SMSTC Geometry & Topology 1 Assignment 1 Matt Booth

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1 SMSTC Geometry & Topology 1 Assignment 1 Matt Booth Question 1 i) Let be the space with one point. Suppose X is contractible. Then by definition we have maps f : X and g : X such that gf id X and fg id. Note that the map g simply picks out a point of X. So gf is constant on X. Hence the identity map on X is nullhomotopic. Conversely, if id X is nullhomotopic, let c : X X be the constant map it s homotopic to. Observe that we can factor c into two maps X f g X. Then gf = c id X. Note that fg = id so in particular fg id, and hence X is contractible. ii) Let X, Y be spaces, f : X Y a map and M f its mapping cylinder. Let ι : X M f be the inclusion. Let p : M f Y be the projection onto Y. Let w : Y M f be the inclusion. Then we know that w is a homotopy inverse of p since M f deformation retracts onto w(y ). Moreover the map f factors as f = pι. Suppose f is a homotopy equivalence. Then f has some homotopy inverse g. Consider the map gp : M f X. We see that gpι = gf id X and that ιgp wpιgp = wfgp wp = id Mf. Hence ι is a homotopy equivalence, with homotopy inverse gp. Conversely suppose that ι : X M f is a homotopy equivalence. Then since f is the composition of two homotopy equivalences, f is a homotopy equivalence. iii) Let f, g : X Y be homotopic maps. Let C f and C g be the mapping cones of f and g respectively. Note that we obtain C f by attaching the cone CX to Y along X via f. Pick a homotopy H : X I Y from f to g. Consider the space Z defined by attaching CX I to Y along X I via H. Clearly Z contains C f and C g as subspaces. Moreover we see that Z deformation retracts to C f by the map collapsing I to {0}. Similarly we see that Z deformation retracts to C g by the map collapsing I to {1}. Hence C f Z C g as required. Furthermore, these homotopies are relative to the space Y. 1

2 Question 2 i) Let i : RP 1 RP 2 be the inclusion map. Think of RP 2 as the 2-disc D 2 with antipodal points on D 2 identified via the quotient map q : D 2 RP 2. Then the inclusion of S 1 as D 2 induces the inclusion i of RP 1 in RP 2. We can easily define a continuous map φ : RP 1 D 2 RP 2 by doing i on RP 1 and q on D 2. Note that if x D 2 then φ(x) = q(x) = iπ 1 (x) = φ(π 1 (x)). Hence φ defines a continuous map f 2 : RP 1 D 2 (x π 1(x)) RP2 on the quotient. We see that the domain of f 2 is exactly the space X := RP 1 π1 D 2 obtained from RP 1 by attaching a 2-cell along π 1. We want to prove that f 2 is a bijection. It s obviously surjective since φ was surjective. To show injectivity, note that set-theoretically both X and RP 2 are the disjoint union of the sets RP 1 and int(d 2 ), and that the function f 2 respects this disjoint union. So it suffices to show that f 2 is injective when restricted to the subspaces RP 1 and int(d 2 ). It s clear that f 2 RP 1 = i and f 2 int(d2 ) = q int(d2 ). Now it s clear that f 2 is injective, since both i and q int(d2 ) are injective. ii) We ve constructed a continuous bijection f 2 : X RP 2. First note that RP 1 is compact since it is the quotient of a compact space. Note the general fact that attaching a disc to a space Y by a map ψ will preserve compactness, since Y ψ D m is a quotient of a disjoint union of two compact spaces and hence compact. So we see that X is compact. Moreover, note that RP 2 is Hausdorff. So the map f 2 is a continuous bijection from a compact space to a Hausdorff space and hence a homeomorphism. iii) The argument from parts i) and ii) can be easily generalised to get a homeomorphism f n+1 : RP n πn D n+1 RP n+1. Identifying RP n+1 with a quotient of D n+1 we can again define a continuous surjection φ : RP n D n+1 RP n+1 that descends to a continuous surjection f n+1 as above. Once again the set-theoretic decomposition of both spaces into RP n and int(d n+1 ) shows that f n+1 is injective. Finally, since RP n πn D n+1 is compact and RP n+1 is Hausdorff, we see that f n+1 is actually a homeomorphism. iv) We have a homeomorphism f n+1 : RP n πn D n+1 RP n+1. Hence we can quotient out both sides by the copy of RP n lying inside and get a homeomorphism RPn πn D n+1 RP = RP n+1 n RP. It s clear that the left hand side is homeomorphic to the space α D n+1 = S n+1, where α is the unique map S n. n Hence composing with this homeomorphism, f n+1 induces a homeomorphism g n+1 : S n+1 RPn+1 RP. n 2

3 Question 3 i) For a function ɛ : {1,..., n + 1} {1, 1} let f ɛ : S n S n be the function (x 1,..., x n+1 ) (ɛ(1) x 1,..., ɛ(n + 1) x n+1 ). Let f be the map induced by the map 1 1, i 1 for i > 1. It s clear that f ɛ is a bijection since it s an involution: f ɛ f ɛ = id. So to prove that f ɛ is a homeomorphism it suffices to show that f ɛ is an open map. Take an open set U S n. Note that we give S n R n+1 the subspace topology. Fix x U. Since U is open, find an open ball B R n+1 containing x such that B S n U. Note that f ɛ (B) is a ball in R n+1. It s clear that f ɛ (B) S n f ɛ (U), and that f ɛ (x) f ɛ (B). Since f ɛ is surjective, for each y f ɛ (U) we ve found an open set containing y and contained in f ɛ (U). Hence f ɛ (U) is open. Hence f ɛ is an open map. ii) Suppose ɛ 1 ( 1) = k. We wish to show that f ɛ is homotopic to f ( 1) k. Let ɛ i be the map sending i 1 and all other elements to 1. Let f i := f ɛi. Let α 1 α k be the elements of ɛ 1 ( 1). Then we can factor f ɛ = f α1 f αk. So to prove that f ɛ is homotopic to f ( 1) k it suffices to prove that each of the f i is homotopic to f = f 1 and observe that (f ) k = f ( 1) k. Certainly f 1 is homotopic to f 1, and by symmetry we can reduce proving that f i f 1 for i 1 to proving that f 2 f 1. We can show that the maps f 1 and f 2 are homotopic by constructing an explicit homotopy between them. Define a map H : S n I R n+1 by H(x 1, x 2, x 3,, x n+1, t) = (x 2 sin(πt) x 1 cos(πt), x 2 cos(πt)+x 1 sin(πt), x 3,, x n+1 ). It s not hard to see that the image of H actually lies in S n. It s clear that H is continuous, and that H(x, 0) = f 1 (x) and H(x, 1) = f 2 (x). Hence H is a homotopy from f 1 to f 2. iii) Let the map c n : RP n S n be the composition of the projection to RP n /RP n 1 with the homeomorphism to S n. Observe that the map c n consists of quotienting RP n by the image under π n of S n 1, followed by a homeomorphism to identify the quotient with S n. Consider the map α n := c n π n from S n to itself. Let ν : S n S n S n be the composition of the quotient map S n S n /S n 1 with the homeomorphism to S n S n. Note that the equivalence relation on S n identifying the antipodal points defines an equivalence relation 1 on S n S n since the space S n 1 is a union of equivalence classes of. Let q : S n S n (S n S n )/ 1 be the associated quotient map. By interchanging the quotients we may factor α n as q ν by first quotienting S n by S n 1 and then quotienting the resulting space by the equivalence relation 3

4 1. So we can identify (S n S n )/ 1 with S n. Note that since q is a map from S n S n, we can write q as q 1 q 2 for some maps q i : S n S n. Using Exercise 4.4 we see that [α n ] = [(q 1 q 2 ) ν] = [q 1 ] + [q 2 ] in π n (S n ). So to compute the homotopy class of α n we simply need to know the homotopy classes of the q i. These depend on our identification of (S n S n )/ 1 with S n. Observe that the equivalence relation 1 identifies the two copies of S n in S n S n via the map that reflects in the basepoint : if we take our basepoint to be 1 in the first sphere and 1 in the second then the point x in the first copy of S n is identified with the point x in the second copy. Hence if we take the first copy of S n in S n S n to be the set of equivalence class representatives of 1 then we see that q 1 is the identity and q 2 is the antipodal map. Since the antipodal map is f ɛ for ɛ the constant map 1, we see that the antipodal map is homotopic to f ( 1) n+1. Noting that f + = id, we get that [α n ] = [f + ] + [f ( 1) n+1]. Suppose that n is even. Then the map f ( 1) n+1 is just f. So to show that α n is nullhomotopic it suffices to show that [f ] = [id]. Thinking of S n as the space I n / I n, we see that f is the map (x 1, x n ) (1 x 1,, x n ). But this map is the homotopy inverse of the map (x 1, x n ) (x 1,, x n ), i.e. the identity. Hence [f ] = [id]. On the other hand if n is odd it s clear that [α n ] = 2[id]. Question 4 i) Let ι 1, ι 2 : S 1 S 1 S 1 be the inclusions of the component circles in the first and second factors respectively. The torus T 2 has a CW structure with 1-skeleton S 1 S 1 and a single 2-cell, attached by wrapping around the first component S 1, then the second, then back around the first S 1 in the opposite direction and finally around the second S 1 in the opposite direction. Call this attaching map φ : S 1 S 1 S 1. We wish to find the image [φ] of φ in the fundamental group π 1 (S 1 S 1 ), which we know is a free group with generators [ι 1 ] and [ι 2 ]. But from the description above it s clear that [φ] = [ι 1 ][ι 2 ][ι 1 ] 1 [ι 2 ] 1. ii) Now consider the suspension S(φ) : S 2 S(S 1 S 1 ). We wish to show that S(φ) is nullhomotopic, equivalently that [S(φ)] = 1 in the (multiplicatively written!) homotopy group π 2 (S(S 1 S 1 )). Write σ for the suspension homomorphism π 1 (S 1 S 1 ) π 2 (S(S 1 S 1 )). Then [S(φ)] = σ[φ]. But note that [φ] is a commutator in π 1 (S 1 S 1 ) so that σ[φ] is a commutator in π 2 (S(S 1 S 1 )). Since π 2 (S(S 1 S 1 )) is abelian, σ[φ] is the identity. Hence [S(φ)] is the identity, as required. 4

5 iii) Now we wish to determine the homotopy type of the space X := S(T 2 ) the suspension of the 2-torus. Since T 2 is a CW complex we can give X a CW structure. More precisely, T 2 is a CW complex with one 0-cell two 1-cells with attaching maps α, β : S 0 one 2-cell with attaching map φ Hence (using Exercise 4.2) X is a CW complex with two 0-cells one 1-cell with attaching map id : S 0 S 0 two 2-cells with attaching maps S(α), S(β) one 3-cell with attaching map S(φ) Since S(φ) is nullhomotopic, X is homotopy equivalent to the CW complex with the same cells and 2-skeleton as X, but where the attaching map of the 3-cell is a constant map. Attaching a 3-cell along a constant map is the same as taking a wedge sum with a 3-sphere. So we can say that X X (2) S 3. To determine the 2-skeleton X (2) of X, observe that the maps S(α), S(β) : S 1 X (1) = [0, 1] are just the projection of the circle onto one coordinate. So the 2-skeleton of X is a line with two 2-spheres glued along it. Shrinking this line down it s not hard to see that X (2) is homotopy equivalent to the wedge sum of two 2-spheres. Hence X S 2 S 2 S 3. Question 5 i) Let X be obtained from A by attaching an (n + 1)-cell along a map f. We want to prove that the inclusion A X admits a retraction if and only if f is nullhomotopic. Note that X is the mapping cone of f. If f is nullhomotopic, find a constant map g that it s homotopic to. Applying Question 1, part iii) we see that C f and C g are homotopy equivalent rel A. Note that since g is constant, C g is homotopy equivalent rel A to the wedge sum A S n+1. Certainly this space admits a retraction r onto A: just collapse S n+1 to the basepoint of the wedge sum. Let h : X A S n+1 be the composition of the homotopy equivalences X C f C g A S n+1. Then the map h is a homotopy equivalence rel A. The composition of h and r is then a retraction X A. 5

6 Conversely suppose that the inclusion i : A X admits a retraction r. Consider the map if : S n X. Thinking of X as C f, this map is nullhomotopic since it lands in the subspace C(S n ) which is contractible (intuitively, we can slide if up the cone). Since r is a retraction, f = rif. So f is the composition of r with a nullhomotopic map and hence is nullhomotopic. ii) Let n > 0. Note that CP n+1 is obtained from CP n by attaching a (2n + 1)- cell along the projection map π 2n+1. So by the above CP n CP n+1 admits a retraction if and only if π 2n+1 is nullhomotopic. Note that we have a fibre bundle S 1 S 2n+1 π2n+1 CP n. This gives us a long exact sequence of homotopy groups π i+1 CP n π i S 1 π i S 2n+1 p π i CP n π i 1 S 1 where p is the map induced by π 2n+1. Note that S 1 has contractible universal cover so the homotopy groups π i S 1 are zero for i > 1. Hence for i > 2 the long exact sequence breaks up into exact sequences 0 π i S 2n+1 p π i CP n 0. In particular since 2n + 1 > 2 we have an isomorphism π 2n+1 S 2n+1 p π 2n+1 CP n. Note that π 2n+1 S 2n+1 = Z so the map p is certainly not the zero map. So π 2n+1 is not nullhomotopic and hence CP n CP n+1 does not admit a retraction. iii) We can follow a similar line of reasoning to show that RP n RP n+1 does not admit a retraction for n > 0: it suffices to show that for some i the map p : π i S n π i RP n induced by the projection π n is not the zero map. We have a covering space S 0 S n RP n and this gives us a long exact sequence of homotopy groups π i+1 RP n π i S 0 π i S n p π i RP n π i 1 S 0 Note that for i > 0 the homotopy groups π i (S 0 ) all vanish, and hence we get exact sequences 0 π i S n p π i RP n, reducing to isomorphisms if i > 1. Setting i = n as before will give us an exact sequence 0 Z p π n RP n Exactness at Z then implies that p is an injection Z π n (RP n ) and so is certainly not the zero map. iv) Lemma. Let X be a CW complex with no cells above dimension m 1. Suppose that the CW complex Y is obtained from X by attaching m-cells to X along maps φ i. Then S(Y ) is obtained from S(X) by attaching (m + 1)-cells along the maps S(φ i ). Proof. Note that Y has a CW structure with k-skeleton X (k) for k < m and m-cells attached via the maps φ i. Hence S(Y ) has a CW structure with two 0-cells, (k + 1)-skeleton S(X k ) for k < m and (m + 1)-cells attached via the maps S(φ i ). But this is simply the space S(X) with (m + 1)-cells attached via the maps S(φ i ). 6

7 Suppose now that n is an odd natural number. We wish to show that the inclusion S(RP n ) S(RP n+1 ) does not admit a retraction. Applying the Lemma we see that S(RP n+1 ) is obtained from S(RP n ) by attaching an (n + 2)- cell along S(π n ). So to show that the inclusion does not admit a retraction it suffices to show that the map S(π n ) is not nullhomotopic. Letting c n be the map from Question 3, we know that since n is odd, [c n π n ] = 2[id S n] in π n (S n ). Applying the suspension homomorphism to this identity we obtain the identity [S(c n π n )] = 2[S(id S n)] in π n+1 (S n+1 ). Noting that suspension is a functor and that the suspension of the identity map on S n is the identity map on S n+1, we see that [S(c n ) S(π n )] = 2[id S n+1] 0 in π n+1 (S n+1 ) = Z. If S(π n ) were nullhomotopic then the map S(c n ) S(π n ) would be nullhomotopic and hence [S(c n ) S(π n )] = 0. But this is not the case; hence S(π n ) is not nullhomotopic and hence the inclusion S(RP n ) S(RP n+1 ) does not admit a retraction. 7

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