Algebraic Topology Exam 2006: Solutions

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1 Algebraic Topology Exam 006: Solutions Comments: [B] means bookwork. [H] means similar to homework question. [U] means unseen..(a)[6 marks. B] (i) An open set in X Y is an arbitrary union of sets of the form U V, where U X is open and V Y is open. (ii) An open set in X/ is a set U such that q (U) X is open, where q : X X/ is the map x [x]. (iii) An open set in X is one of the form U X where U X is open. (b)[3 marks. B] (i) α is continuous if and only if both π X α and π Y α are continuous, where π X and π Y are the projection maps. (ii) β is continuous if and only if β q is continuous. (iii) γ is continuous if and only if i γ is continuous, where i : X X is the inclusion map. (c)[5 marks. H] Suppose there is another topology satisfying the criterion. Denote by W = (X Y, T ) this topological space. Since the identity map on W is continuous, π X = π X id X : W X and π Y : W Y are continuous. Let φ : X Y W be the identity map on the set X Y. Since π X φ = π X : X Y X is continuous and π Y φ = π Y is continuous, φ is continuous. Similarly φ is continuous, so φ is a homeomorphism. (d)[7 marks. Variation on bookwork.] X/ is the image of X under the continuous map q, so is compact. The given condition implies that θ induces an injective map X/ H, which is continuous by (b)(ii). Let H be the image of this; since it is a subspace of H, it is Hausdorff. By (b)(iii), θ induces a continuous bijection ψ : X/ H. Let C X/ be closed. Then C is compact. [Proof: If U i : i I} is an open cover of C, where U i = V i C for some open V i X/, then V i : i I} X/ \C} is an open cover of X/, and so has a finite subcover V j : j J} X/ \C}. Then U j : j J} is a finite subcover of U i : i I}.] ψ(c) is a continuous image of a compact space, so is compact, and therefore closed in H since H is Hausdorff. Thus ψ is continuous, so ψ is a homeomorphism. (e) (i)[ marks. B] A = [0, ), B = z C : z = }, σ : A B given by σ(t) = e πit. (ii)[ marks, U] A = [0, ] with the indiscrete topology, B = [0, ] with the usual topology. σ : A B is the identity map..(a)[b, 5 marks](i) If α, β : X Y are continuous maps, a homotopy from α to β is a continuous map H : X I Y such that, for all x X, H(x, 0) = α(x) and H(x, ) = β(x).

2 (ii+iii) A homotopy equivalence is a continuous map α : X Y so that, for some continuous γ : Y X, α γ id Y and γ α id X. Then γ is a homotopy inverse of α. (b)[b, 5 marks] If α : X Y is continuous, H(x, t) = α(x) gives a homotopy α α. So is reflexive. If H : α β is a homotopy then H : β α, where H (x, t) = H(x, t), is a homotopy. So is symmetric. If H : α β and H : β γ are homotopies, then H : α γ, where H(x, t) = H (x, t) if 0 t H (x, t ) if t is a homotopy. So is transitive. (c)[b, 3 marks] Let H : α α and H : β β be homotopies. Then H(x, t) = H (H (x, t), t) gives a homotopy H : β α β α. (d)[u, marks](i) If α and β are homotopy inverses for α, β : X X, then (α β) (β α ) = α (β β ) α α id X α = α α id X, and similarly (β α ) (α β) id X. So α β is a homotopy equivalence with homotopy inverse β α. (ii) Let X be R and let α : X X be a constant map. Then α is a homotopy equivalence, since X is contractible, but α is not invertible. (iii) Let α be a homotopy inverse to α. Then β α α α id X and similarly α β id X. (iv) Multiplication is well-defined by (c) and (d)(i). id X G(X), so [id X ] is an identity element in Ḡ(X). Multiplication is associative, since composition of maps is. If α is a homotopy inverse to α, then [α ] is an inverse to [α]. (v) If X is contractible, then Ḡ(X) is a trivial group, since all maps X X are homotopic. If X = S, then the homotopy class of a map α is determined by its degree deg(α) Z, and deg(α β) = deg(α)deg(β). So α is a homotopy equivalence if and only if deg(α) = ±, so Ḡ(X) is cyclic of order. 3.(a)[B, marks.] Let P X (x 0, x 0 ) be the set of paths θ : I X with θ(0) = θ() = x 0. Define an equivalence relation on P X (x 0, x 0 ) where θ θ if and only if there is a homotopy H : θ θ with H(0, t) = H(, t) = x 0 for all t I. As a set, π (X, x 0 ) = P X (x 0, x 0 )/.

3 For θ, φ P X (x 0, x 0 ), define θ φ P X (x 0, x 0 ) by θ(t) if 0 t (θ φ)(t) = φ(t ) if t. Multiplication in π (X, x 0 ) is given by [θ][φ] = [θ φ]. The identity element is [e], where e : I X is given by e(t) = x 0 for all t I. The inverse of [θ] is [θ ], where θ (t) = θ( t) for t I. Proofs. Lemma: Let f : I I be continuous with f(0) = 0, f() =. Then for any θ P X (x 0, x 0 ), θ θ f. [Proof: Define Hθ θ f by H(s, t) = θ(( t)s + tf(s)). Then H(0, t) = θ(0) = x 0 = θ() = H(, t) for any t I.] Associativity: Define f : I I by f(s) = s if 0 s 4 s + 4 if 4 s s+ if s Then (θ φ) ψ = (θ (φ ψ)) f, so ([θ][φ])[ψ] = [θ]([φ][ψ]. Identity: Define f : I I by s if 0 s f(s) = if s Then θ e = θ f, so [θ][e] = [θ]. Similarly [e][θ] = [θ]. Inverses: Define H : e θ θ by θ(ts) if 0 s H(s, t) = θ(t( s)) if s. Then H(0, t) = θ(0) = x 0 and H(, t) = θ(0) = x 0, so [e] = [θ][θ ]. (b)[b, ++3 marks.](i) α ([θ]) = [α θ]. (ii) β (α ([θ])) = β ([α θ]) = [β α θ] = (β α) ([θ]). (iii) Let θ P X (x 0, x 0 ). Define h : I I Y by h(s, t) = H(θ(s), t). Then, for s I, h(s, 0) = H(θ(s), 0) = (α θ)(s) and h(s, ) = H(θ(s), ) = (α θ)(s), so h : α θ α θ. For t I, h(0, t) = h(, t) = H(x 0, t), which is y 0, since H is basepoint-preserving. Thus [α θ] = [α θ], so α = α. (c)(i)[b, 3 marks.] [φ] π (X, x 0 ) is sent to [θ φ θ] π (X, x ). This is a group homomorphism since [θ φ θ θ φ θ ] = [θ φ φ θ] and is an isomorphism since it has an inverse [ψ] [θ ψ θ ]. (ii)[u, 3 marks.] If [θ ] = [θ ], then [θ φ θ ] = [θ φ θ ]. No. Let x 0 = x, where π (X, x 0 ) has non-trivial centre, and let [θ ], [θ ] be two elements of the centre of π (X, x 0 ). Then they both induce the identity map.

4 4.(a) [B, 5 marks] A continuous map such that X is also path-connected, and such that X has a covering U i : i I} by open sets such that, for each i, p (U i ) is a disjoint union of spaces W i,j such that the restriction of p to W i,j is a homeomorphism W i,j U i. The Homotopy Lifting Property states that if p : X X is a covering, H : Y I X is a homotopy between continuous maps α, β : Y X, and α : Y X is a continuous map such that p α = α, then there is a unique continuous map β : Y X and a unique homotopy H from α to β such that p H = H. (b) [B, 6 marks] Suppose θ is a loop in X based at x 0 such that p [θ] =. Then there is a basepoint preserving homotopy H from p θ to the trivial loop. By the homotopy lifting property, this lifts uniquely to a homotopy H from θ to a lifting of the trivial loop. Since H(0, t) = x 0 for all t I, H(0, t) p (x 0 ) for all t I. But since p (x 0 ) is discrete, and H(0, 0) = x 0, H(0, t) = x0 for all t. Similarly H(, t) = x 0 for all t and H(s, ) = x 0 for all s. Hence H is a basepoint preserving homotopy from θ to the trivial loop. Thus [θ]=. (c) [B, 4 marks] p is a regular covering if the image of p is a normal subgroup of π (X, x 0 ). It is a universal covering if X is simply connected. If p is universal, then π ( X, x 0 ) is the trivial group, so the image of p is the trivial subgroup of π (X, x 0 ), which is normal. (d) [B, 5 marks] An action of a group G on a topological space Y is free if each point y Y has an open neighbourhood U y such that the sets gu y : g G} are pairwise disjoint. If G acts freely on a space X, then the natural map from X to the quotient space X/G is a regular covering space. Conversely if p : X X is a regular covering, then there is a free action of the quotient group G = π (X, x 0 )/p π ( X, x 0 ) on X such that X is homeomorphic to the quotient space X/G. (e) [U, mark] Every subgroup of an abelian group is normal. (f) [U, 4 marks] If G = S 3 then there is a free action of G on Ỹ so that the quotient space Ỹ /G is homeomorphic to Y. Let H be a (nonnormal) subgroup of G of order. Then the natural map ˆp : Ỹ /H Y is a covering of Y such that the image of the fundamental group of Ỹ /H in π (Y, y 0 ) is a non-normal subgroup of order. 5.(a) [B, 8 marks] (i) A sequence X := X n+ X n X n... of homomorphisms d X n : X n X n of abelian groups, such that d X n d X n = 0 for all n Z.

5 (ii) A chain map α : X Y between two chain complexes is a sequence of maps α n : X n Y n such that α n d X n = d Y n α n for all n Z. (iii) The nth homology group H n (X ) is the quotient group ker(d X n )/im(d X n+). (iv) A sequence of abelian groups and homomorphisms 0 X α Y β Z 0, such that α is injective, β is surjective, and im(α) = ker(β). (iv) A sequence of chain complexes and chain maps such that α 0 X β Y Z 0, α 0 X n β n n Yn Zn 0 is a short exact sequence of abelian groups for each n Z. (b) [B, 9 marks] There is a long exact sequence H n+ (Z ) H n (X ) Hn(α ) H n (Y ) Hn(β ) H n (Z )... of homology groups where (if for a complex C we denote by x the element of H n (C ) represented by an element x ker(d C n )), and H n (α )( x) = α n (x) H n (β )(ȳ) = β n (y), and the map H n+ (Z ) H n (X ) is constructed as follows: Let z ker(d Z n+). Pick y Y n+ such that β n+ (y) = z. Since β n d Y n+(y) = d Z n+β n+ (y) = d Z n+(z) = 0, d Y n+(y) ker(β n ) = im(α n ), and so we can choose x X n such that α n (x) = d Y n+(y). Then α n d X n (x) = d Y n α n (x) = d Y n d Y n+(y) = 0, and so, since α n is injective, d X n (x) = 0, and the map we want to construct sends z to x. (c) [U, 8 marks] Considering the columns as chain complexes, we have a short exact sequence of chain complexes, so we get a long exact sequence of homology groups 0 H A H B H C H D H E H F H G H H H I 0, where H A is the homology of the first column at A, etc. (i) Two out of every three terms in the long exact sequence are zero, so the remaining terms are also zero. (ii) H B = H E = H H = 0, so the long exact sequence becomes 0 H A 0 H C H D 0 H F H G 0 H I 0,

6 and in particular gives an isomorphism H F H G. But H F = 0 iff the third column is exact at F, and H G = 0 iff the map D G is surjective.