) (1) x 0, so h : I Y is a continuous function such that (h )(0) (h )(1) y 0. Therefore [h ] 1 (Y, y 0 ). Define h * : 1 (X, x 0 ) 1 (Y, y 0 ) by h * ([ ]) [h ]. In order to show that h * is well-defined, it is sufficient to show that if, (X, x 0 ) and p, then h p h. Suppose

8.2 The Fundamental Group 277, (X, x 0 ) and p. Then there is a continuous function F: I I X such that F(x, 0) (x) and F(x, 1) (x) for all x I and F(0, t) F(1, t) x 0 for all t I. Define G: I I Y by G(x, t) (h F)(x, t). Then it is easy to see that G(x, 0) (h )(x) and G(x, 1) (h )(x) for all x I and G(0, t) G(1, t) y 0. Therefore h p h, and so h * is well defined. Now we show that h * is a homomorphism. Let [ ], [ ] 1 (X, x 0 ). Then h * ([ ] [ ]) h * ([ * ]) [h ( * )] and h * ([ ]) h * ([ ]) [h ] [h ] [(h ) * (h )]. Since (h ( * ))(x) { (h )(2x), if 0 x 1 2 (h )(2x 1), if 1 2 x 1 ((h ) * (h ))(x), h * is a homomorphism. In the remainder of this text, whenever we speak of the homomorphism induced by a continuous function, we mean the homomorphism defined in the proof of Theorem 8.10. Definition. If (X, A) and (Y, B) are topological pairs and f, g: (X, A) (Y, B) are maps such that f A g A, then f is homotopic to g relative to A, denoted by f g rel A, if there is a continuous function H: X I Y such that H(x, 0) f(x) and H(x, 1) g(x) for all x X and H(a, t) f (a) for all a A and t I. If x 0 X, then we write f g rel x 0 rather than f g rel {x 0 }. THEOREM 8.11. Let (X, ) and (Y, ) be topological spaces, let x 0 X and y 0 Y, and let h, k: (X, x 0 ) (Y, y 0 ) be maps such that h k rel x 0.Then h * k *. Proof. Since h k rel x 0, there is a continuous function H: X I Y such that H(x, 0) h(x) and H(x, 1) k(x) for all x X and H(x 0, t) y 0 for all t I. Let [ ] 1 (X, x 0 ). Then h * ([ ]) [h ] and k * ([ ]) [k ]. Define G: I I Y by G(x, t) H( (x), t) for all (x, t) I I. By Theorem 2.27, G is continuous. Note that G(x, 0) H( (x), 0) (h )(x) and G(x, 1) ( (x), 1) (k )(x) for each x I and G(0, t) H( (0), t) H(x 0, t) y 0 and G(1, t) H( (1), t) H(x 0, t) y 0 for each t I. Therefore h p k, and hence [h ] [k ]. Thus h * k *. THEOREM 8.12. Let (X, ), (Y, ), and (Z, ) be topological spaces, let x 0 X, y 0 Y, and z 0 Z, and let h: (X, x 0 ) (Y, y 0 ) and k: (Y, y 0 ) (Z, z 0 ) be maps. Then (k h) * k * h *. Proof. Let [ ] 1 (X, x 0 ). Then (k h) * ([ ]) [(k h) ] and (k * h * )[ ]) k * ([h ]) [k (h )]. Since the composition of functions is associative (see Exercise 21 in Appendix C), (k h) k (h ). Therefore (k h) * k * h *.

278 Chapter 8 The Fundamental Group and Covering Spaces Definition. Let (X, ) and (Y, ) be topological spaces and let x 0 X and y 0 Y. Then (X, x 0 ) and (Y, y 0 ) are of the same homotopy type if there exist maps f : (X, x 0 ) (Y, y 0 ) and g: (Y, y 0 ) (X, x 0 ) such that f g i Y rel y 0 and g f i X rel. x 0 THEOREM 8.13. Let (X, ) and (Y, ) be topological spaces and let x 0 X and y 0 Y. If (X, x 0 ) and (Y, y 0 ) are of the same homotopy type, then 1 (X, x 0 ) is isomorphic to 1 (Y, y 0 ). Proof. Suppose (X, x 0 ) and (Y, y 0 ) are of the same homotopy type. Then there exist maps f : (X, x 0 ) (Y, y 0 ) and g: (Y, y 0 ) (X, x 0 ) such that f g i Y rel y 0 and g f i X rel x 0. By Theorem 8.11, (f g) * : 1 (Y, y 0 ) 1 (Y, y 0 ) and (g f) * : 1 (X, x 0 ) 1 (X, x 0 ) are the identity homomorphisms. By Theorem 8.12, (f g) * f * g * and (g f) * g * f *. By Theorem C.2, f * : 1 (X, x 0 ) 1 (Y, y 0 ) is an isomorphism. THEOREM 8.14. Let (X, ) and (Y, ) be topological spaces, let h: X Y be a homeomorphism, let x 0 X, and let y 0 h(x 0 ). Then 1 (X, x 0 ) is isomorphic to 1 (Y, y 0 ). Proof. Since h is a homeomorphism, h 1 : Y X is a homeomorphism. Also note that x. Since h 1 h i and h h 1 0 h 1 (y 0 ) X i Y, (X, x 0 ) and (Y, y 0 ) are of the same homotopy type. Therefore by Theorem 8.13, 1 (X, x 0 ) is isomorphic to 1 (Y, y 0 ). EXERCISES 8.2 1. Let (X, ) be a topological space, let x 0, x 1 X, let : I X be a continuous function such that (0) x 0 and (1) x 1, and let, (X, x 0 ). (a) Prove that if, then ( * ) * p ( * ) *. (b) Prove that ( * ( * )) * p (( * ) * ) * (( * ) * ). (c) Prove that if ( * ) * p ( * ) *, then p. (d) Prove that ( * ( * ( * ))) * p. 2. A pathwise connected space (X, ) is simply connected provided that for each x 0 X, 1 (X, x 0 ) is the trivial group. Let (X, ) be a simply connected space and let, : I X be paths such that (0) (0) and (1) (1). Prove that p. 3. Prove that every contractible space is simply connected.

8.3 The Fundamental Group of the Circle 279 4. A topological space (X, ) is weakly contractible if there exists x 0 X and a continuous function H: X I X such that H(x, 0) x and H(x, 1) x 0 for each x X. (a) Give an example of a weakly contractible space that is not contractible. (b) Prove that every weakly contractible space is simply connected. 5. A subspace A of a topological space (X, ) is a retract of X if there exists a continuous function r: X A such that r(a) a for each a A. The function r is called a retraction of X onto A. Let A be a subspace of a topological space (X, ), let r: X A be a retraction of X onto A, and let a 0 A. Prove that r * : 1 (X, a 0 ) 1 (A, a 0 ) is a surjection. Hint: Consider the map j: (A, a 0 ) (X, a 0 ), defined by j(a) a for each a A. 6. Let (X, ) be a pathwise connected space, let x 0, x 1 X, let (Y, ) be a topological space, let h: X Y be a continuous function, let y 0 h(x 0 ) and y 1 h(x 1 ), and for each i 0, 1, let h i denote the map from the pair (X, x i ) into the pair (Y, y i ) defined by h i (x) h(x) for each x X. Prove that there are isomorphisms : 1 (X, x 0 ) 1 (X, x 1 ) and : 1 (Y, y 0 ) 1 (Y, y 1 ) such that (h 1 ) * (h 0 ) *. 8.3 The Fundamental Group of the Circle Our goal in this section is to show that the fundamental group of the circle is isomorphic to the group of integers. In order to do this we need several preliminary results. We begin with the definition of the function p that maps onto S 1. Define p: S 1 by p(x) ( cos 2 x, sin 2 x). One can think of p as a function that wraps around S 1. In particular, note that for each integer n, p is a one-to-one map of [n, n 1) onto S 1. Furthermore, if U is the open subset of S 1 indicated in Figure 8.7, then p 1 (U) is the union of a pairwise disjoint collection of open intervals. ( U ( Figure 8.7

280 Chapter 8 The Fundamental Group and Covering Spaces To be specific, let U {(x, y) S 1 : x 0 and y 0}. Then if x p 1 (U), cos 2 x and sin 2 x are both positive, so p 1 (U) n (n, n 1 4). Moreover, for each n, p is a one-to-one function from the closed interval [n, n 1 [n, n 1 onto 4] 4] U. Thus, since [n, n 1 is compact, p is a homeomorphism of [n, n 1 [n, n 1 4] 4] onto 4] U. Therefore p is a homeomorphism of (n, n 1 (n, n 1 4) onto U. Throughout the 4) remainder of this section, p denotes the function defined above. The following result is known as the Covering Path Property. You may wish to consider Example 5 before reading the proof. THEOREM 8.15. Let : I S 1 be a path and let x 0 such that p(x 0 ) (0). Then there is a unique path : I such that (0) x 0 and p. Proof. Since is continuous, for each t I there is a connected neighborhood U t of t such that (U is a proper subset of S 1 t ). Since I is compact, the open cover {U t : t I} of I has a finite subcover {U 1, U 2,..., U m }. If U i I for some i, then (I) is a connected proper subset of S 1. If U i I for any i, choose U i so that 0 U i. There exists t I such that t 0, [0, t ] U i, and t U i U j for some j i. If U i U j I, ([0, t ] ) and ([t, 1]) are connected proper subsets of S 1. If U i U j I, choose t I such that t t, [t, t ] U j and t U j U for some k (i k j). Since {U 1, U 2,..., U m } is a finite cover of I, after a finite number of steps, we will obtain t 0, t 1, t 2,..., t n such that 0 t 0, t n 1, t i 1 t i and ([t i 1, t i ]) is a connected proper subset of S 1 for each i 1, 2,..., n. For each i 0, 1, 2,..., n, let P i be the statement: There is a unique continuous function i : [0, t i ] such that i (0) x 0 and p i [0, ti ]. In order to prove the theorem, it is sufficient to show that P i is true for each i. It is clear that P 0 is true. Suppose 0 i n and P is true. Let V denote the component of p 1 i 1 i 1 ( [t i 1, t i ])) that contains i 1 (t i 1 ), and let p i 1 p Vi 1. Then p i 1 : V i 1 ([t i 1, t i ]) is a homeomorphism. Define i : [0, t i ] by Then i is continuous, i (0) x 0, and p i [0, ti ]. We must show that i is unique. Suppose i : [0, t i ] is a continuous function such that i (0) x 0 and p i [0, ti ]. If 0 t t i 1, then i (t) i (t) since P i 1 is true. Suppose t i 1 t t. Since (p i ) (t) (t) ([t i 1, t i ]), i (t) p 1 i ( ([t i 1, t i ])). Therefore since is continuous and V is a component of p 1 i (t) V i 1 i i 1 ( ([t i 1, t i ])). Therefore i (t), i (t) V i 1 and (p i 1 i )(t) (p i 1 i )(t). Hence i (t) i (t) since is a homeomorphism. p i 1 i (t) { i 1(t), if 0 t t i 1 (p i 1 ) 1 ( (t)), if t i 1 t t i. EXAMPLE 5. Let : I S 1 be a homeomorphism that maps I onto {(x, y) S 1 : x 0 and y 0} and has the property that (0) (1, 0) and (1) (0, 1) and let x 0 2. Then the function : I given by Theorem 8.15 is

8.3 The Fundamental Group of the Circle 281 a homeomorphism of I onto [2, 2.25], which has the property that (0) 2 and (1) 2.25. The following result is known as the Covering Homotopy Property. THEOREM 8.16. Let (X, ) be a topological space, let f : X be a continuous function, and let H: X I S 1 be a continuous function such that H(x, 0) (p f)(x) for each x X. Then there is a continuous function F: X I such that F(x, 0) f(x) and (p F)(x, t) H(x, t) for each (x, t) X I. Proof. For each x X, define x : I S 1 by x (t) H(x, t). Now (p f)(x) x (0), and hence, by Theorem 8.15, there is a unique path x : I such that x (0) f(x) and p x x. Define F: X I R by F(x, t) x (t). Then (p F)(x, t) (p x )(t) H(x, t) and F(x, 0) x (0) f(x). We must show that F is continuous. Let x 0 X. For each t I, there is a neighborhood M t of x 0 and a connected neighborhood N t of t such that H(M t N t ) is contained in a proper connected subset of S 1. Since I is compact, the open cover {N of I has a finite subcover. Let M n t : t I} {N t1, N t2..., N tn } i 1 M ti. Then M is a neighborhood of x 0, and, for each i 1, 2,..., n, H(M N ti ) is contained in a proper connected subset of S 1. Thus there exist t 0, t 1,..., t m such that t 0 0, t m 1, and, for each j 1, 2,..., m, t j 1 t j and H(M [t j 1, t j ]) is contained in a proper connected subset of S 1. For each j 0, 1,..., m, let P j be the statement: There is a unique continuous function G j : M [0, t j ] such that G j (x, 0) f(x) and (p G j )(x, t) H(x, t). We want to show that P j is true for each j. It is clear that P 0 is true. Suppose 0 j m and P is true. Let U be a connected proper subset of S 1 j 1 j 1 that contains H(M I j 1 ), let V denote the component of p 1 j 1 (U j 1 ) that contains G j 1 (M {t j 1 }), and let p j 1 p Vj 1. Then p j 1 : V j 1 U j 1 is a homeomorphism. Define G j : M [0, t j ] by G j (x, t) { G j 1(x, t), if 0 t t j 1 (p j 1 ) 1 (H(x, t)), if t j 1 t t j. Then G j is continuous, G j (x, 0) f(x), and (p G j )(x, t) H(x, t). We must show that G j is unique. Suppose G j : M [0, t j ] is a continuous function such that G j (x, 0) f(x) and (p G j )(x, t) H(x, t). If 0 t t j 1, then G j (x, t) G j (x, t) because P j 1 is true. Suppose t j 1 t t j. Since (p G j )(x, t) H(x, t) U j 1, G j (x, t) p 1 (U j 1 ). Therefore, since G j is continuous and V j 1 is a component of p 1 (U j 1 ), G j (x, t) V j 1. Hence G j (x, t) and G j (x, t) are members of V j 1 and (p j 1 G j )(x, t) (p j 1 G j )(x, t). Thus, since p j 1 is a homeomorphism, G j (x, t) G j (x, t). We have proved that there is a unique continuous function G: M I such that G(x, 0) f(x) and (p G)(x, t) H(x, t). Therefore G F (M I). Since M is a neighborhood of in X, F is continuous at (x 0, t) for each x 0

282 Chapter 8 The Fundamental Group and Covering Spaces t I. Since x 0 is an arbitrary member of X, F is continuous. This proof also shows that F is unique. If is a loop in S 1 at (1, 0), then p(0) (0), and hence, by Theorem 8.15, there is a unique path : I 1 such that (0) 0 and p. Since (p )(1) (1) (1, 0), (1) p 1 (1, 0), and hence (1) is an integer. The integer (1) is called the degree of the loop, and we write deg( ) (1). EXAMPLE 6. Define : I S 1 by (x) ( cos 4 x, sin 4 x). Then wraps I around S 1 twice in a clockwise direction. The unique path : I such that (0) 0 and p given by Theorem 8.15 is defined by (x) ˇ2x for each x I. Therefore deg( ) (1) 2. THEOREM 8.17. Let and be loops in S 1 1 2 at (1, 0) such that 1 p 2. Then deg( 1 ) deg( 2 ). Proof. Since there is a continuous function H: I I S 1 1 p 2, such that H(x, 0) 1 (x) and H(x, 1) 2 (x) for each x I and H(0, t) H(1, t) (1, 0) for each t I. By Theorem 8.15, there is a unique path 1 : I such that 1 (0) 0 and (p 1 ) 1. Thus, by Theorem 8.16, there is a unique continuous funtion F: such that (p F)(x, t) H(x, t) for each (x, t) I I and F(x, 0) 1 (x) for each x I. Define : I by (t) F(0, t) for each t I. Then is continuous, and (p )(t) (p F)(0, t) H(0, t) (1, 0) for each t I. Therefore (I) p 1 (1, 0), and, since (I) is connected and 1 (1, 0) is a discrete subspace of, is a constant function. Thus F(0, t) (t) (0) F(0, 0) 1 (0) 0 for each t I. Define 2 : I by 2 (x) F(x, 1) for each x I. Then 2 (0) F(0, 1) 0 and (p 2 )(x) (p F)(x, 1) H(x, 1) 2 (x) for each x I. By definition, deg( 1 ) 1 (1) and deg( 2 ) 2 (1). Now define a path : I by (t) F(1, t) for each t I. Again (p )(t) (p F)(1, t) H(1, t) (1, 0) for each t I, and hence (I) p 1 (1, 0). Therefore is a constant function, and hence F(1, t) (t) (0) F(1, 0) 1 (0) 0 for each t I. Therefore 2 (1) F(1, 1) F(1, 0) 1 (1), and hence deg( 1 ) deg( 2 ). We are ready to prove that the fundamental group of the circle is isomorphic to the group of integers. Since S 1 is pathwise connected, the fundamental group of S 1 is independent of the base point. THEOREM 8.18. 1 (S 1, (1, 0)) is isomorphic to the group of integers. Proof. Define : 1 (S 1, (1, 0)) by ([ ]) deg( ). By Theorem 8.17, is well-defined. Let [ 1 ], [ 2 ] 1 (S 1, (1, 0)). By Theorem 8.15, there are unique paths 1, 2 : I such that 1 (0) 2 (0) 0, p 1 1, and p 2 2. By definition, deg( 1 ) 1 (1) and deg( 2 ) 2 (1). Define : I by (x) { 1(2x), if 0 x 1 2 1 (1) 2 (2x 1), if 1 2 x 1.

8.4 Covering Spaces 283 Since 2 (0) 0, is continuous. Now (0) 1 (0) 0, and, since p( 1 (1) 2 (2x 1)) p( 2 (2x 1)), (p )(x) { (p 1)(2x), if 0 x 1 2 (p 2 )(2x 1), if 1 2 x 1 { 1(2x), if 0 x 1 2. 2 (2x 1), if 1 2 x 1 ( 1 * 2 )(x) Therefore ([ 1 ] [ 2 ]) ([ 1 * 2 ]) deg( 1 * 2 ) (1) 1 (1) 2 (1) deg( 1 ) deg( 2 ) ([ 1 ]) ([ 2 ]), and so is a homomorphism. Now we show that maps 1 (S 1, (1, 0)) onto. Let z, and define a path by for each. Then and so p : I S 1 1 : I (t) zt t I (0) 0 (1) z, is a loop in S 1 at (1, 0). Therefore [p ] 1 (S 1, (1, 0)), and, by definition, deg(p ) (1) z. Therefore ([p ]) deg(p ) z. Finally, we show that is one-to-one. Let [ 1 ], [ 2 ] 1 (S 1, (1, 0)) such that ([ 1 ]) ([ 2 ]). Then deg( 1 ) deg( 2 ). By Theorem 8.15, there are unique paths 1, 2 : I such that 1 (0) 2 (0) 0, p 1 1, and p 2 2. By definition, deg( 1 ) 1 (1) and deg( 2 ) 2 (1). So 1 (1) 2 (1). Define F: I I by F(x, t) (1 t) 1 (x) t 2 (x) for each (x, t) I I. Then F(x, 0) 1 (x) and F(x, 1) 2 (x) for each x I and F(0, t) 0 and F(1, t) 1 (1) for each t I. Therefore p F: I I S 1 2 (1) is a continuous function such that (p F)(x, 0) (p 1 )(x) 1 (x) and (p F)(x, 1) (p 2 )(x) 2 (x) for each x I and (p F)(0, t) p(0) (1, 0) and (p F)(1, t) (p 1 )(1) 1 (1) (1, 0) for each t I. Thus 1 p 2, so [ 1 ] [ 2 ]. There are no exercises in this section. The sole purpose of the section is to calculate the fundamental group of a familiar figure and thus provide you with a concrete example. In the next section, we generalize some of the theorems in this section and give some exercises. 8.4 Covering Spaces In this section we generalize Theorems 8.15 and 8.16 by replacing the function p: S 1 defined by p(x) (cos 2 x, sin 2 x) with a function called a covering map from an arbitrary topological space (E, ) into another topological space (B, ). These two generalized theorems are used in Section 9.4. Definition. Let (E, ) and (B, ) be topological spaces, and let p: E be a continuous surjection. An open subset U of B is evenly covered by p if p 1 (U) can be written as the union of a pairwise disjoint collection {V : } of open sets such that for each, p is a homeomorphism of V onto U. Each V is called a slice of p 1 (U).

284 Chapter 8 The Fundamental Group and Covering Spaces Definition. Let (E, ) and (B, ) be topological spaces, and let p: E B be a continuous surjection. If each member of B has a neighborhood that is evenly covered by p, then p is a covering map and E is covering space of B. Note that the function p: S 1 in Section 8.3 is a covering map. Covering spaces were introduced by Poincaré in 1883. Definition. Let (E, ), (B, ), and (X, ) be topological spaces, let p: E B be a covering map, and let f: X B be a continuous function. A lifting of f is a continuous function f : X E such that p f f (see Figure 8.8). E f' p X f Figure 8.8 B Notice that Theorem 8.15 provides a lifting of a path in S 1, where p is the specific function discussed in Section 8.3 rather than an arbitrary covering map. THEOREM 8.19. Let (E, ) and (B, ) be topological spaces, let p: E B be a covering map, let e 0 E, let b 0 p(e 0 ), and let : I B be a path such that (0) b 0. Then there exists a unique lifting : I E of f such that (0) e 0. Proof. Let {U : ˇ } be an open cover of B such that for each, U is evenly covered by p. By Theorem 4.4, there exists t 0, t 1,..., t n such that 0 t 0 t 1... t n 1 and for each i 1, 2,..., n, ([t i 1, t i ]) U for some. We define the lifting inductively. Define (0) e 0, and assume that (t) has been defined for all t [0, t i 1 ], where 1 i n. There exists such that ([t i 1, t i ]) U. Let {V : } be the collection of slices of p 1 (U ). Now (t i 1 ) is a member of exactly one member V of For define (t) (p V ) 1. t [t i 1, t i ], ( (t)). Since p V : V U is a homomorphism, is continuous on [t i 1, t i ] and hence on [0, t i ]. Therefore, by induction, we can define on I. It follows immediately from the definition of that p. The uniqueness of is also proved inductively. Suppose is another lifting of such that (0) e 0, and assume that (t) (t) for all t [0, t i 1 ], where 1 i n. Let U be a member of the open cover of B such that ([t i 1, t i ]) U, and let V be the member of chosen in the definition of. Since is a lifting of, ([t i 1, t i ]) p 1 (U ) V. Since is a collection of pairwise disjoint open sets and ([t i 1, t i ]) is connected, ([t i 1, t i ]) is a subset of one member of.

8.4 Covering Spaces 285 Since (t i 1 ) (t i 1 ) V, ([t i 1, t i ]) V. Therefore, for each t [t i 1, t i ], (t) V. But V p 1 p 1 ( (t)) ( (t)) { (t)}, and so (t) (t). Therefore (t) (t) for all t [0, t i ]. By induction,. THEOREM 8.20. Let (E, ) and (B, ) be topological spaces, let p: E B be a covering map, let e 0 E, let b 0 p(e 0 ), and let F: I I B be a continuous function such that F(0, 0) b 0. Then there exists a lifting F : I I E of F Such that F (0, 0) e 0. Moreover, if F is a path homotopy then F is also. Proof. Define F (0, 0) e 0. By Theorem 8.19, there exists a unique lifting F : {0} I E of F {0} I such that F (0, 0) e 0 and a unique lifting F : I {0} E of F I {0} such that F(0, 0) e 0. Therefore, we assume that a lifting F of F is defined on ({0} I) (I {0}) and we extend it to I I. Let {U : } be an open cover of B such that for each, U is evenly covered by p. By Theorem 4.4, there exists s 0, s 1,..., s m and t 0, t 1,..., t n such that 0 s 0 s 1... s m 1, 0 t 0 t 1... t n 1, and for each i 1, 2,..., m and j 1, 2,..., n, F([s i 1, s i ] [t j 1, t j ]) U for some. For each i 1, 2,..., m and j 1, 2,..., n, let U i J j [s i 1, s i ] [t j 1, t j ]. We define F inductively on the rectangles I i J j in the following order: I 1 J 1, I 2 J 1,..., I m J 1, I 1 J 2, I 2 J 2,..., I m J 2, I 1 J 3,..., I m J n. I 1 J n I 2 J n I m J n I 1 J 2 I 2 J 2 I m J 2 I 1 J 1 I 2 J 2 I m J 1 Suppose 1 p m, 1 q n, and assume that F is defined on C ({0} I) (I {0}) m i 1 q 1 j 1 (I i J j ) p 1 i 1 (I i J q ). We define F on I p J q. There exists such that F(I p J q ) U. Let {V : } be the collection of slices of p 1 (U ). Now F is already defined on D C (I p J q ). Since D is connected, F (D) is connected. Since is a collection of pairwise disjoint open sets, F (D) is a subset of one member V of. Now p V is a homeomorphism of V onto U. Since F is a lifting of F C, ((p V ) F )(x) F(x) for all x D. For x I define F (x) (p V ) 1 p I q, (F(x)). Then F is a lifting of F (C (Ip J q )). Therefore, by induction we can define F on I I. Now suppose F is a path homotopy. Then F(0, t) b 0 for all t I. Since F is a lifting of F, F (0, t) p 1 (b 0 ) for all t I. Since {0} I is connected, F is continu-

286 Chapter 8 The Fundamental Group and Covering Spaces ous, and p 1 (b 0 ) has the discrete topology as a subspace of E, F ({0} I) is connected and hence it must be a single point. Likewise, there exists b 1 B such that F(1, t) b for all, so F (1, t) p 1 1 t I (b 1 ) for all t I, and hence F ({1} I) must be a single point. Therefore F is a path homotopy. THEOREM 8.21. Let (E, ) and (B, ) be topological spaces, let p: E B be a covering map, let e 0 E, let b 0 p(e 0 ), let b 1 B, let and be paths in B from b 0 to b 1 that are path homotopic, and let and be liftings of and respectively such that (0) (0) e 0. Then (1) (1) and p. Proof. Let F: I I B be a continuous function such that F(x, 0) (x) and F(x, 1) (x) for all x I and F(0, t) b 0 and F(1, t) b 1 for all t I. By Theorem 8.20, there exists a lifting F : I I E of F such that F (0, t) e 0 for all t I and F ({1} I) is a set consisting of a single point, say e 1. The continuous function F (I {0}) is a lifting of F (I {0}) such that F (0, 0) e 0. Since the lifting of paths is unique (Theorem 8.19), F (x, 0) (x) for all x I. Likewise, F (I {1}) is a lifting of F (I {1}) such that F (0, 1) e 0. Again by Theorem 8.19, F (x, 1) (x) for all x I. Therefore (1) (1) e 1 and p. EXERCISES 8.4 (B, ) 1. Let (E, ) and (B, ) be topological spaces, and let p: E B be a covering map. Show that p is open. 2. Let (E 1, 1 ), (E 2, 2 ), (B 1, 1 ), and be topological spaces, and let p 1 : E 1 B 1 and p 2 : E 2 B 2 2 2 be covering maps. Prove that the function p: E 1 E 2 B 1 B 2 defined by p(x, y) (p 1 (x), p 2 (y)) is a covering map. 3. In this exercise, let p 1 denote the function p that is used throughout Section 8.3, and define p: S 1 S 1 by p(x, y) (p 1 (x), p 1 (y)). By Exercise 2, p is a covering map. Since the torus is homeomorphic to S 1 S 1, is a covering space of the torus. Draw pictures in and describe verbally the covering map p. 4. Let (X, ) be a topogical space, let Y be a set, let be the discrete topology on Y, and let 1 : X Y X be the projection map. Prove that 1 is a covering map. 5. Let (E, ) be a topological space, let (B, ) be a connected space, let n, and let p: E B be a covering map such that for some b 0 B, p 1 (b 0 ) is a set with n members. Prove that for each b B, p 1 (B) is a set with n members. 6. Let (E, ) be a topological space, let (B, ) be a locally connected, connected space, let p: E B be a covering map, and let C be a component of E. Prove that p C : C B is a covering map.

8.5 Applications and Additional Examples of Fundamental Groups 287 7. Let (E, ) be a pathwise connected space, let (B, ) be a topological space, let p: E B be a covering map, and let b B. (a) Prove that there is a surjection : 1 (B, b) p 1 (b). (b) Prove that if (E, ) is simply connected, then is a bijection. 8. The projective plane, which we have previously introduced, can also be defined as the quotient space obtained from S 2 by indentifying each point x of S 2 with its antipodal point x. Let p: S 2 P 2 denote the natural map that maps each point of S 2 into the equivalence class that contains it. Prove that P 2, defined in this manner, is a surface and that the function p is a covering map. 8.5 Applications and Additional Examples of Fundamental Groups We know that the fundamental group of a contractible space is the trivial group (see Exercise 8 of Section 8.1 and Exercises 2 and 3 Section 8.2) and that the fundamental group of the circle is isomorphic to the group of integers. In this section, we give some applications and some theorems that are useful in determining the fundamental group of a space. We begin by showing that S 1 is not a retract (see Exercise 5 of Section 8.2) of B 2 {(x, y) : x 2 y 2 1}. EXAMPLE 7. We show that B 2 is contractible. Define H: B 2 I B 2 by H((x, y), t) (1 t)(x, y). Then H is continuous, H((x, y), 0) (x, y) and H((0, 0), t) (0, 0) for each t I. S 1 THEOREM 8.22. is not a retract of. The diagram in Figure 8.9 is helpful in following the proof. B 2 1 (B 2 ) j * r * 1 (S 1 ) id Figure 8.9 1 (S 1 ) Proof. Suppose S 1 is a retract of B 2. Then there is a continuous function r: B 2 S 1 such that r(x) x for each x S 1. Define j: S 1 B 2 by j(x) x for each x S 1. Then r j: S 1 S 1 is the identity function, and hence ( j) * : 1 (S 1, (1, 0)) 1 (S 1, (1, 0)) is the identity homomorphism. By Theorem 8.12, (rj) * r * j *.Thus

288 Chapter 8 The Fundamental Group and Covering Spaces r maps onto. This is a contradiction, because 1 (B 2 1 (S 1 1 (B 2 *, (1, 0)), (1, 0)), (1, 0)) is the trivial group and 1 (S 1, (1, 0)) is isomorphic to the group of integers. Recall that a topological space (X, ) has the fixed-point property if for each continuous function f: X X there exists x X such that f(x) x. The following theorem is due to the Dutch mathematician L.E.J. Brouwer (1881 1966). THEOREM 8.23. (Brouwer Fixed-Point Theorem). B 2 has the fixed-point property. Proof. Suppose there is a continuous function f : B 2 B 2 such that f(x) x for any x B 2. For each x B 2, let L x denote the line segment that begins at f(x) and passes through x, and let y x L x S 1 { f(x)}. (Note that y x is uniquely determined see Figure 8.10.) L x f(x) x y x Figure 8.10 It is a straightforward but tedious exercise (see Exercise 4) to show that the function r : B 2 S 1 defined by r(x) y for each x B 2 x is continuous. Since r(x) x for each x S 1, r is a retraction of B 2 onto S 1. This contradicts Theorem 8.22. Now we introduce a property of subspaces that is stronger than retract. Definition. A subspace A of a topological space (X, ) is a deformation retract of X provided there is a continuous function H: X ˇI X such that H(x, 0) x and H(x, 1) A for all x X and H(a, t) a for all a A and t I. The homotopy H is called a deformation retraction. EXAMPLE 8. Let X {(x, y) : 1 x 2 y 2 4}. Define H: X I X by H(x, t) (1 t)x t x x for each (x, t) X I. Then H is continuous, H(x, 0) x and H(x, 1) x x S 1 for all x X, and H(x, t) x for all x S 1 and t I. Therefore H is a deformation retraction, and hence S 1 is a deformation retract of X. Using Example 8 and Theorem 8.18, the following theorem tells us that the fundamental group of the space X in Example 8 is isomorphic to the group of integers.

8.5 Applications and Additional Examples of Fundamental Groups 289 THEOREM 8.24. If A is a deformation retract of a topological space (X, ) and a A, then 1 (X, a 0 ) is isomorphic to 1 (A, a 0 ). Proof. Let H: X I X be a continuous function such that H(x, 0) x and H(x, 1) A for all x X and H(a, t) a for all a A and t I. Define h: X A by h(x) H(x, 1) for each x X, and let h * : 1 (X, a 0 ) 1 (A, a 0 ) be the homomorphism, given by Theorem 8.10, induced by h. This means that h * is defined by h * ([ ]) [h ] for all [ ] 1 (X, a 0 ). Now we show that h * is one-to-one. Suppose [ ], [ ] 1 (X, a ) and h * ([ ]) h * ([ ]). Then [h ] [h ]. Since A X, h may be considered as a path in X. We complete the proof that [ ] [ ] by showing that p h in X. Define F: I I X by F(x, t) H( (x), t) for all (x, t) I I. Then F(x, 0) H( (x), 0) (x) and F(x, 1) H( (x), 1) (h )(x) for all x I, and F(0, t) H( (0), t) H(a 0, t) a 0 and F(1, t) H( (1), t) H(a 0, t) a 0 for all t I. Since [ ] is an arbitrary member of 1 (X, a 0 ), this also shows that p h in X. Therefore, in 1 (X, a 0 ), we have [ ] [h ] [h ] [ ], and so h * is one-to-one. Now let [ ] 1 (A, 0 ). Then : I A and since A X, we may also consider to be a path in X. Let [ ] X denote the member of 1 (X, a 0 ) that contains. Since h(a) H(a, 1) a for all a A, (h )(x) (x) for all x I. Therefore h * ([ ] X ) [h ] [ ], and so h * maps 1 (X, x 0 ) onto 1 (A, a 0 ). Therefore h * is an isomorphism. By Example 8, S 1 is a deformation retract of X {(x, y) : 1 x 2 y 2 4}. Therefore, by Theorems 8.24 and 8.18, the fundamental group of X is isomorphic to the group of integers. EXAMPLE 9. Let p (p and let S {(x, y) :(x p 1 ) 2 1, p 2 ), R (y p 2 ) 2 1}. In Exercise 1, you are asked to show that S is a deformation retract of {p}. Thus it follows that the fundamental group of {p} is isomorphic to the group of integers. Since the torus is homeomorphic to S 1 S 1, we can determine the fundamental group of the torus by proving a theorem about the fundamental group of the Cartesian product of two spaces. The definition of the direct product of two groups is given in Appendix I. THEOREM 8.25. Let (X, ) and (Y, ) be topological spaces and let x 0 X and y 0 Y. Then 1 (X Y, (x 0, y 0 )) is isomorphic to 1 (X, x 0 ) 1 (Y, y 0 ). Proof. Let p: X Y X and q: X Y Y be the projection maps (since 1 is used as part of the symbol to denote the fundamental group of a space, we now use different symbols to denote the projection maps, but p and q are the functions that we have previously denoted by 1 and 2 ), and let p * : 1 (X Y, (x 0, y 0 )) 1 (X, x 0 ) and q * : 1 (X Y, (x 0, y 0 )) 1 (Y, y 0 ) denote the induced homeomorphisms. Define : 1 (X Y, (x 0, y 0 )) 1 (X, x 0 ) 1 (Y, y 0 ) by ([ ]) ([p ],

290 Chapter 8 The Fundamental Group and Covering Spaces [q ]). Since [p ] p * ([ ]) and [q ] q * ([ ]), by Exercise 12 of Appendix I, is a homomorphism. We use Exercise 13 of Appendix I in order to prove that is one-toone. Let [ ] 1 (X Y, (x 0, y 0 )) such that ([ ]) is the identity element of 1 (X, x 0 ) 1 (Y, y 0 ). This means that [p ] is the identity element of 1 (X, x 0 ) and [q f ] is the identity element of 1 (Y, y 0 ). Then p p e x0 and q p e y0, where e x0 and e y0 are the constant paths defined by e x0 (t) x 0 and e y0 (t) y 0 for all t I. Let F: I I X and G: I I Y be the homotopies such that F(x, 0) (p )(x) and F(x, 1) x 0 for all x I, F(0, t) F(1, t) x 0 for all t I, G(x, 0) (q )(x) and G(x, 1) y 0 for all x I, and G(0, t) G(1, t) y 0 for all t I. Define H: I I X Y by H(x, t) (F(x, t), G(x, t)) for all (x, t) I I. Then H(x, 0) (F(x, 0), G(x, 0)) ((p )(x), (q )(x)) (x) and H(x, 1) (F(x, 1), G(x, 1)) (x 0, y 0 ) for all x X, and H(0, t) (F(0, t), G(0, t)) (x 0, y 0 ) and H(1, t) (F(1, t), G(1, t)) (x 0, y 0 ) for all t I. Therefore p e, where e is the constant path defined by e(t) (x 0, y 0 ) for all t I. Therefore by Exercise 13 of Appendix I, is one-to-one. Now we show that maps 1 (X Y, (x 0, y 0 )) onto 1 (X, x 0 ) 1 (Y, y 0 ). Let ([ ], [ ]) 1 (X, x 0 ) 1 (Y, y 0 ). Define : I X Y by (t) ( (t), (t)) for all t I. Then [ ] 1 (X Y, (x 0, y 0 )), and ([ ]) ([p ], [q ]) ([ ], [ ]), and so is an isomorphism. If A is a subset of a set X, the inclusion map i: A X is the function defined by i(a) a for each a A. A homomorphism from a group G into a group H is called the zero homomorphism if for each g G, (g) is the identity element of H. The following theorem is a special case of a famous theorem called the Van Kampen Theorem. THEOREM 8.26. Let (X, ) be a topological space, let U and V be open subsets of X such that X U V and U V is pathwise connected. Let x 0 U V, and suppose the inclusion maps i: U X and j: V X induce zero homomorphisms i * : 1 (U, x 0 ) 1 (X, x 0 ) and j * : 1 (V, x 0 ) 1 (X, x 0 ). Then 1 (X, x 0 ) is the trivial group. Proof. Let [ ] 1 (X, x 0 ). By Theorem 4.4, there exist t 0, t 1,..., t n such that t 0 0, t n 1, and, for each k 1, 2,..., n, t k 1 t k and ([t k 1, t k ]) is a subset of one of U or V. Among all such subdivisions t 0, t 1,..., t n of I, choose one with the smallest number of members. We prove by contradiction that for each k 1, 2,..., n 1, f(t k ) U V. (Note that if (I) is a subset of one of U or V, then n 1, and we have nothing to prove.) Suppose k {1, 2,..., n 1} and (t k ) U. Then neither ([t k 1, t k ]) nor ([t k, t k 1 ]) is a subset of U, so ([t k 1, t k ]) ([t k, t i 1 ]) V. Thus we can discard t k and obtain a subdivision with a smaller number of members that still have the desired properties. Therefore for each k 1, 2,..., n 1, (t k ) U. The same argument shows that (t k ) V for each k 1, 2,..., n 1. Since (t 0 ) (t n ) x 0 U V, (t k ) U V for each k 0, 1,..., n.