1 Introduction When are two spaces X; Y homeomorphic, that is, when is there a continuous map f : X ! Y with continuous inverse? Let f; g : X ! Y be continuous maps. f is homotopic to g if there exists a homotopy, H : X × I ! Y such that H(x; 0) = f(x) and H(x; 1) = g(x). Put Ht : X ! Y by Ht(x) = H(x; t), then H0 = f; H1 = g. Notation: f; g; ft with f0 = f; f1 = g from Hatcher. If f is homotopic to g, then we write f ' g. Similarly for spaces. Let X; Y be two spaces. They have the same homotopy type if there are continuous f : X ! Y and g : Y ! X such that g◦f : X ! X and f ◦g : Y ! Y are homotopic to the appropriate identity maps. Examples: In Rn, we have Bn = Dn = fx 2 Rn : kxk ≤ 1g. Bn has the homotopy type of a point, for example, f0g. Let i : f0g ! Bn be the inclusion, and r : Bn ! f0g by r(x) = 0. n n n Then r◦i = idf0g and i◦r : B ! B is homotopic to idBn by H : B ×I ! I by H : Bn × I ! Bn, H(x; t) = (1 − t)x. But for n > 1, Sn−1 = fx 2 Rn : kxk = 1g are not homotopy equivalent to a point. 2 Fundamental Group Definition 2.1 (Path). Let X be a topological space. A path in X is f : I ! X a continuous map. Two paths from x0 to x1 are homotopic if there exists H : I × I ! S such that H(s; 0) = f(s), H(s; 1) = g(s), H(0; t) = x0 and H(1; t) = x1. In Euclidean space (in fact, any convex subspace of Euclidean space) every path is homotopic to a straight line path, by a straight line homotopy. Let f be a path from x0 to x1 and g be a path from x1 to x2, then f · g is a f(2t) 0 ≤ t ≤ 1=2 new path from x to x defined by (f · g)t) = 0 2 g(2t − 1) 1=2 ≤ t ≤ 1 If x0 2 X, then let cx0 : I ! X be the constant path cx0 (t) = x0. Easy: If f0 ' f1 from x0 to x1 and g0 ' g1 from x1 to x2, then f0 ·g0 ' f1 ·g1. Definition 2.2 (Loop). Fix x0 2 X and call x0 the base point. A loop at x0 is a path f in X such that f(0) = f(1) = x0. Let [f] be all loops at x0 which are path homotopic to f. We denote this relation by f 'p g. Definition 2.3. We define π1(X; x0) = f[f]: f is a loop at x0g (We can also define π0(X; x0) = all path components of X with base point being path component of x0.) Example: O(2) has two path components, the base point is x0 = I. 2 What is π1(O(2);I)? SO(2) is the set of all rotations of R , and it is home- omorphic to the circle. We claim that π1(X; x0) is a group, and call it the Fundamental Group of X at x0. 1 Theorem 2.1. π1(X; x0) with composition as above forms a group. −1 Proof. [f][g] = [f · g] is the product. Take the identity to be [cx0 ]. For [f] , ¯ ¯ we take [f] where f = f(1 − t). We must check (f · g) · g 'p f · (g · h), ¯ ¯ f · cx0 'p f 'p cx0 · f and f · f 'p f · f 'p cx0 . We can do these using reparameterizations. If f : I ! X is any path, and γ : I ! I is continuous with γ(0) = 0 and γ(1) = 1, then f(γ(t)) is a path which is homotopic to f. F (s; t) = f((1 − t)γ(s) + ts). Now, see page 27. Lemma 2.2 (Pasting Lemma). Let X be a topological space such that X = A1 [ ::: [ A`, each Ai is closed. Let f : X ! Y be a function to another topological space. If 8i, fjAi : Ai ! Y is continuous, then so is f. Proof. Let C be a closed set in Y . Then f −1(C) = fj−1(C) [ ::: [ fj−1(C). A1 A` Each term in this union is closed in the appropriate Ai, and as Ai is closed in X, each term is itself closed in X. As the finite union of closed sets is closed, f −1(C) is closed in X, and so f is continuous. Let f; g; h be paths in X such that (f · g) · h is defined. We want to show that this composition is associative. That is, that (f · g) · h 'p f · (g · h). The right hand side is also equal to [(f · g) · h] ◦ γ, where 8 1 1 2 t 0 ≤ t ≤ 2 < 1 1 3 γ = t − 2 2 ≤ t ≤ 4 : 3 2t − 1 4 ≤ t ≤ 1 8 f( 4s ) 0 ≤ s ≤ (t + 1)=4 < t+1 And so, we define H(s; t) = g(4s − t − 1) (t + 1)=4 ≤ s ≤ (t + 2)=t : 4s−2−t h( 2−t )(t + 2)=4 ≤ s ≤ 1 Change of Base Point Let X be a topological space, and x0; x1 2 X such that there is a path h ¯ from x0 to x1. Then 9βh : π1(X; x1) ! π1(X; x0) given by βh([f]) = [h · f · h] which is an isomorphism of groups. And so, if X is path-connected, we can write π1(X). Definition 2.4 (Simply-Connected). Call X simply connected if X is path connected and π1(X) is trivial. Quotient Topology I = [0; 1], and we want to identify 0 ∼ 1. ≈ So I= ∼ is a space, and we believe it is homeomorphic (!) the circle S1 ⊂ C. So we define p : I ! S1 such that p(t) = e2πit. p(0) = p(1) = 1 2 S1 ⊂ C. Definition 2.5 (Quotient Topology). More generally, let X be a topological space and ∼ be an equivalence relation. Let g : X ! X= ∼ take x to [x]. Call V ⊂ X= ∼ open iff g−1(V ) is open in X. Then we get a topology on X= ∼ and we get that 2 g ... X ................................................................................................... X= ∼ ..... ..... 8f ..... ..... ..... .. ...... ...... .......... ....... 9h Y 1 Proposition 2.3. π1(S ; 1) ' Z. 1 Proof. We want to define a homomorphism Φ : Z ! π1(S ) be Φ(n) = [!n] 2πins where !n(s) = e . We need to show that this is a homomorphism. Let p : R ! S1 by p(s) = 2πis e . Note that !n = p ◦ !~n, where! ~n : I ! R is given by! ~n(s) = ns. R ........ ...... ..... ..... ..... ..... ..... !~n ..... ..... ..... p ..... ..... ..... ..... ..... ..... ..... ..... ..... ... !n ... I ............................................................................................................. S1 ~ ~ Φ(n) = [p ◦ f] whenever f is any path in R from 0 to n. Then! ~n must be homotopic to f~. And so, Φ is a homomorphism. Let τm : R ! R be a translation, τm : x 7! x + m. Then! ~m · τm!~n is a path in R from 0 (to m) to n + m. Thus Φ(n + m) = [p ◦ (! ~m · τm!~n] = [p ◦ !~m] · [p(τm!~n)] = Φ(m) · Φ(n). We need to show that Φ is surjective and that ker Φ = f0g. We will need two facts. 1 −1 1. If f : I ! S starting at x0 2 X and letx ~0 2 p (x0), then 9 unique lift ~ f : I ! R starting atx ~0. 1 This implies that Φ is surjective, as we let [f] 2 π1(S ; 1) and taking ~ ~ x0 = 1; x~0 = 0, we apply it and get f. Then p(f(1)) = f(1) = 1. So ~ f~(1) = m 2 Z, e2πif(1) = 1 so Φ(m) = [p ◦ f~] = [f]. 1 1 2. Let F : I×I ! S be a homotopy of paths, all starting at x0 2 S . Choose −1 ~ somex ~0 2 p (x0). Then, 9 a unique lifted homotopy F : I × I ! R of paths starting atx ~0. (R; x~0) ....... ..... ..... ....... ... ..... .. ~ .. F ..... p . .... ... ..... .. .. ..... F .... 1 I × I ............................................................................ (S ; x0) ~ We note that p ◦ F~ = F , and so p ◦ ft = ft. The second fact implies that Φ is injective. Suppose that Φ(n) = e = [c1] 2 (S1; 1). Assume that n 6= 0. Then [!n] = e = [c1]. Then !n is path homotopic to the constant map, and so, by the second fact, we can lift this homotopy 3 to! ~n 'p !~0 = c0, but these have different endpoints, and so cannot be path homotopic, which is a contradiction. All that remains is to prove these two facts. 1 Observe the property of p : R ! S : there exists an open cover fUαg 1 −1 of S such that p (Uα) = [iVαi where the Vαi are disjoint open sets and pj : V ! U is a homeomorphism. Vαi αi α One such cover is the covering of S1 by two open sets with appropriate overlap. We will first prove fact 1: (R; x¯0) ....... ..... ... ..... .. ~ .. f ..... p .... ... ..... .. .. ........ ..... f ... .... 1 (I; 0) ............................................................................ (S ; x0) The Lebesgue Covering Lemma states that if X is a compact metric space and fWαg is an open cover of X, then there exists λ > 0 such that every subset of X with diameter less than λ lies in some Wα. So, there exists a partition 0 = t0 < t1 < : : : ; t` = 1 such that f([ti−1; ti]) ⊂ ~ some Uα. We construct a lifting f on [0; tk] by induction on k. We start with ~ ~ ~ t0, [0; 0], f(0) =x ~0 Assume k < `, have f such that p ◦ f = f. We need ~ ~ to define f on [tk; tk+1].