Fundamental group. Chapter The loop space Ω(X, x 0 ) and the fundamental group

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1 Chapter 6 Fundamental group 6. The loop space Ω(X, x 0 ) and the fundamental group π (X, x 0 ) Let X be a topological space with a basepoint x 0 X. The space of paths in X emanating from x 0 is the space P (X, x 0 ):={ω X I : ω(0) = x 0 }. The space of loops based at x 0 is the subspace Ω(X, x 0 ):={ω X I : ω(0) = ω() = x 0 }. There is a natural addition of loops: for ω, ω 2 Ω(X, x 0 ), their concatenation ω ω 2 = ω is given by { ω (2t), 0 t ω(t) =, 2 ω 2 (2t ), t. 2 Two loops ω 0 and ω are homotopic if there is a path in Ω(X, x 0 ) connecting ω 0 to ω, i.e., a homotopy H : I I X satisfying H(t, 0) = ω 0 (t), H(t, ) = ω (t); H(0,s)=x 0, H(,s)=x 0. The set of homotopy classes of loops in X with basepoint x 0 is denoted by π (X, x 0 ). The following proposition justifies calling this the fundamental group of X (with basepoint x 0 ). Proposition 6.. The loop addition induces a group structure on π (X, x 0 ).

2 602 Fundamental group Proof. () Associativity. (ω ω 2 ) ω 3 and ω (ω 2 ω 3 ), though not strictly equal (as functions I X), are homotopic. ω ω 2 ω 3 t ω ω 2 ω ω ω 2 ω 3 (2) Identity. The constant path ω 0 : I X (with ω 0 (t) =x 0 for every t I) satisfies ω ω 0 ω ω 0 ω for every ω Ω(X, x 0 ). ω ω 0 ω t ω 0 ω t ω ω ω ω 0 0 ω (3) Inverse. For ω : I X, let ω be the loop defined by ω(t) = ω( t), t I. Then ω ( ω) ω 0 ( ω) ω.

3 6. The loop space Ω(X, x 0 ) and the fundamental group π (X, x 0 ) 603 ω 0 = ω ω ω t ω [, t] ω t ω [ t,] t ω [0,t] ω t ω [t,0] ω ω 0 ω 0 = ω Proposition 6.2. π ( ) is a covariant functor from the category of spaces (with basepoints and basepoint preserving maps) to the category of groups (and homomorphisms). If f,g :(X, x 0 ) (Y,y 0 ) are homotopic rel endpoints, then f = g. Proposition 6.3. If f :(X, x 0 ) (Y,y 0 ) is a homotopy equivalence, then f : π (X, x 0 ) π (Y,y 0 ) is an isomorphism. 6.. Conjugacy of fundamental groups Let x 0, x X be in the same path component, i.e., there is a path α : I X with α(0) = x 0 and α() = x. Such a path induces an isomorphism α : π (X, x 0 ) π (X, x ), namely, α ([ω]) = [ω ], with α( 3t), 0 t, 3 ω(t) = ω(3t ), 3 t 2, 3 2 α(3t 2), t. 3 We shall simply write π (X) for π (X, x 0 ) when there is no danger of confusion of the basepoint, or when only the isomorphism class of the fundamental group is relevant. A connected space X is simply connected if π (X) =0. Clearly, a contractible space is simply connected. Also, S n is simply connected when n 2.

4 604 Fundamental group 6.2 The fundamental group of the circle The first nontrivial example of fundamental group is that of the circle. We take as basepoint of S the complex number. Theorem 6.4. π (S, ) is isomorphic to the additive group of integers. Proof. Consider the exponential map exp : R S given by exp(t) =e 2πit,t R. Clearly, exp () = Z. A loop in Ω(S, ) is given by a map ω : I S satisfying ω(0) = ω() =. Such a loop is the image under exp of a unique path ω : I R with initial point ω(0) = 0. Note that ω() is an integer, which we denote by deg(ω), and call the degree of ω. () deg(ω) depends only on the homotopy class of ω. In other words, if ω Ω(S, ) is homotopic to ω rel endpoints, then deg(ω ) = deg(ω). Therefore we have a function deg : π (S, ) Z. (2) Clearly, every integer n is equal to deg(ω) for some loop ω in S. If n 0, the path f n : I R given by f n (t) =nt clearly projects under exp into a loop in S (which winds around the circle n times, counterclockwise or clockwise according as n is positive or negative). If n = 0, we simply take the constant loop. This shows that deg is surjective. (3) On the other hand, deg is injective. This follows from the fact that any two paths in R with the same endpoints are homotopic rel endpoints. Therefore, deg : π (S, ) Z is a bijection. (4) Consider two loops ω and ω 2 in Ω(S, ). If these are covered under exp by paths ω, ω 2 with ω () = m and ω 2 () = n, then the loop ω ω 2 is covered by the path which is the concatenation of ω and ω 2 : { ω (2t), 0 t ω(t) =, 2 m + ω 2 (2t ), t. 2 Clearly, deg([ω ] [ω 2 ]) = deg([ω ω 2 ]) = ω() = m+n =deg([ω ])+deg([ω 2 ]). This means that deg : π (S, ) is bijective, it is an isomorphism. Z is a homomorphism. Since deg

5 6.3 Fundamental theorem of algebra Fundamental theorem of algebra Theorem 6.5. Every nonconstant polynomial with complex coefficients has a zero in C. Proof. We deduce a contradiction by assuming a polynomial p(z) =z n + a z n + + a n z + a n without zero. Considered the polynomial as a map p : S C {0}. Write p(z) =z n + f(z). We show that for sufficiently large r>0,if z = r, z n + tf(z) is nonzero for t [0, ]. This follows from z n + tf(z) > z n t f(z) > z n f(z) n > z n a k z n k k= ( n ) > z n a k z n assuming z > ( = r k= ) n a k r n, k= which is nonzero if we choose r =max( n k= a k, ). Thus, via the homotopy p t : S [0, ] C {0} given by p t (z) = (rz) n + tf(rz), the maps z p(rz) and z (rz) n are homotopic. Regarded as a map S S, the latter map clearly has degree n. This is a contradiction since the map z p(rz) is homotopic to the constant map z a n via the homotopy z p(trz).

6 606 Fundamental group

7 Chapter 7 Covering spaces 7. Covering maps The exponential map exp : R S is an example of a covering map. We say that a map p : E B is a covering map if each b B has a neighborhood U such that p (U) is the disjoint union of open sets each of which is mapped homeomorphically onto U by p. p (U) p U U p (b) π Here are some examples of covering maps. () The exponential map exp : R S : exp(t) =e 2πit. (2) The plane R 2 covering the torus: R 2 S S. (3) S n RP n is a double covering. Proposition 7. (Path lifting property). Let ω be a path in B with initial point b and e E covering b, i.e, p(e) =b. There is a unique lifting ω of ω with initial point e E. Proposition 7.2 (Homotopy lifting property). Let h t : I B be a homotopy of free paths in B, and ω a path in E covering the ititial points h t (0). Then there is a unique homotopy h t covering h t.

8 608 Covering spaces Proof of path lifting property Let {U b : b B} be a family of coordinate neighborhoods. Consider a path ω : I B. The family {ω (U b ): b B} is an open cover of I, say with Lebesque number ε. Choose n so that <ε, and subdivide I equally into n equal subintervals [t k,t k+ ] for 0 k n. n Each [t k,t k+ ] is contained in some ω (U b ). In particular, [t 0,t ] ω (U ω(0) ). There is a unique neighborhood W of e homeomorphic to U ω(0), and subsequently a unique lifting of ω [t0,t ] to W : W ω [t 0,t ] U ω(0) ω [t0,t ] Iterating, we obtain a unique lifting over each [t k,t k+ ] for k =,...,n. These together give a unique lifting of ω over I. W p W 7.2 Covering spaces and fundamental groups Proposition 7.3. (a) Monodromy. If ω and ω 2 are homotopic paths (rel endpoints), and ω and ω 2 their liftings with the same initial point, then ω and ω 2 have the same endpoint. (b) If B is path-connected, all p (b) have the same cardinality. (c) Let b 0 B and F = p (b 0 ). The fundamental group π (B,b 0 ) acts as a group of permutations of F. Proof. (a) Let h t be a homotopy of paths ω ω 2 rel endpoints b 0 and b. By the homotopy lifting property, there is a homotopy h t covering h t. The path h has image in the discrete set p (b ). It must be constant. This means ω and ω 2 have the same endpoint. (b) Let ω be a path in B connecting b 0 and b. By the unique path lifting property, this induces a map ω : p (b 0 ) p (b ) given by ω (e) = ω e (), where ω e is the unique lifting of ω with initial point e p (b 0 ). This is a bijection with inverse ω similarly defined.

9 7.3 Universal covering 609 (c) Let F = p (b 0 ) and ω be a loop with ω(0) = ω() = b 0. By (a), for each e F, ω e () depends only on the homotopy class of ω in π (B,b 0 ). There is a map φ : F π (B,b 0 ) F defined by φ(e, [ω]) = ω e (). This satisfies (i) φ(e, [ω ] [ω 2 ]) = φ(φ(e, [ω ]), [ω 2 ]), (ii) φ(e, [ω 0 ]) = e for the constant path ω 0. In particular, if e 0 F, the map φ 0 : π (B,b 0 ) F given by φ 0 ([ω]) = φ(e 0, [ω]) is onto. This is because for every e F and a path ω in E joining e 0 to e,ifweputω = p ω, this is a loop at b 0 and clearly φ 0 ([ω]) = φ(e 0, [ω]) = ω e0 () = e. 7.3 Universal covering A covering map p : E connected. B is universal if the total space E is simply Proposition 7.4. If p : E B is a universal covering, there is a one-to-one correspondence π (B,b 0 ) F. Proof. If E is simply connected, we show that φ 0 is one-to-one. Suppose φ 0 ([ω 0 ]) = φ 0 ([ω ]). This means that the liftings of ω 0 and ω (with common initial point e 0 ) have the same endpoint e. Since E is simply connected, there is a homotopy f t : I E satisfying f t (0) = e 0 and f t () = e for t [0, ]. It is clear that p f t is a homotopy of ω 0 and ω. This shows that φ 0 is a one-to-one correspondence. Remark. If E is simply connected, the one-to-one correspondence φ 0 makes F into a group with identity e 0.Fore,e 2 F, let ω be a loop corresponding to e. Find a lifting ω with initial point e 2. The end point of this lifting is the product e e 2. Proposition 7.5. (a) π (S S )=Z Z. (b) π (RP n )=Z 2 for n 2.

10 60 Covering spaces 7.4 Borsuk-Ulam theorem for S 2 Theorem 7.6 (Borsuk-Ulam). There is no continuous map f : S 2 S satisfying f( x) = f(x). Proof. We deduce a contradiction by assuming an antipodal map f : S 2 S. Such a map induces g : RP 2 S making the diagram f S 2 S p 2 p RP 2 S commutative. Let ω be a path in S 2 whose endpoints are antipodal. Since f is antipodal, the endpoints of f ω are antipodal on S. Note that p 2 ω and p f ω are loops in RP 2 nd S respectively. Now [p 2 ω] and [p f ω] are nontrivial elements in the fundamental groups since they acts nontrivially on p 2 (x 0) and p (y 0). This contradicts the commutativity of the diagram above since g : π (RP 2 ) π (S ) is trivial. Corollary 7.7. (a) Let f : S 2 R 2 be a continuous map satisfying f( x) = f(x). Then there exists x S 2 such that f(x) =0. (b) Let f : S 2 R 2 be a continuous map. There exists x S 2 such that f(x) =f( x). g

11 Chapter 8 Calculation of fundamental groups 8. Van Kampen theorem Consider a space X with basepoint x 0. Suppose X = X X 2, both X, X 2 containing the basepoint x 0. Theorem 8. (van Kampen theorem). If X X 2 is path-connected and X = int X int X 2, then i 2 π (X X 2,x 0 ) π (X 2,x 0 ) i π (X,x 0 ) π (X, x 0 ) j j 2 is a pushout diagram. This means π (X, x 0 ) is isomorphic to the quotient F/R where (i) F is the free product of π (X,x 0 ) and π (X 2,x 0 ), and (ii) R is the normal subgroup of F generated by words of the form i (x)i 2 (x), x π (X X 2,x 0 ). In particular, if X X 2 is simply connected, then π (X, x 0 ) is the free product of π (X,x 0 ) and π (X 2,x 0 ).

12 62 Calculation of fundamental groups Examples () Consider the one-point union of two circles, S S. The fundamental group π (S S ) is the free group on 2 generators. (2) π (S n )=0for n Fundamental group from triangulations