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Let $X$ be a finite CW complex. We can compute the cohomology groups of $X$ via sheaf cohomology of the constant sheaf. Are the homology groups of $X$ with $\mathbb{Z}$ coefficents the cohomology groups of some sheaf of abelian groups $S$ on $X$?

$$H_i(X, \mathbb{Z}) \cong H^i(X, S)$$

(If necessary, feel free to impose additional conditions on $X$, eg that $X$ is a topological manifold, ... )

More specifically, let $\mathbb{Z}_X$ denote the usual constant sheaf on $X$, and $\mathbb{Z}_X^{\vee} = Hom ( \mathbb{Z}_X, \mathbb{Z}_X)$ the sheaf-theoretic Hom. Is there an isomorphism $H_i(X, \mathbb{Z}) \cong H^i(X, \mathbb{Z}_X^{\vee})$?

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Your $\mathbb Z_X^\vee$ is often just $\mathbb Z_X$. –  Mariano Suárez-Alvarez Mar 13 '13 at 1:06
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1 Answer

There is such a sheaf $S$ (or at least a complex of sheaves), but your first guess as for what it might be is not correct (see Mariano's comment).

In the case when $X$ is a compact topological $n$-manifold, then $S$ is given by the orientation sheaf (shifted into cohomological degree $-n$). Then the isomorphism $H_n(X, \mathbb Z) \cong H^n(X, S)$ is Poincare duality.

For a finite CW complex (or maybe some more general compact topological space), $S$ is something called the dualizing complex $\omega _X$. This is part of the more general story of Verdier duality.

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