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Let X be a compact space.

Recall that its Čech cohomology $H^\bullet(X,\mathbb Z)$ is given by the colomit $\mathrm{colim}_U\big(H^*(C^\bullet(U;\mathbb Z),\delta)\big)$, where $U=(U_i)$ runs over all open covers of X, ordered by refining. For completeness, let us also recall that the n-cochains $C^n(U;\mathbb Z)$ are the group of continuous ℤ-valued functions on $\bigsqcup U_{i_1}\cap\ldots\cap U_{i_{n+1}}$.
Since X is compact, we may restrict ourselves to finite covers, without modifying the answer.

•  Definition:  A closed cover $V=(V_i)$ of X is a finite collection of closed subsets $V_i$ whose union is X.

We may now consider the modified Čech cohomology $\tilde H^\bullet(X,\mathbb Z)$, where we use closed covers instead of open covers.

•  Question: Are $\tilde H^\bullet(X,\mathbb Z)$ and $H^\bullet(X,\mathbb Z)$ isomorphic?


PS: I know how to show that $\tilde H^1(X,\mathbb Z)$ and $H^1(X,\mathbb Z)$ are isomorphic, by using the fact that they both classify ℤ-principal bundles.

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Andre, for n=2, can't you use the long exact sequence in sheaf cohomologies and just compute $H^1(X,U(1))$ with respect to both covers instead? Since $U(1)$ is a Lie group, Palais' theorem should still apply. –  David Carchedi May 25 '10 at 1:05
    
Interesting thought. In order for that argument to work, I would need to know that the sheaf of R-valued functions is acyclic w.r.t. closed covers. –  André Henriques May 25 '10 at 12:28
    
The sheaf of continuous functions is not soft or fine, like it is in the open topology, but instead it is flasque, which is enough. –  Ben Wieland May 27 '10 at 19:32
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3 Answers

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As Angelo says, there is a Mayer-Vietoris (spectral) sequence for closed covers. That comes from an exact sequence of sheaves, which also shows that closed covers are covers in the sense of a Grothendieck topology. Probably it's true for proper surjective maps in general.

I think that means that there is a geometric morphism from the topos of closed covers to the usual topos. This yields a comparison map $H^i_{closed}\to H^i_{open}$. If the space is Hausdorff, one can use partitions of unity to refine open covers by closed covers. That shows that the comparison map is surjective: any cohomology class is defined on an open cover, so one can refine it to a closed cover, so it comes from the closed cohomology. If one is careful about the bookkeeping, probably one can arrange a splitting of the Čech complexes, as in André's comment. But we can do better without mentioning this other topos.

We can compute the usual cohomology using the Čech process for open covers. But we could also do it for both open covers and closed covers. This won't change the answer, since closed covers are covers for the usual Grothendieck topology. But now we can get rid of the open covers, since the closed covers are cofinal.

I'm not sure what happens in the non-Hausdorff case. It is certainly not possible to refine open covers by closed covers, but it seems to me that the line with the doubled origin works anyway.

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When I said "compact", I certainly meant "Hausdorff and compact". Also, I should have phrased my question using sheaf cohomology as opposed to Cech cohomology... You correctly mention that open covers can be refined by closed covers, and so one get a comparison homomorphism H^i_{open}(X) --> H^i_{closed}(X). My question is: is that map an isomorphism? –  André Henriques May 18 '10 at 6:37
    
heavily edited in reply. the main point is that we also have a map in the other direction. –  Ben Wieland May 18 '10 at 20:22
    
I am not sure I understand where the comparison map $H^i_{closed}\to H^i_{open}$ should come from. Could you please explain with a little more detail? –  Angelo May 19 '10 at 6:52
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One way to see it is to look at the edge map in the spectral sequence you mentioned in your answer. One edge map is from the cohomology of the total space to the cohomology of the disjoont union of the closed subspaces. The other gives a map from the Čech cohomology of the particular closed cover to sheaf cohomology. In the limit of all closed covers, it is the desired map. –  Ben Wieland May 19 '10 at 18:35
    
Oh, I see. This gives a map from closed Čech cohomology to usual cohomology, which seems to be surjective, as you say. You also have a map from usual coholomogy to closed cohomology. However, it is not clear to me that the map from closed Čech cohomology to closed sheaf cohomology (which is what André really wants) is an isomorphism. Even for the usual cohomology this is a nontrivial theorem. (I must say that I am rather confused by this business.) –  Angelo May 19 '10 at 19:42
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If $V$ is a finite closed cover of a space $X$, with the property that the higher cohomology of all finite intersections vanishes, then the Čech cohomology of $V$ coincides with that of $X$. This is because there is a Mayer-Vietoris spectral sequence for finite closed covers, completely analogous to that for open covers, which is is easy to obtain from the long exact sequence of sheaves $$ 0 \to \mathbb Z_X \to \bigoplus \mathbb Z_{V_i} \to \bigoplus \mathbb Z_{V_i \cap V_j} \to \cdots $$ (essentially, this comes from the nerve of the cover). Here $\mathbb Z_Y$ denotes the constant sheaf on $Y$, pushed forward to $X$.

On the other hand, I would be a little nervous about what happens going to the limit, even for very nice spaces. I have never thought about it, but it seems to me that there could be finite nasty closed covers of a manifold that don't admit nice refinements; so I would deem it likely that one has to restrict the covers to a mild class (I could be completely wrong, of course).

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Thank you Angelo for your answer. But I really care about the general case: I don't want to assume local contractibility, or anything similar. –  André Henriques May 10 '10 at 15:54
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In this case, you should have the usual spectral sequences linking Čech cohomology and ordinary (sheaf) cohomology. This should imply that the $H^1$'s are the same; but it would surprise me if you can prove degeneracy without any local condition. On the other hand, it is known that the Čech to sheaf cohomology spectral sequence for open covers degenerates for paracompact spaces (I forget what the result is called, you can find it in Godement's book on sheaves). This is also very surprising to me. Maybe analyzing the proof of this result one could get some ideas for your case too. –  Angelo May 10 '10 at 16:47
    
Actually, I care about sheaf cohomology for the Grothendieck topology generated by closed covers (even more generally, generated by surjective maps between compact sets). I just phrased my question in that way because then, it's understandable by a wider audience. I expect that all those cohomology theories will be the same. –  André Henriques May 10 '10 at 16:53
    
Oh, ok. Hypercovers won't work? You alway have cohomological descent for those. –  Angelo May 10 '10 at 17:00
    
How do you plan to use cohomological descent to show my isomorphism? –  André Henriques May 11 '10 at 12:16
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Not in general. You need your cover to be by sets whose interiors cover the space. See this question, Tom Church's answer, and the references therein.

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Could you please elaborate? In the post that you point out, the problematic "cover" consists of an open set and of its closed complement. Clearly, you'll get nonsensical answers in you try to do cohomology using those kinds of "covers". But that was not my question: I'd like to cover my space using closed sets only. –  André Henriques May 10 '10 at 14:48
    
Have you followed the references and read what was in them? –  Andrew Stacey May 10 '10 at 17:39
    
If you have a nice compact space (like a cube $I^n$) then one could get away with covers by (closed) subcubes that have edges parallel to the edges of the cube, such that the subcubes only intersect on their boundaries, to calculate $H^1$ with any coefficients. I suspect that compactness is enough, or else something that looks like compactness for closed sets, because then one can use a coverage (in the sense in the Elephant, or in Baez-Hoffnung's smooth spaces paper) where covers consist of a finite number of closed sets. –  David Roberts May 11 '10 at 7:14
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