1
$\begingroup$

In some notes I came across the following equality (or to be more precise: isomorphism) that $\check{H}^2(M,\mathbb{T}) \cong H^3(M,\mathbb{Z})$ where $\check{H}$ is Cech cohomology and $\mathbb{T}$ is the circle. I would like to understand why is it true. EDIT: Here $M$ is assumed to be smooth manifold (this is enough for my purposes).

  • 1
    Am I right in thinking that there is a short exact sequence of sheaves, $ 0 \to \mathbb Z \to \mathbb R \to \mathbb T \to 0$? Your $H^2(\mathbb T) \to H^3(\mathbb Z)$ is one of the morphisms in the associated long exact sequence. Can you give us any more information about the space $M$? – 2017-02-24
  • 0
    I made an edit to my post, sorry for being not precise. – 2017-02-24
  • 0
    Is that all we know about $M$? What if $M = S^3$? Then the LES I mentioned has a bit that looks like $0 \to H^2(\mathbb T) \to H^3 (\mathbb Z) \to H^3 (\mathbb R)$, and the final morphism maps the generator in $H^3(\mathbb Z) \cong \mathbb Z$ to something non-trivial in $H^3 (\mathbb R) \cong \mathbb R$... which means that $H^2(\mathbb T) = 0$! Did I make a mistake somewhere? – 2017-02-24
  • 0
    Why you claim that $H^2(M\mathbb{T})$ is zero? – 2017-02-24
  • 0
    What do you mean by $\check{H}^2(M,\mathbb{T})$. Is $\mathbb{T}$ identified with the sheaf of continuous function $M\rightarrow\mathbb{T}$ ? – 2017-02-24
  • 3
    @Kenny We're not taking $\Bbb R$ coefficients, we're taking the cohomology of the sheaf of real-valued continuous functions. (The sheaf of locally constant functions would compute cohomology with real coefficients.) This is a fine sheaf, so has trivial cohomology in all positive degrees. – 2017-02-24
  • 1
    @MikeMiller I see! Thanks for pointing this out! So $\mathbb T$ is meant to represent the sheaf of circle-valued continuous functions (as opposed to circle-valued locally constant functions). Then I agree that the LES implies that $H^2(\mathbb T) = H^3(\mathbb Z)$. – 2017-02-24
  • 0
    Since I'm a bit uncertain about the stuff concerning sheafs (once I've learned that Cech cohomology is isomorphic with the singular cohomology I though I could relax and not bothering knowing the details about sheaf theory) I would be very grateful for a canonical answer. – 2017-02-24
  • 1
    Mike Miller's answer is the canonical one. Write the long exact sequence associated to $0\rightarrow\mathbb{Z}\rightarrow\mathcal{C}\rightarrow\mathbb{T}\rightarrow 0$ and use the fact that the sheaf of continuous function $\mathcal{C}$ is fine so has vanishing higher cohomology. – 2017-02-24

1 Answers 1

2

Here is an alternative approach to the one given in the comments. I will blithely assume the following fact, although I am not sure in what generality it is true: if $A$ is a topological abelian group and $X$ is reasonable, then the Cech cohomology $\check{H}^n(X, A)$ should agree with the representable cohomology given by homotopy classes of maps from $X$ to the classifying space $B^n A$.

Assuming this, it now suffices to observe that the classifying space $B^2 S^1$ can be identified with the classifying space $B^3 \mathbb{Z}$, since $S^1 = B \mathbb{Z}$.

In this language, the long exact sequence associated to the short exact sequence of topological groups

$$0 \to \mathbb{Z} \to \mathbb{R} \to S^1 \to 0$$

deloops to the fiber sequence of classifying spaces

$$B^2 \mathbb{Z} \to B^2 \mathbb{R} \to B^2 S^1$$

and since $\mathbb{R}$ is contractible, so is $B^2 \mathbb{R}$. This fiber sequence therefore identifies $B^2 \mathbb{Z}$ with the loop space of $B^2 S^1$, and hence identifies $B^2 S^1$ with the delooping $B(B^2 \mathbb{Z}) \cong B^3 \mathbb{Z}$.