Topology Hmwk 1 All problems are from Allen Hatcher Algebraic Topology (online) ch 3.2

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1 Topology Hmwk 1 All problems are from Allen Hatcher Algebraic Topology (online) ch 3.2 Andrew Ma March 1, 214 I m turning in this assignment late. I don t have the time to do all of the problems here myself without help, so I got solutions online just so that at least I get familiar with the concepts on this problem set. Note to self: I m not proud of my work here, but I think the bright side to this assignment is that I did nearly an entire 2-week assignment in 2 days without help modulo the internet, so that at the very least I can still produce answers even under high pressure conditions. 1 1 I don t have a good answer for this question. This is the last problem that I need to do, it s 4:47am and while I have some ideas, I don t think I can get anything solid and working in the next hour or so. I think it s better in this situation to just ask Prof. Kent I got help from reading a solution from Stanford to get me on the right track with this proof. Proof. To begin, we can compute the cup product on H k (M g ) for k 2 easily. Use the fact that H (M g ) = Z, H 1 (M g ) = Z 2g, H 2 (M g ) = Z, and H k (M g ; R) = for k 3 [H, section 2.2, example 2.36 page 141]. By the universal coefficient theorem we may conclude that H k (M g ) = for k 3, hence any cup product involving H k (M g ), k 2 is trivial (unless of course it involves H (M g ) but the generator of H (M g ) by definition, acts as the identity). All that is left is to understand the cup product on H 1 (M g ; R) H 1 (M g ; R). Here we will use the quotient map q : M g g 1 T i. First note that this quotient map induces a map q : H 1 (M g ) H 1 ( g 1 T i) so that if A i, B i are generators for the 1

2 ith Z 2 pair in M g then a i, b i generators corresponding to the ith torus in H 1 ( g 1 T i) s.t. A a i and B i b i. Then all of the homologies are free groups, by the universal coefficient theorem, the cohomologies are literally the dual of the homologies. Meaning, ai Ai and bi Bi where these are the generators of the duals i.e. H 1 (M g ; R) and H 1 ( g 1 T i; R) respectively. Thus understanding the cup product on H 1 (M g ; R) amounts to understanding the cup product on generators of H 1 ( g 1 T i; R). These are well understood since H ( g 1 T i) = g 1 H (T i ; R) [H, section 3.2 example 3.14 page 213]. Then by doing analysis similar to that at the end of example 3.7 [H, section 3.2. page 27], we get that for H 1 (T i ; R) generators a i, b i we have that a i b i = c i, a i a i = b i b i = Thus for generators a i, b j H 1 ( g 1 T i; R) we get a i b j = δ ij c i, a i a j = b i b j = where a i, b i correspond to generators for the T i. Finally, H 1 (M g ; R) inherits this cup product by the explanation in the first paragraph so that for generators Ai, B i H 1 (M g ; R) we have A i B j = δ ij c i, A i A j = B i B j = I got help from reading a solution online to get me on the right track with this proof. I also thought of doing some kind of induction by multiplying H 1 elements together to show that an isomorphism on H 1 gets an isomorphism on H k but this involved knowing how the map on H 2 (Y) H 2 (X) worked i.e. I needed to know the structure of the map. So ultimately I would conclude something like this proof. a Recall that H (RP n ; Z 2 ) = Z 2 [α]/(α n+1 ) [H, section 3.2 Thm 3.19 page 22] where α is the generator of H 1 (RP n ; Z 2 ). It will also be useful to keep in mind that in the cohomology ring, generator generator = generator. Claim 1. All maps f : RP n RP m induce the trivial map on H 1 (RP m ; Z 2 ) H 1 (RP n ; Z 2 ) Proof. Assume that f : H 1 (RP m ; Z 2 ) = Z 2 H 1 (RP n ; Z 2 ) = Z 2 is a non-trivial map (where the equalities are seen because H (RP n ; Z 2 ) may be seen as a direct sum 2

3 of copies of Z 2 generated by α i terms). Then f sends the generator to the generator hence f : Z 2 [α]/(α m+1 ) Z 2 [α]/(α n+1 ) α α But then = f (α m+1 ) = α m+1 however = α n+1 Z 2 [α]/(α n+1 ). This is a contradiction. Claim 2. All maps f : CP n CP m induce the trivial map on H 1 (CP m ; Z) H 1 (CP n ; Z) Proof. This proof is similar to the previous proof. Again using thm 13.3 of section 3.2. b I got this from a solution online again. Proof. Using the function g : S n S n 1 get an induced map ĝ : RP n RP n 1. By part (a) this induces a trivial map H 1 (RP n 1 H 1 (RP n ) thus a it induces a trivial map on H 1 (RP n ) H 1 (RP n 1 ) too (by the isomorphism provided from the universal coefficient theorem). Hence g also induces a trivial map on π 1 (RP n ) (since π 1 (RP n ) = Z 2 = H 1 (RP n )). This now satisfies the lifting criterion so that we may now lift ĝ to a map g : RP n S n 1. Next consider the maps g and gπ n where π n : S n RP n. These are both lifts of the map π n 1 g : S n RP n 1 (notice that the lifting criterion here holds because RP n 1 RP n π 1 (RP n 1 ) π 1 (RP n )). Thus by the uniqueness of the lifting, both maps should be equal if they agree at a point. Wlog, assume that g and gπ n agree at point - we may assume this because given a fixed point x S n, π n (x ) = g([x ]) = g(x ) or g( x ) = g(x ) and we may replace g with g in the latter case. Hence the maps ought to be equal, but for a general point x S n, g( x) = g(x) while gπ n (x) = π n ( x) so that they may not be the same map. A contradiction I got this solution from Evan Dummit s online solutions Claim 3. The Lefschetz number of f is τ f = 1 + d + d d n 3

4 Proof. First recall that H k (CP n ; M) = M for k even adn otherwise (I got this from topospaces wiki but I m sure it s in Hatcher somewhere too). Then the induced map f : H (CP n ; Z) = Z[α]/(α n+1 ) H (CP n ; Z) = Z[α]/(α n+1 ) is a ring homomorphism. Since f also maps H 2 (CP n ; Z) H 2 (CP n ; Z) we have that f (α) = dα for d Z. It follows that f (α k ) = d k α k for α k H 2k (CP n ; Z) i.e. f acts as multiplication by d k. By the universal coefficient theorem, H 2k (CP n ; Z) is the dual of H 2k (CP n ) and so, for a map f (α) = cα there is a corresponding map f that is also multiplication by c. Hence we have a f on H 2k (CP n ) is multiplication by d k. Hence tr( f : H 2k H k ) = d k and τ f = 1 + d + d d n (Notice that H i (CP n ) = for i not even). Claim 4. When n is even there is a fixed point. Proof. Using the previous claim, τ f = 1 + d + + d n and by the rational root theorem the only integer roots are ±1. If n is even, neither of these roots make τ f = and by Lefschetz fixed point theorem there is a fixed point. Another solution is to note that (d 1)τ f = d n+1 1 which also won t have an integer root because these are the n + 1 roots of unity. Claim 5. When n is odd there is a fixed point unless f (α) = α Proof. Similar to the previous claim, but this time, if d = 1 then τ f = and there may not be a fixed point. This case d = 1 is the situation when f (α) = α Proof. We have H (RP 3 ; Z 2 ) = Z 2 [x]/(x 4 ) by example 3.12 [H, section 3.2 page 213]. Additionally we have H (RP 2 S 3 ; Z 2 ) = H (RP 2 ; Z 2 ) H (S 3 ; Z 2 ) [H, section 3.2 example 3.14 page 213] = Z 2 [y]/(y 3 ) H (S 3 ; Z 2 ) = Z 2 [y]/(y 3 ) Z 2 where the last equality is by H k (S 3 ) = Z for k = 3 and otherwise, in addition to the universal coefficient theorem. Notice that Z 2 [x]/(x 4 ) = Z2 [y]/(y 3 ) Z 2 since if there existed an isomorphism then x (y a, ) in the first component (this is a ring hom. which must send indeterminant to indeterminant) but then x 3 (, ) 4

5 or (, 1). The first case can t happen because an isomorphism is an injection and if the second case were true then x (1, ) since the zero divisors must correspond, but this map is also not possible. In Evan Dummit s solutions he notices that Z 2 [x]/(x 4 ) is a PID, but Z 2 [y]/(y 3 ) Z 2 = Z 2 [y]/(y 3 ) Z 2 [z]/(z 2 ) is not. I like this faster, easier proof Admittedly, I didn t have insight into how to approach this problem - I think because I don t have experience with cellular cohomology enough. I will probably ask about this later. I will present a proof I found online. (I will try to cite the source later) Proof. There is a straightforward cellular chain complex for CP 2 given by adding cells at even dimension i.e., 2, 4 (which I found online - I will try to add a citation later) The space X is a CP 2 an additional 3-cell so it will have a cellular chain complex Z }{{} dim Z }{{} dim 2 p Z }{{} dim 3 Z }{{} dim 4 This happens to be the same chain complex for Y as well and the cellular cochain complex is p Z Z Z Z Using this chain complex, we may compute cohomologies using ker /Im to conclude that with Z coefficients, the cohomology groups are Z for dimension 4, Z p for dimension 3, and for everything else. Therfore all cup products on the cohomology ring will be trivial in this case. However, with Z p coefficients, the cohomology groups (computed from the chain) are all Z p for dimensions, 2, 3, 4. I think this is because in this case the chain is Z p Z p p i.e. map Z p Z p Because there are limited non-trivial cohomology groups, most of the cup products are still trivial, but we will show the cup product on the 2nd cohomology groups differ for X and Y. Claim 6. The cup product x x is trivial for the generator x H 2 (Y; Z p ) 5

6 Proof. There exists a projection map p : Y M(Z p, 2) which induces a map p : H 2 (M; Z p ) H 2 (Y; Z p ). Given x H 2 (Y; Z p ), x H 2 (M; Z p ) s.t. p (x ) = x Currently I don t see why this map is surjective though. Then x x = p (x ) p (x ) = p (x x ). However x x = because H 4 (M; Z p ) =. Hence cup product x x is trivial. Claim 7. The cup product x x is non-trivial for the generator x H 2 (X; Z p ) Proof. Since X is a modified CP 2 there is a canonical map i : CP 2 X which induces an isomorphism on H ( ; Z p ) for dimensions 2 and 4. Now for a generator y H 2 (X; Z p ) there is a y H 2 (CP 2 ; Z p ) s.t. i (y ) = y and then by commutativity of induced maps with cup products we get y y = i (y ) i (y ) = i (y y ) but y y is the generator of H 4 (CP 2 ; Z p ) since generators cup product to generators. Then i is an isomorphism sending the generator y y to the generator of H 2 (X; Z p ). Hence the cup product y y maps to the generator and is non-trivial. By the above two claims we get that X and Y have non-isomorphic cohomology rings For the ease of computation I will assume that these are tensors over Z. Proof. By the assumption that H n (X, Z) is free then by the universal coefficient theorem we have H n (X; Z p ) = hom(h n (X); Z p ) = Z p α A H n (X; Z) = hom(h n (X); Z) = Z α A Where α A index the generators of the free group H n (X)(potentially an arbitrary product, notice that if this were a direct sum, we could simply state that tensors commute with arbitrary direct sums and Z Z Z p = Z p ). Hence we may view the cohomology rings as H (X; Z) Z p = ( H n (X; Z)) Z p = Z Z p H (X; Z p ) = ( H n (X; Z p ) ) = Z 6

7 We may now construct a bilinear map f : Z Z p Z x β, y (x β y) I think it is easy to check that this map is bilinear and it is clearly surjective, hence it induces a map on tensor products f : Z ( Z p Z ) x β y (x β y) This homomorphism will again be surjective so all that is left is to check that this is injective and we will have an isomorphism. ) Consider a simple tensor which is a generator of the tensor ring s.t. ( x β 1 (note that any simple tensor may be put in this form). This implies that x β = Z p, β B, but this implies that β, x β = px β. Then x β 1 = px β 1 = p x β 1 = = ( x β ( x β where this last tensor is the tensor, hence only maps to and the homomorphism is injective. Thus we have an isomorphism. ) ) p I got help on this from the Standford solutions online because it reminded me to look at the Kunneth formula specifically it help me understand that each of the generators of H k+l (S k S l ; Z) is generated by a cup product of elements in H k (S k ; Z) and H l (S l ; Z) 7

8 Proof. We have the following diagram H k (S k ; Z) H l (S l ; Z) H k+l (S k S l ; Z) f H k+l (S k+l ; Z) where the first equality is provided by the Kunneth Formula and projections p 1 : S k S l S k and p 2 : S k S l S l. Aside: Usually, the Kunneth formula is a statement { about cohomology rings, however in this case, the homology groups H n (S i ; Z) = and so, by the universal Z n = i o.w. coefficient theorem, the statement of the Kunneth formula reduce the statement above. It follows that for any generator c H k+l (S k S l ; Z) there must exist generators a H k (S k ; Z and b H l (S l ; Z) s.t. p1 (a) p 2 (b) = c. However, by the commutivity of induced maps with the cup product f (c) = f (p 1(a) p 2(b)) = f (p 1(a)) f (p 2(b)) but p 1 (a) Hk (S k S l ; Z) s.t. f p 1 (a) Hk (S k+l ; Z) but this is a trivial group (seen by again applying universal coefficient theorem to the homology groups of S k+l )) so f p 1 (a) = and similarly for f p 2 (b). Therefore f (c) = f (p 1 (a)) f (p 2 (b)) = and f is the trivial map. By the isomorphism provided in the universal coefficient theorem, f being a trivial map implies that f is also a trivial map (specifically because taking duals is a functor and the set of maps from H n ( ) Z has non-trivial maps). Therefore, the induced map on homologies, f : k+l (S k+l ) H k+l (S k S l ) is also trivial I don t have a good answer for these. Perhaps in the future I will go back and correct this Claim 8. For the closed orientable surface M of genus g 1, for each α H 1 (M; Z) there exists β inh 1 (M; Z) s.t. α β = Proof. This follows from work done in example 3.7 [H, section 3.2 page 27] or work done in problem 1 of this hmwk. The first cohomology group H 1 (M; Z) = Z 2g, has generators α i, β i for 1 i g. s.t. α i β i =. Hence given any element α H 1 (M; Z), we can realize α as a sum with some α i or β i. Suppose α = α i +..., then using the distributive property of the cup product we get that α α i =. 8

9 I don t see how to get the deduction. I see that H (X Y) splits as a direct product and that we can apply the Kunneth formula to the product, but beyond this I m not sure how to produce an α in this cohomology ring that will serve as a counter example to a proposed isomorphism. I think that the second problem will be similar to the first, following the work in the example 3.7 again. 9