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30

Fundamental reason why cohomology is more powerful is that singular cohomology has a natural ring structure over them. If $\varphi$ and $\psi$ are $R$-valued $m$ and $n$-cochains on a topological space $X$, then one can define an $(m+n)$ cochain $\varphi \smile \psi$ by $$(\varphi \smile \psi)([v_0, \cdots, v_{m+n}]) = \varphi([v_0, \cdots, v_m]) ...


16

Quote from Elements of Algebraic Topology by J. R. Munkres: One answer is that cohomology appers naturally when one studies the problem of classifying, up to homotopy, maps of one space in other. Another is that cohomology is involved when one integrates differential forms on manifolds. Still another answer is [...] that the cohomology groups have an ...


5

Others have given the first reasons that arose historically, and that relate to a first course in algebraic topology. But a further point comes up from a more advanced perspective: sheaf cohomology has a natural definition using injective resolutions, and these exist in any category of sheaves. You can define sheaf homology by projective resolutions but ...


5

Not in general. Suppose that all of the $C_n = \oplus_{i = 0}^{\infty} \mathbb{k}$, with differentials $0$, and let $G = \mathbb{k}$. Then $Hom(C_n, \mathbb{k}) = \Pi_{i = 0}^{\infty} \mathbb{k}$. Taking the $\mathbb{k}$ dual again gives something containing as a subspace $\Pi_{i = 0}^{\infty} \mathbb{k}$, with zero as the differentials. The cohomology ...


5

The Hodge duality gives the non degeneracy of the pairing. The Hodge star define a duality on de Rham forms, that is, if $\alpha$ is a non-zero $k$-form then $\star \alpha$ is a non-zero $(n-k)$-form. The defining property of the $\star$ operator is indeed given by $\beta \wedge \star \alpha = \langle \alpha, \beta \rangle \operatorname{vol}$, where ...


4

If $G = \mathbb R$ (or any field), and $C$ is the simplical chain complex associated to a countably infinite discrete space (e.g., the integers), then $H_0 = Z_0$ will be an infinite direct sum of copies of the reals, but $Z^0$ will be the dual space of that, and the double dual will not be isomorphic to the original (because of the inifinite dimensions), ...


4

Since $X$ is contractible, $X/A$ is naturally homotopy equivalent to the suspension $\Sigma A$. (Explicitly, contractibility of $X$ allows you to extend the inclusion $A\to X$ to a cone $CA$ on $A$, and then you get an induced map $\Sigma A=CA/A\to X/A$. From the long exact sequences on cohomology it is immediate that this map induces isomorphisms on ...


4

Here is an answer without Mayer–Vietoris – maybe it's possible to use the technique you describe, but the following approach seems easier to me. Homology commutes with (some) directed colimits. You can e.g. find a precise statement in Hatcher's book Algebraic Topology, Proposition 3.33. Your space $X$ can naturally be seen as a directed colimit ...


4

A topological space $X$ is called $k$-connected if $\pi_i(X) = 0$ for $0 \leq i \leq k$. The Hurewicz theorem states that if $X$ is $(n-2)$-connected, then the Hurewicz homomorphism $h_* : \pi_n(X) \to H_n(X)$ is surjective. In your example, while $\pi_2(\mathbb{RP}^5) = 0$, $\pi_1(\mathbb{RP}^5) = \mathbb{Z}/2\mathbb{Z} \neq 0$, so $\mathbb{RP}^5$ is not ...


3

Let $\alpha$ be an $n$-cell of $X$ and $\beta$ be an $n$-cell of Y. Then $f(\alpha)=\sum_{\beta\in J_n'} y_{\alpha \beta} \beta$. Here $J_n'$ denotes the set of $n$-cells of $Y$. We wish to determine the values of $y_{\alpha \beta}\in \mathbb Z$. Let $\varphi_\alpha:S^{n-1}\to X^{n-1}$ be the attaching map of $\alpha$. Let $\overline{f}:X^n/X^{n-1}\to ...


3

Yes, this is true over $\mathbb{Z}$ by the universal coefficient theorem over $\mathbb{Z}$. For the question with "homology" and "cohomology" switched see this math.SE question.


3

Here's a Mayer-Vietoris argument. Write $X$ as a union $\cup_{i=0}^\infty T_i$ where each $T_i$ is a torus with two boundary components (except $T_0$, which has only one boundary component). Let $A= \sqcup_{i=0}^\infty T_{2i}$ and $B= \sqcup_{i=0}^\infty T_{2i+1}$. Then $A \cap B$ is a countably infinite disjoint union of circles $\sqcup_{i=1}^\infty ...


3

I would tend to think of this as "Attaching the cone $C(\partial M)$ to $M$" as opposed to "collapsing the boundary" - I can visualize the former far better than the latter. The first thing to note is that this is not actually a manifold. Actually, the cone $C(N)$, $N$ a manifold, is a manifold if and only if $N$ is a sphere. (Proof: the local homology of ...


3

The introduction to the book linked below (it is all about mod 2 (co)homology) gives a nice example of a result that was first proved with mod 2 homology. It provided the tools to generalize Poincaré duality to all closed manifolds, regardless of orientibility. The author also mentions that they are used to develop the theory of spin structures and in some ...


3

The (reduced) Künneth theorem applied to $X = M(G,n) \wedge M(H,n)$ tells you that: $$\tilde{H}_i(X) = \begin{cases} G \otimes_{\mathbb{Z}} H & i = 2n \\ \operatorname{Tor}_1^{\mathbb{Z}}(G,H) & i = 2n+1 \\ 0 & i \neq 2n, 2n+1 \end{cases}$$ Clearly if $\operatorname{Tor}(G,H) = 0$ then $M(G,n) \wedge M(H,n) \simeq M(G \otimes H, 2n)$. If ...


3

These conditions are equivalent if $X$ is a levelwise finite CW complex (finitely many cells of each dimension), since this condition ensures that each homology group is finitely generated. You can prove this using the universal coefficient theorem as described here; it ensures that the torsion subgroup of $H_k(X, \mathbb{Z})$ is isomorphic to the torsion ...


3

Take $C_*=0\to \mathbb{Z}\stackrel{2}\to\mathbb{Z}\to 0$ (with the $\mathbb{Z}$s in degree $0$ and $1$ and every other term in the complex $0$) and let $D_*=0\to\mathbb{Z}/2\mathbb{Z}\to 0$ (with the $\mathbb{Z}/2\mathbb{Z}$ in degree $0$ and every other term in the complex $0$). Then there is a unique nonzero chain map $f:C_*\to D_*$, and this induces an ...


3

One has the following quotient map $q : (Y \times I, Y \times \partial I) \to (Y \times S^1, Y \times s_0)$ obtained from the equivalence relation $\sim$ on $Y \times I$ defined on $\partial I \times Y$ by $(y, 0) \sim (y, 1)$. One can prove that $q_*$ on cohomology is an isomorphism as follows. $$\require{AMScd} \begin{CD} H^k(Y \times I, Y \times ...


2

You can compute this from the described pathspace spectral sequence, using the fact that $P$ is contractible so every nontrivial term in the spectral sequence except the $(0,0)$ term must eventually disappear. Since $H^*(K(\mathbb{Z},1);\mathbb{Z})$ vanishes unless $*=0$ or $*=1$, the only possible nontrivial differentials on the spectral sequence are the ...


2

Actually, the answer is almost yes on the level of forms themselves. You've provided an inner product on $\Omega^k(M)$, and once you pass to its Hilbert space completion (the so-called space of $L^2$ forms) it is true, by the Riesz representation theorem, that any continuous linear functional $\Omega^k_{L^2}(M) \to \Bbb R$ is uniquely represented by $\alpha ...


2

The answer is yes for the following reason: $\omega_n$ is the Euler class mod 2 Now use e.g. Theorem 4.7 here which says $e(\nu_N)([N])$ counts number of intersections, or you argue that the Thom class of $\nu_N$ in $M$ is the Poincaré dual of $N$ (follows from this exercise). Hence, by pulling back to the cohomology of $N$ the result follows. You ...


2

This is true if you assume $H_n(X)$ is finitely generated for all $n$ (all coefficients in this post will be $\mathbb{Z}$). In particular, this holds if $X$ has the homotopy type of a CW-complex with finitely many cells in each degree. To prove this, we invoke the classification of finitely generated abelian groups, which says that $H_n(X)$ is a finite ...


2

What you wrote is in general completely false. Consider $A = [0,1]$ and $B = \{0,1\}$. Then $H_1(A) = 0 = H_1(B)$, but $A/B$ is a circle and $H_1(A/B) = \mathbb{Z}$. In general what is true is that if the inclusion $B \subset A$ is a cofibration (for example if $B$ is a sub-CW-complex of $A$), then the reduced homology group $\tilde{H}_k(A/B)$ is isomorphic ...


2

Your mistake is in the equality $\tilde{H}^2(S^2 \times (D^2, S^1)) = \tilde{H}^2(S^2 \times S^2)$ you wrote. It might be a bit easier to see why this is false in one dimension lower. Consider the trivial bundle $S^1 \times D^1 \to S^1$ instead. In the proof you would consider $(S^1 \times D^1, S^1 \times S^0)$, and it might be tempting to write $H^1(S^1 ...


1

For the hurewiczs homomorphism, ALL the homotopy groups $\pi_i, i\leq n$ have to be zero to have a surjective morphism $\pi_{n+1}(M)\rightarrow H_n(M)$. In your case, $\pi_1(RP^5)$ is not zero.


1

Equivariant cohomology are defined as ordinary cohomology of $V \times_G EG$. There is a projection $p: V \times_G EG \rightarrow BG$.This projection is fiber bundle with fiber $V$. If $V$ is contractible, then this projection is homotopy equivalence, hence induces isomorphism on cohomology $H^{*} (BG) \rightarrow H^{*} ( V \times_G EG)$. I.e. isomorphism ...


1

As explained in the other answer, if $(X, A)$ is a CW-pair, i.e., $X$ is a CW-complex and $A$ is a subcomplex, then one has the long exact sequence of homology groups $$\cdots \longrightarrow \tilde{H}_{k-1}(X/A) \longrightarrow \tilde{H}_k(A) \stackrel{i_*}\longrightarrow \tilde{H}_k(X) \stackrel{q_*} \longrightarrow \tilde{H}_k(X/A) \longrightarrow ...


1

You're right that $\iota_{\mathbb{F}_2,0}$ is not the multiplicative unit. However, it still is its own square. Indeed, for any space $X$, every element of $H^0(X,\mathbb{F}_2)$ is its own square. Even more strongly, this holds on the chain level: any singular $0$-cochain on a space with coefficients in $\mathbb{F}_2$ is its own square. Indeed, a ...


1

Ok it seems I've found what I was looking for. It is a consequence of the axioms of a multiplicative cohomology theory. The question I've written may seem a little confused, but I think there is a splendid explanation on Tammo tom Dieck's Algebraic Topology at page 413, Proposition 17.3.1 clearly, $\sigma$ is the suspension isomorphism he defined ...


1

Cohomology appears naturaly. Lots of problems are solved locally and then the local solutions glued together to construct global solutions, and cohomology is the precise description of how to do that.



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