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Consider the tower of spaces (I take $S^1 = K(\mathbb{Z}, 1)$ for simplicity, but any $K(\mathbb{Z}, n)$ would do): $$\ldots \to S^1 \xrightarrow{\cdot^p} S^1 \xrightarrow{\cdot^p} S^1$$ Where I'm considering $X_s = S^1 = \{ z \in \mathbb{C} : |z| = 1 \}$, $p$ is a prime number, and $X_s \to X_{s-1}$ is raising to the $p$th power (you may have to replace ...

2

Have a look at the statement of the universal coefficient theorem, and pay attention to the difference between $R$ and $G$, $$0 \to \operatorname{Ext}_R^1(\operatorname{H}_{k-1}(X; R), G) \to H^k(X; G) \to \operatorname{Hom}_R(H_k(X; R), G)\to 0.$$ In the context of your question, $G = \Bbb Z_2$ and there are two ways to apply the universal coefficient ...

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You got confused with the coefficients that are involved in the universal coefficient theorem. Let $K$ be a principle ideal domain and $M$ a $K$-module. We have a short exact sequence $$0 \to Ext_K(\ H_{l-1}(RP^n, K),\ M) \to H^l(RP^n, M) \to Hom_{K}(\ H_l(RP^n, R),\ M) \to 0 \,.$$ You looked at $Ext_K(\ H_{l-1}(RP^n, M),\ M)$ and $Hom_{M}( H_l(RP^n, ... 0 The universal coefficient theorem that you're trying to use only works for chain complexes whose terms are free abelian groups. But when you take homology with coefficients in$\Bbb Z_2$, your chain groups aren't at all free abelian groups. Rather, they are free as$\Bbb Z_2$-modules. The version of the universal coefficient theorem you want to use here is ... 2 Let$G$be the Galois group of$K$over$k$. The image is the kernel of a map$Br(K)^G \to H^3(G, K^*)$. Look in books on Galois cohomology, preferably one written in French... 4 There is a paper, that may partially answer your question. The paper is entitled "Every finite abelian group is the Brauer group of a ring" by T. J. Ford. You can find it here. 3$H_i(M,\mathbb{Z}_2)$is a finite-dimensional vector space.$H^i(M,\mathbb{Z}_2)$is its dual, and finite-dimensional vector spaces over any field are isomorphic to their duals. 0 I suggest that you read two of the early papers which gave the first interesting examples of infinite dimensional bounded cohomology. In the first paper Brooks computes 2-D bounded cohomology of free groups, and in the second paper Brooks and Series compute 2-D bounded cohomology of surface groups. Brooks, Robert. "Some remarks on bounded cohomology". ... 0 The de Rham cohomology is functorial in the sense that if you have a smooth map$f: M\to N$between two manifolds then you have an induced map on the$p$-th cohomology groups $$f^*: H^p(M) \to H^p(N)$$ given by the pullback, which is defined on every 1-form$\omega$by the following formula $$f^*\omega(X(p)) = \omega(f_*(X(p)))$$ Lemma. Let$M$and$N$... 2 Hint: Use the Mayer-Vietoris sequence to prove the first part. For the second part, consider any space whose cohomology ring has nontrivial products and check what the dimension of the products would be. For example$H^*(\mathbb{CP}^2) = \mathbb{Z}[x]/(x^3)$with$|x| = 2$. This means that if$X = \Sigma \mathbb{CP}^2$, then: $$H^i(X) = \begin{cases} ... 1 The answer to your question is that you don't view those things as kG-modules. To construct the Bockstein, you identify \operatorname{Ext}_{\mathbb{F}_pG}(\mathbb{F}_p, \mathbb{F}_p) with \operatorname{Ext}_{\mathbb{Z}G}(\mathbb{Z}, \mathbb{F}_p). You have a short exact sequence of \mathbb{Z}G-modules with trivial action$$ 0 \to \mathbb{F}_p \to ... 1 With regard to your first question: in that section, the assumption given at the start (that$M$is a projective$R$-module) is in force throughout. This makes$ -\otimes_R M$exact, since projective modules are flat. Otherwise you are right that there is no reason to think the sequence will remain exact after tensoring with$M$. For your second question, ... 0 First of all, you need to put a simplicial complex structure on$\mathbb{R}$in order to define simplicial cohomology on it. Suppose you take the integers for your$0$-skeleton and your 1-simplices to be$[n,n+1]$for$n \in \mathbb{Z}$. The second thing is that the proper definition for an$n$-cochain$f$with compact support is that there exists a finite ... 1 I've learned the answer to my question, which I write here for future questioners: First, we assume that$\Lambda$is Artinian, so that any finitely generated$\Lambda$-module has a composition series, and also any submodule of a finitely generated module is itself finitely generated. We first show that any finitely generated module has a unique projective ... 4 The Euler-PoincarĂ© characteristic$\chi(V)=\chi(V,\mathcal O_V)$of the structure sheaf of a smooth projective variety is called the Hirzebruch arithmetic genus and it is multiplicative : $$\chi(V\times W)=\chi(V)\times \chi(W)$$ This immediately solves your problem:$\chi(\mathbb P^1\times \mathbb P^1)=\chi(\mathbb P^1)\times \chi(\mathbb P^1)=1\times ...

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I think you just need to write everything out explicitly here; $$<i^*(c),y> = i^*(c)(y) = c(i(y)) = c(i_*(y)) = <c,i_*(y)>$$ Be careful to understand what is going on at each step.

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Every f.g. abelian group is the direct sum of finitely many finite cyclic groups and a f.g. torsion-free (hence free) abelian group. $\mathrm{Ext}_\mathbb{Z}^1 (-, \mathbb{Z})$ preserves finite direct sums, so it suffices to show that $\mathrm{Ext}_\mathbb{Z}^1 (M, \mathbb{Z})$ is a finite group if $M$ is a finite cyclic group. In fact, for all positive ...

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The line bundle $\eta$ has a nowhere vanishing section ($x \mapsto x$ is a section if you use the standard embedding $S^n \hookrightarrow \mathbb{R}^{n+1}$), therefore it is trivial. Stiefel-Whitney classes are invariant under isomorphism of bundles, therefore $w_i(\eta) = 0$ for all $i \geq 1$.

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