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Let $\mathcal{F}$ be a sheaf over $X$ and $\mathcal{U}=\{U_i\}_{i\in I}$ a covering of $X$.

I say that $\mathcal{U}$ is acyclic for $\mathcal{F}$ if $H^k(U_{i_0 \ldots U_n}, \mathcal{F}|_{U_{i_0 \ldots U_n}})=0$ for every $i_0 \ldots i_n \in I$ and every $k>0$, where $U_{i_0 \ldots U_n} = U_{i_0} \cap\cdots \cap U_{i_n}$ and the $H^k$ are the cohomology groups defined via resolutions (ie, using derived catergories).

My question is: given a sheaf $\mathcal{F}$, is it always possible to find an acyclic covering for it? My guess is that the answer is yes, but I don't know if my proof is ok.

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To build a counterexample, it is enough to exhibit a space $X$ with a point $x$ such that no neighborhood of $x$ in $X$ is acyclic. $\mathbb R^2\setminus\{1/n:n\in\mathbb N\}$ should do (use the constant sheaf $\underline{\mathbb Z}$)

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  • $\begingroup$ What if $X$ is paracompact? Does there exist a counterexample? $\endgroup$ Dec 21, 2011 at 21:20
  • $\begingroup$ @Lucas, my example is paracompact, since it is a metric space. $\endgroup$ Dec 21, 2011 at 21:24
  • $\begingroup$ Of course, you are right. $\endgroup$ Dec 21, 2011 at 21:38
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    $\begingroup$ I was trying to see why cech and derived functor cohomology agree. I managed to show that for acyclic coverings we have $\check{H}^i(\mathcal{U},\mathcal{F}) \simeq H^i(X,\mathcal{F})$ but i dont know what happens in the limit without using that every covering admits an acyclic subcovering. Do you know how to do this? $\endgroup$ Dec 21, 2011 at 22:01
  • $\begingroup$ @LucasKaufmann I'm interested in that question too. Have you found an answer to this? $\endgroup$
    – mathcourse
    Jun 21, 2019 at 12:59

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