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2

The answer is yes; if $A \to B$ is any morphism of abelian groups then it induces a natural transformation $H_n(-, A) \to H_n(-, B)$, which implies that the map induced by $f$ on $H_n(-, \mathbb{Z}_2)$ is the reduction $\bmod 2$ of the map induced on $H_n(-, \mathbb{Z})$.


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I think for differential topology, and geometry it is probably best to learn de Rham cohomology. Bott and Tu's book is the canonical reference.


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Although Amazon says that Massey's Singular Homology Theory is a sequel to Massey's Algebraic Topology: An Introduction, the earlier book is not a logical prerequisite for Singular Homology Theory. There are plenty of online lecture notes that might be a right fit for you, for example: ...


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This is true for the dumbest of possible reasons. If $G$ is a compact, connected Lie group, then after inverting finitely many primes in the coefficient ring, $H^*(G) = \Lambda P$ is an exterior algebra. Over characteristics other than 2, exterior algebras are free commutative graded algebras (CGAs), so any surjection $H^*(E) \to H^*(G)$ splits. One simply ...


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This is not a full answer, but maybe it can helps... As mt_ said, $\overline{f}$ and $\tilde{f}$ does not mean anything here. However, there is a map $H^2(G,\mathbb{Z}^\times)\rightarrow H^2(G,R^\times)$ (at least if $G$ acts trivialy on $-1\in R^\times$). So that if $f\in H^2(G,\mathbb{Z}^\times)$ there is a corresponding element in $f_R\in ...


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You may just apply Borsuk-Ulam theorem directly. Define a function $f$ from $S^n$ to $\mathbb{R}^n$ as follows: If $x$ is a point on $S^n$, then there is a great $S^{n-1}$ that is orthogonal to the point $x$. The $S^{n-1}$ divides $S^n$ into two regions. Let's call them $U$ and $V$, where $U$ is the region containing $x$. If $\mu$ is the measure given, ...


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(Posting an answer to get this off the unanswered queue.) Ralph Mellish mentions that proof of the kind requested can be found as Cor 1.26 in Vick's Homology Theory. I note the same proof appears as theorem 2.28 in Chapter 2.2 of Hatcher's Algebraic Topology.


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You are missing a hypothesis they are assuming, which is that $H_*(M)[\pi^{-1}]$ (which is just $H_*(M)[m_1^{-1}]$ in this case) can be constructed by right fractions. This implies that $H_*(M_\infty)=H_*(M)[m_1^{-1}]$, as the colimit that computes $H_*(M_\infty)$ is exactly the right fractions for $H_*(M)[m_1^{-1}]$. Since element of $M$ is homotopic to ...


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See this blog post. It's true in general that the cohomology of a smooth hypersurface of degree $n$ in $\mathbb{CP}^d$ only depends on $d$ and $n$.


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Keep in mind the following picture: This is supposed to represent a homotopy of paths $[0,1] \to \Delta^2$ between the path that goes $0 \to 1 \to 2$ and the path that goes straight $0 \to 2$. If you want a precise definition, let: $$\tilde{H}(s,t) = \begin{cases} (1-s) \cdot (1-2t, 2t, 0) + s \cdot (1-t, 0, t), & 0 \le t \le 1/2, \\ (1-s) \cdot (0, ...


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I believe I've found the answer myself. Let me now speak of $\omega$ on the level of forms, recalling that $E_{2} = H_{\delta} H_{d}(C^{*}(\pi^{-1} \mathfrak{U}, \Omega^{*}))$. By this I mean $\omega_{\alpha_0 \ldots \alpha_p} = \sum_{i = 1}^{n} a^{\alpha_0 \ldots \alpha_p}_{i} \sigma_i$ represents something in $H_{\delta} H_d$ cohomology, so $d\omega = 0$ ...


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As Lee Mosher has pointed out in the comments, the correct representation for the fundamental group should be $F_3/\langle aabcb^{-1}c^{-1} =1\rangle$, or in other words, $\langle a, b, c|aabcb^{-1}c^{-1} = 1\rangle$. $H_1$ is precisely abelianization of this group, and the abelianization just attaches the relators $[a, b] = [b, c] = [a, c] = 1$ to the ...


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A little bit late... (1 year later). You can also use @Seirios idea inductively, i.e. $$H_k(B^n,S^{n-1})\overset{\sim}{\longrightarrow}H_{k-1}(S^{n-1})$$ by useing $H_k(B^n,S^{n-1})\simeq \overset{\sim}{H}_k(B^n/S^{n-1},pt.)\simeq \overset{\sim}{H}_k(S^{n})$ (because $B^n/S^{n-1}\cong S^{n}$). I think the basic idea can be found in Hatcher's book. But I ...


2

(Edit after incorrect first answer) Let $f$ be a cocycle (central extension with abelian quotient): writing additively, the cocycle condition is $f(a,b)+f(a+b,c)=f(a,b+c)+f(b,c)$. Then $g:(a,b)\mapsto f(a,b)-f(0,0)$ is also a cocycle. When $c=0$ the cocycle relation yields $g(a,0)=0$ for all $a$, and similarly $g(0,a)=0$ for all $a$. When $c=a$ the ...


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Note that the squaring operation $H^1(-;\mathbb{Z}/2)\to H^2(-;\mathbb{Z}/2)$ coincides with the Bockstein. It follows your cohomological condition is equivalent to $H_1(M;\mathbb{Z}_{(2)})=\mathbb{Z}/2$. One example of such a manifold is the Enriques surface, which has fundamental group $\mathbb{Z}/2$ and universal cover the K3 surface. More generally, ...


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Using the long exact sequence, we know that $H^0(G,\mathbb{Z})=\mathbb{Z}^G=\mathbb{Z}$, so it suffices calculate the kernel of the map $H^0(G,\mathbb{Z})\to H^1(G,IG)$. I think an easier way to go about this problem (easier because the map above is not obvious, at least to me) is just with a direct calculation, which is done below. If ...


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Your lecturer defined the Čech cohomology of the cover $\mathcal U$ of $X$. Since the cover uses small enough balls it is a good cover, and gives the same result as any other good cover. The Wikipedia article begins the same way, at least now. I do not know what it said when you posted this question. And since good covers are cofinal in all covers of $X$ ...


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In order to mark this question as answered, the above reasoning is correct. (Thanks to Qiaochu Yuan for the confirm)


4

I should have found a proof of the statement. First I state the following claims that are well known results about co-h-spaces (for a reference see the literature cited in the comments). Claim 1: if X is a co-h-space then $\pi_1(X)$ is free. Claim 2: if X is a co-h-space then taken $\alpha \in H^{p_1}(X; G_1)$, $\beta \in H^{p_2}(X; g_2)$ the cup product ...


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As you probably know the homology of the mapping cone is related to the homotopy of the chain complexes $C$ and $D$ by the following exact sequence: $$ \cdots\to H_n(D)\to H_n(E(\varphi))\to H_{n-1}(C)\stackrel{\varphi_*}\to H_{n-1}(D)\to\cdots $$ If $\varphi_*:H(C)\to H(D)$ is an isomorphism, then $H(E(\varphi))=0$. Since $E(\varphi)$ is projective (in ...



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