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MAT 531: Topology&Geometry, II Spring 2011

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MAT 531: Topology&Geometry, II Spring 2011 MAT 531: Topology&Geometry, II Spring 2011 Solutions to Problem Set 7 Problem 1 (15pts) Let X be a path-connected topological space and (S∗(X),∂) the singular chain complex of continuous simplices into X with integer coefficients. Denote by H1(X; Z) the corresponding first homology group. (a) Show that there exists a well-defined surjective homomorphism h: π1(X,x0) −→ H1(X; Z). (b) Show that the kernel of this homomorphism is the commutator subgroup of π1(X,x0) so that h induces an isomorphism Φ: π1(X,x0) π1(X,x0),π1(X,x0) −→ H1(X; Z). This is the first part of the Hurewicz Theorem. The motivation for this result is that π1(X,x0) is generated by loops based at x0 ∈X, i.e. continuous maps α: I −→X such that α(0)=α(1)=x0, while H1(X; Z) is generated by formal linear combinations of 1-simplicies, i.e. continuous maps f :∆1 =I −→ X. In particular, a loop (as well as any path) in X is a 1-simplex. However, the equivalence relations on paths and 1-simplicies used to define π1(X,x0) and H1(X; Z) and the groups structures are quite different. So we will need to show that equivalent paths are equivalent as 1-simplicies and a product of two paths corresponds to the sum of the two 1-simplicies. We will denote the path-homotopy equivalence class of a path α (loop or not) by [α] and the image of a 1-simplex in S1(X)/∂S2(X) by {α}. It will be essential to distinguish between a point x0 ∈ X and k the k-simplex taking the entire standard k-simplex ∆ to x0. Denote the latter by fk,x0 . Lemma 0: If α: I −→X is a loop, ∂α=0. Lemma 1: If x0 ∈X, f1,x0 ∈∂S2(X). Lemma 2: If α,β : I −→X are path-homotopic, then α−β ∈∂S2(X). Lemma 3: If α,β : I −→X are paths such that α(1)=β(0), then α+β−α∗β ∈∂S2(X). Lemma 4: If α : I −→X andα ¯ : I −→X is its inverse, then α+α ¯ ∈∂S2(X). Lemma 5: If F :∆2 −→X is a 2-simplex, then 2 2 2 (F ◦ι0)∗(F ◦ι1)∗(F ◦ι2) = [id] ∈ π1 X, F (1, 0) . 1 2 First, recall the maps ιj and ιj used to define the boundaries of 1- and 2-simplicies: 1 0 1 1 1 ιj :∆ −→ ∆ , ι0(0) = 1, ι1(0) = 0; 2 1 2 2 2 2 ιj :∆ −→∆ , ι0(s)=(1−s,s), ι1(s)=(0,s), ι2(s)=(s, 0) ∀ s∈I; 2 ι0 2 2 ι1 1 ι0 1 ι1 2 0 0 1 0 1 ι2 0 1 1 0 1 2 1 2 Figure 1: The boundary maps ιj :∆ −→∆ and ιj :∆ −→∆ . see Figure 1. These maps respect the orders of the vertices. By the above, if α is a loop based at x0, 1 1 ∂α = α ◦ ι0 − α ◦ ι1 = f0,α(1) − f0,α(0) = f0,x0 − f0,x0 = 0. For Lemma 1, note that 2 2 2 ∂f2,x0 = f2,x0 ◦ ι0 − f2,x0 ◦ ι1 + f2,x0 ◦ ι2 = f1,x0 − f1,x0 + f1,x0 = f1,x0 , 2 since f2,x0 ◦ ιj maps all of I to x0. For Lemma 2, choose a path-homotopy from α to β, i.e. a continuous map F : I ×I −→ X s.t. F (s, 0) = α(s), F (s, 1) = β(s), F (0,t)= F (1,t) ∀ s,t∈[0, 1]. There is a quotient map q : I ×I −→ ∆2 s.t. q(s, 0)=(s, 0), q(s, 1) = q(0,s), q(0,t)=(0, 0), q(1,t)=(1−t,t), i.e. q contracts the left edge of I ×I and maps the other three edges linearly onto the edges of ∆2. Since F is constant along the fibers of q, F induces a continuous map 2 F¯ :∆ −→ X s.t. F = F¯ ◦ q =⇒ F¯(s, 0) = α(s), F¯(0,t)= β(t), F¯(s, 1−s)= x1 ∀ s,t∈I ¯ ¯ 2 ¯ 2 ¯ 2 =⇒ ∂F = F ◦ ι0 − F ◦ ι1 + F ◦ ι2 = f1,x1 − β + α; see Figure 2. Along with Lemma 1, this implies Lemma 2. β x0 x1 α q F 2 F¯ β x1 X α 0 1 Figure 2: A path homotopy gives rise to a boundary between the corresponding 1-simplices. For Lemma 3, define α(x+2y), if x+2y ≤1; F :∆2 −→ X by F (x,y)= (β(x+2y−1), if x+2y ≥1. 2 2 β β F ∗ X α 0α 1 Figure 3: A boundary between the 1-simplex corresponding to a composition of paths and the sum of the 1-simplices corresponding to the paths. see Figure 3. This map is well-defined and continuous, since it is continuous on the two closed sets and agrees on the overlap, where it equals α(1)=β(0). Furthermore, 2 2 F ι0(s) = F (1−s,s)= β(s), F ι2(s) = F (s, 0) = α(s); 2 α(2s), if 2s≤1; F ι1(s) = F (0,s)= (β(2s−1), if 2s≥1; 2 2 2 =⇒ ∂F = F ◦ ι0 − F ◦ ι1 + F ◦ ι2 = β − α∗β + α. For Lemma 4, note that α+α ¯ = α+α ¯−α∗α¯ + α∗α¯−f1,α(0) + f1,α(0). Since α∗α¯ is path-homotopic to the constant path f1,α(0), each of the three expressions above belongs to ∂S2(X) by Lemmas 1-3. This implies Lemma 4. For Lemma 5, choose a continuous map q : I ×I −→∆2 such that (1−2s, 2s), if s∈[0, 1/2]; q(s, 0) = (0, 3−4s), if s∈[1/2, 3/4]; q(s, 1) = q(0,t)= q(1,t)=(1, 0) ∀ s,t∈I. (4s−3, 0), if s∈[3/4, 1]; 2 2 2 Then, F◦q is a path-homotopy from (F◦ι0)∗ (F ◦ι1)∗(F◦ι2) to the constant loop f1,F (1,0); see Figure 4. F (1, 0) 2 F F (1 2 1 0) ◦ q ι F , ι 2 , ◦ 0 0) X (1 F F 2 2 2 2 0F ◦ι2 1 (F ◦ι0)∗ (F ◦ι1)∗(F ◦ι2) Figure 4: Boundary of a 2-simplex is loop homotopic to the constant loop. (a) We now define the homomorphism h: π1(X,x0) −→ H1(X; Z) by h [α] = {α}∈ H1(X; Z). By Lemma 0, ∂α=0 and thus {α}∈H1(X; Z). By Lemma 2, the map h is well-defined, i.e. [α]=[β] =⇒ {α} = {β}. 3 By Lemma 3, h is indeed a homomorphism: h [α]∗[β] = h [α∗β] = {α∗β} = {α}+{β} = h [α] + h [β] . To show that h is surjective, for each x∈X choose a path γx :(I, 0, 1)−→ (X,x 0,x ) from x0 to x. If N c = aiσi ∈S1(X), i=1 X a1 aN let αc = γσ1(0) ∗σ1 ∗γ¯σ1(1) ∗ ... ∗ γσN (0) ∗σN ∗γ¯σN (1) . This is a product of loops at x0. It is essential that ai ∈Z, i.e. we are dealing with integer homology. The loop αc is not uniquely determined by c, even if the paths γx are fixed, as it depends on the ordering of the σi’s. This is irrelevant, however, at this point. Since h is a homomorphism, N N h [αc] = aih [γσi(0) ∗σi ∗γ¯σi(1)] = ai γσi(0) ∗σi ∗γ¯σi(1) i i X=1 X=1 N N = ai {γσi(0)}+{σi}−{γσi(1)} = {c} + ai {γσi(0)} −{γσi(1)} . i=1 i=1 X X The third equality above follows from Lemma 3 (not from h being a homomorphism). If c∈ker ∂, N N ai f1,σi(1) −f1,σi(0) = ∂c =0 =⇒ ai {γσi(0)} −{γσi(1)} = 0 i=1 i=1 X X =⇒ h [αc] = {c}∈ H1(X; Z) ∀ c ∈ ker ∂. This shows that h is surjective. (b) Since the group H1(X; Z) is abelian, h must vanish on the commutator subgroup of π1(X; x0). Since this subgroup is normal, h induces a group homomorphism Φ: Abel π1(X,x0) ≡π1(X,x0) π1(X,x0),π1(X,x0) −→ H1(X; Z). We will show that this map is an isomorphism by constructing an inverse Ψ for Φ. If α is a loop based at x0, denote its image (and the image of [α]) in Abel(π1(X,x0)) by hαi. For each 1-simplex σ ∈S1(X), let g(σ)= hασi∈ Abel π1(X,x0) . Since S1(X) is a free abelian group with a basis consisting of 1-simplic ies σ and Abel(π1(X,x0)) is abelian, g extends to a homomorphism g : S1(X) −→ Abel π1(X,x0) . If F :∆2 −→X is a 2-simplex, 2 2 2 g(∂F )= g(F ◦ι0) − g(F ◦ι1)+ g(F ◦ι2) 2 2 2 = γ 2 ∗(F ◦ι )∗γ¯ 2 − γ 2 ∗(F ◦ι )∗γ¯ 2 + γ 2 ∗(F ◦ι )∗γ¯ 2 F (ι0(0)) 0 F (ι0(1)) F (ι1(0)) 1 F (ι1(1)) F (ι2(0)) 2 F (ι2(1)) 2 2 −1 2 = γF (1,0) ∗(F ◦ι0)∗γ¯F (0,1) ∗ γF (0,0) ∗(F ◦ι1)∗γ¯F (0,1) ∗ γF (0 ,0) ∗(F ◦ι2)∗γ¯F (1,0) 2 2 2 = γF (1,0) ∗ (F ◦ι0)∗(F ◦ι1)∗(F ◦ι2) ∗ γ¯F (1,0) . 4 2 2 2 By Lemma 5, (F ◦ι0)∗(F ◦ι1)∗(F ◦ι2) is path-homotopic to the constant loop at F (1, 0) and thus 2 2 2 γF (1,0) ∗ (F ◦ι0)∗(F ◦ι1)∗(F ◦ι2) ∗ γ¯F (1,0) = [id] ∈ π1(X,x0) 2 2 2 =⇒ g(∂F )= γF(1,0) ∗ (F ◦ι0)∗(F ◦ι1)∗(F ◦ι2) ∗ γ¯F (1,0) = 0 ∈ Abel π1(X,x0) .
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