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Let $X$ be a topological space and let $\mathcal{F}$ be a sheaf over $X$. Suppose we have a open covering $\mathcal{U}$ of $X$. When $\check{\mathrm{H}}^n(X,\mathcal{F})\cong \check{\mathrm{H}}^n(\mathcal{U},\mathcal{F}) $ ? (When Čech cohomology is independent of covering ?)

mikis
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1 Answers1

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There is a sufficient condition called Leray Theorem: If the covering $\mathcal{U}$ is acyclic for the sheaf $\mathcal{F}$ in the sense that $$H^{k}(U_{i_1}\cap U_{i_2}\cap...\cap U_{i_p},\mathcal {F})=0 $$ for all $k>0$ and all finite collections $\{i_1,i_2,...,i_p\}$ then $H^{*}(X,\mathcal{F})=H^{*}(\mathcal{U},\mathcal{F}).$ The proof uses spectral sequence mechinary.

Bingo
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  • Is it true for every topological space ? I know that it is true for smooth paracompact manifolds. – mikis Mar 07 '15 at 15:58
  • Not sure whether it is true for beyond smooth paracompact manifolds. – Bingo Mar 07 '15 at 16:42
  • Leray's theorem is true in very great generality. The problem is coming up with a covering with this property! – Zhen Lin Mar 07 '15 at 19:05