Let $S^k$ be the presheaf on a space $X$ that assigns to every open set $U$ the abelian group $S^k(U)$ of singular k- cochains on $U$. This is clearly not a sheaf. Consider the sheafification $F^k$ of each $S^k$. These sheaves form an exact resolution of the constant sheaf of integers.

We can take global sections on this sheaf resolution to obtain a cochain complex $F^*(X)$. Does the cohomology of this cochain complex coincides with ordinary singular cohomology?


You haven't said this explicitly, but your claim is that the sheaf cohomology of the constant sheaf $\Bbb Z_X$ is isomorphic to the singular cohomology of $X$: $$H^\ast(X, \Bbb Z_X) \cong H^\ast_{\text{sing}}(X; \Bbb Z).$$ This is true for $X$ locally contractible. One checks that the complexes $S^\ast(X)$ and $\Gamma(X, S^\ast(X))$ are quasi-isomorphic. More generally, the assertion holds for any abelian group $G$: $$H^\ast(X, G_X) \cong H^\ast_{\text{sing}}(X; G)$$ when $X$ is locally contractible.

I don't have any references on hand, but I would expect that you could find the proof in most books that introduce the concept of sheaf cohomology.

  • $\begingroup$ I just found a proof here www3.nd.edu/~lnicolae/sheaves_coh.pdf thank you! $\endgroup$ – Manuel Jun 7 '13 at 20:56
  • $\begingroup$ Is there an example that the sheaf cohomology is not isomorphic to the singular cohomology when $X$ is not locally contractible? $\endgroup$ – Yuchen Liu Jun 9 '13 at 4:39
  • $\begingroup$ @YuchenLiu I do not know of any specific counterexample. $\endgroup$ – Henry T. Horton Jun 11 '13 at 1:37

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