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You are correct. The first displayed formula on page $95$ is $$\delta^{p-1} : \operatorname{Hom}(\mathsf{C}_{p-1}, G) \to \operatorname{Hom}(\mathsf{C}_p, G)$$ and on the first line of that page (before the displayed formula), $\mathsf{C}^p$ is defined as $\operatorname{Hom}(\mathsf{C}_p, G)$. With this definition, the above can be rewritten as $$\delta^... 7 Following Mike's nice hint, note that \pi_{1}(\mathbb{R}\mathbb{P}^{3}) = \mathbb{Z}/2\mathbb{Z}, and \pi_{1}(S^{2} \times S^{1}) \cong \pi_{1}(S^{2}) \times \pi_{1}(S^{1}) \cong \mathbb{Z}, so the induced map \pi_{1}(\mathbb{R}\mathbb{P}^{3}) \to \pi_{1}(S^{2} \times S^{1}) must be the zero map. Recalling that p \colon S^{2} \times \mathbb{R} \to S^{... 5 The cohomology ring of \Bbb{CP}^n is \Bbb Z[x]/(x^{n+1}), with |x|=2. Any map f: \Bbb{CP}^n \to S^2 induces zero on cohomology when n>1, because if z \in H^2(S^2) is a generator, then 0 = f^*(z^2) = f^*(z)^2, so f^*(z) must be zero. Now use the fact that the universal coefficient theorem is natural to see that the induced map on H_2 is ... 4 Let X=[0,1] and A=\{0,1,1/2,1/3,\cdots\}. Then the quotient X/A is homeomorphic to the Hawaiian earring which has uncountable H_1 (try to prove this). On the other hand, H_1(X,A) is isomorphic to \bigoplus_{i=1}^\infty \Bbb Z, which is countable. 2 Let H\subseteq G be a subgroup. There are two standard maps on the group cohomology associated to this: res_H^G:H^*(G, M)\to H^*(H, M|_H) and tr^G_H:H^*(H, M|_H)\to H^*(G, M). We have the relation tr\circ res(x)=[G:H]x. Now we wish to show that in the case H=S, a Sylow-group of G, we have that res_S^G is an injection. Letting x\in Ker(res_S^G)... 2 They mean to have the locally path-connected assumption, presumably, but didn't feel the need to state it (often "space" can be taken to mean "non-terrible space"). As a counterexample in general, take the Warsaw circle X. This has trivial homotopy groups and homology groups (exercise), but collapsing the 'bad part' gives us a non-null map X \to S^1 =: Y... 1 I've recently struggled with the same sort of things, so I'll try to explain how I understand it. This is all supposing you understand the technical tools such as the isomorphism H_n(X,A) \cong H_n(X/A), naturality, degree theory, and the likes. First, let me say that there are a variety of levels you can make this argument so "the right way to see it" ... 3 A quick proof using Stiefel–Whitney classes: a manifold M is orientable iff the first SW class w_1(M) \in H^1(M;\mathbb{Z}/2\mathbb{Z}) is zero. But by the universal coefficient theorem,$$H^1(M;\mathbb{Z}/2\mathbb{Z}) = \operatorname{Hom}_\mathbb{Z}(H_1(M;\mathbb{Z}), \mathbb{Z}/2\mathbb{Z}) = 0.$$Of course under the hood I don't think there's ... 9 For any connected manifold M, there is a homomorphism \pi_1(M)\to\mathbb{Z}/2 which sends a loop to 0 if going around the loop preserves orientation and sends the loop to 1 if going around the loop reverses orientation. This homomorphism is trivial iff M is orientable. Since \mathbb{Z}/2 is abelian, this homomorphism factors through the ... 0 I think that I can give an answer, at least for the very first part of my question. Assume that GL_{n}(\mathbb{R}) is the general linear over \mathbb{R}. It works almost in the same fashion for \mathbb{C} as well. Then, someone can prove that the "natural map" \rho:V_{n}{\mathbb{R}^{\infty}} \rightarrow G_{n}{\mathbb{R}^{\infty}} from the n-frames ... 4 It is not true that (-(j_U)_* \omega , (j_V)_*\omega)=0 iff \omega=0. Note that (j_U)_*\omega and (j_V)_*\omega here are cohomology classes in H^1_c(U) and H^1_c(V), not just 1-forms. So we have to consider the possibility that they might be coboundaries. A 1-form on U is a coboundary iff its integral is 0, and similarly for V. So ... 2 Rudyak's article The problem of realization of homology classes from Poincare up to the present seems to be a good starting point. The original paper is Thom's Quelques propriétés globales des variétés différentiables. 2 The trouble is that an object of the functor category [\cal B,\cal C] need not be projective even if every object in the image is. I think what you would get would be a weak homotopy (homotopy at each object of \cal B). For more on this subject, we my book titled Acyclic Models, which deals with every version of the theorem I am aware of. 5 Consider an embedding of a torus N inside M=(\mathbb R^3 \text{with a line removed}), such that the torus is the boundary of a regular neighborhood of a curve that goes around the line once. Since M is homotopy equivalent to a circle, the induced map on H_2(N)\to H_2(M) is trivial. Yet on the level of H_1 we have a nontrivial map \mathbb Z\oplus \... 1 Functoriality in the first variable only says that for k = 0 (morphisms have degree 0). If f is a morphism of degree k then it induces a map$$[X[k], Y] \to [W, Y]$$and there are some signs involved in moving this shift [k] around. 1 So Roland put you onto the right track, but just so you have something to check against later (and for those who may come after): Let \sigma: \Delta^2 \to X be defined by first projecting \Delta^2 onto the edge [v_0, v_2] (orthogonally, so that v_1 ends up mapping to the midpoint of [v_0, v_2]), and then mapping via f\cdot g: [v_0, v_2] \to X. ... 1 Yes, the candidate is BG the classifying space of G. You can endow G with the discrete topology, you have the universal bundle EG\rightarrow BG which is a covering and EG is contractible BG is a K(G,1), apply Serre 1.5 p.91 Serre, Jean Pierre. Cohomologie des groupes discrets. http://www.maths.ed.ac.uk/~aar/papers/serrecoh2.pdf 1 When you apply the Kunneth theorem, the Tor is being taken over \mathbb{F}_p, not over \mathbb{Z}, and so it vanishes. Over a field the Kunneth theorem just says that the homology of a product is a graded tensor product of homologies. The assumption that p > 3 does not help you in any way. 3 A. De Rham cohomology. i) some general topology (basic notions: what topological spaces are, compactness, connectedness) ii) some smooth manifold theory (basic notions: what manifolds are, tangent spaces) iii) some linear algebra (basic notions: vector spaces, exact sequences, quotient spaces) B. Sheaf cohomology. i) some sheaf theory (basic notions: what ... 0 You nearly have solved your problem. Observe the connecting homomorphisms \partial_* of the long exact sequence you used$$\require{AMScd} \begin{CD} \cdots @>{\partial_*}>> H_i(X,pt) @>{i_{1*}}>> H_i(X \vee Y,pt) @>{}>> H_i(X \vee Y, X) @>{\partial_*}>> \cdots\\ @.@.@VV\pi_{2*}V @A{\cong}Ak_*A \\ @.@.H_i(Y,pt)@= H_i(Y,...

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In the case of the Floer homology of the cotangent bundle the answer is yes. You should have a look at this: The Viterbo transfer as a map of spectra by Thomas Kragh.

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Following Mike Miller's suggestion, consider the cylinder $X =S^1 \times \mathbb{R}$ (as a Riemann surface, you may view it as either $\mathbb{C} \setminus \{0\}$ or $\mathbb{C}/\mathbb{Z}$). As this deformation retracts onto the base circle and homology is a homotopy invariant, we know that $H_2(X;\mathbb{C}) \cong H_2(S^1;\mathbb{C}) =0$. As $\mathbb{C}$ ...

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All the constructions that you used to define the isomorphism are natural/functorial: Given a map $X \to Y$, you have a natural map that respect inclusions, which gives a starting point for all the applications of naturality to come: $$(\Sigma X, C_+ X, C_- X, X \times \{0\}) \to (\Sigma Y, C_+ Y, C_- Y, Y \times \{0\});$$ The long exact sequence of a pair ...

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