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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 ...

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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 $p \colon S^{2} \times \mathbb{R} \to S^{... 6 You are quite right that$H_D^n$is single-graded. However, there is a filtration$$H_D^n = F_0 \supset F_1 \supset F_2 \supset \ldots \supset F_n \supset 0$$on$H_D^n$, and$H_D^{p, n - p}$is defined to be the quotient$$H_D^{p, n - p} = F_p/F_{p + 1}.$$Note that$H_D^{p, n - p}$is not a subgroup of$H_D^n$. Rather, it is a subquotient of$H_D^n$, a ... 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 \...

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 ...

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Wikipedia is talking about generalized cohomology theories. They satisfy a bunch of axioms formally similar to what e.g. singular cohomology satisfies. But the so-called extraordinary cohomology theories (K-theory, cohomotopy, cobordism and so on) do not come from chain complexes. Indeed, by a theorem of Burdick--Conner--Floyd, in some sense if they did "...

4

There are various generalized cohomology theories which are typically not described in terms of cochain complexes, perhaps the simplest example being topological K-theory. There's a precise technical sense in which topological K-theory cannot be "described in terms of cochain complexes," but it's not easy to spell out.

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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 $... 4 Let$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$, which is countable. 3 A. De Rham cohomology. i) some general topology (basic notions: what topological spaces are, compactness, connectedness) ii) some smooth manifold theory (basic notions: what manifolds are, tangent spaces) iii) some linear algebra (basic notions: vector spaces, exact sequences, quotient spaces) B. Sheaf cohomology. i) some sheaf theory (basic notions: what ... 3 A quick proof using Stiefel–Whitney classes: a manifold$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 ... 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$X \to S^1 =: Y$... 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 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. 2 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$\mathbb{C}$... 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. 2 Our topology professor, Prof. Michael Weiss, defined in his lectures a homology theory for topological spaces (which I think he mentioned to be equivalent to Čech cohomology) by the means of something which he calls mapping cycles. These are elements of the sheafification of the free abelian group of continuous maps between any two topological spaces. ... 1 They are distinguished by their$\pi_1X$action. Your definition of coefficients in a module, is the so-called untwisted homology by: $$C_n(X;M) := M \otimes_R C_n(X;R) = M \otimes_R R^{|n|} = M^{|n|},$$ where$|n|$denotes the number of$n$-cells of$X$with no$\pi_1$action on it. (3) in contrast is the twisted homology. As suggested by the name, (2) ... 1 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 $$\delta^... 1 Functoriality in the first variable only says that for k = 0 (morphisms have degree 0). If f is a morphism of degree k then it induces a map$$[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. 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" ...

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