# Tag Info

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 $H_n(-, \mathbb{Z})$.

1

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.

2

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: ...

0

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 ...

0

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 $f_R\in ... 1 You may just apply Borsuk-Ulam theorem directly. Define a function$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$is the measure given, ... 1 (Posting an answer to get this off the unanswered queue.) Ralph Mellish mentions that proof of the kind requested can be found as Cor 1.26 in Vick's Homology Theory. I note the same proof appears as theorem 2.28 in Chapter 2.2 of Hatcher's Algebraic Topology. 1 You are missing a hypothesis they are assuming, which is that$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$is homotopic to ... 2 See this blog post. It's true in general that the cohomology of a smooth hypersurface of degree$n$in$\mathbb{CP}^d$only depends on$d$and$n$. 1 Keep in mind the following picture: This is supposed to represent a homotopy of paths$[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: $$\tilde{H}(s,t) = \begin{cases} (1-s) \cdot (1-2t, 2t, 0) + s \cdot (1-t, 0, t), & 0 \le t \le 1/2, \\ (1-s) \cdot (0, ... 0 I believe I've found the answer myself. Let me now speak of \omega on the level of forms, recalling that E_{2} = H_{\delta} H_{d}(C^{*}(\pi^{-1} \mathfrak{U}, \Omega^{*})). By this I mean \omega_{\alpha_0 \ldots \alpha_p} = \sum_{i = 1}^{n} a^{\alpha_0 \ldots \alpha_p}_{i} \sigma_i represents something in H_{\delta} H_d cohomology, so d\omega = 0 ... 1 As Lee Mosher has pointed out in the comments, the correct representation for the fundamental group should be F_3/\langle aabcb^{-1}c^{-1} =1\rangle, or in other words, \langle a, b, c|aabcb^{-1}c^{-1} = 1\rangle. H_1 is precisely abelianization of this group, and the abelianization just attaches the relators [a, b] = [b, c] = [a, c] = 1 to the ... 0 A little bit late... (1 year later). You can also use @Seirios idea inductively, i.e.$$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 ... 2 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, ... 0 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 ... 3 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 ... 1 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 ... 1 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 ...

Top 50 recent answers are included