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Let $E=\mathbb{C}/L$ be an elliptic curve. Then $\mathbb{C}$ is contractible, and $L$ is the fundamental group of $E$. What's interesting is that we can find the cohomology of $E$, which is the same as the sheaf cohomology $H^i(E,\mathbb{Z})$ for the constant sheaf $\mathbb{Z}$, by taking the group cohomology of $L$ with trivial action on $\mathbb{Z}$. What's more interesting is that if $\mathcal{O}^\times$ is the sheaf of invertible holomorphic functions on $E$, then we can find the sheaf cohomology of this by considering the group cohomology of the action of $L$ on the group of invertible holomorphic functions on $\mathbb{C}$ defined by $(\omega f)(z)=f(z+\omega)$ for $\omega \in L$. In the case of $H^0$, this is obvious, since the holomorphic functions which are fixed by $L$ ($H^0$ in group cohomology), which is the same as invertible holomorphic functions on $E$, which is the same as the global sections of the sheaf on $E$ (and this is true for many other sheaves). In the case of $H^1$, one can see that both are the group of invertible line bundles on $E$.

My question is: why is the group ($L$-module) of holomorphic functions on $\mathbb{C}$ the natural one to consider? In this case, it's clearer, since we know what holomorphic functions are. If in general we have a covering map $U \to X$ for some space $X$, with $U$ a contractible universal cover, then given a sheaf on $X$, how do we know what sheaf on $U$ to consider? Specifically, which sheaf on $U$ gives the sheaf cohomology when we take the group cohomology of its global sections?

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Actually, is it just the sheafification of the pullback of our sheaf on $E$ to a presheaf on $U$? –  David Corwin Jul 5 '10 at 11:18
There is a nice appendix on all of this (with some minor typos) at the end of chapter 1 of Mumford's book on abelian varieties. –  BCnrd Jul 5 '10 at 14:44
To expand on Brian's comment: The pullback of the sheaf $G_m$ to the universal cover has vanishing higher cohomology, and the results in Mumford then imply that group cohomology agrees with sheaf cohomology. –  David Corwin May 2 '13 at 20:50

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I doubt that in general one can construct a reasonable sheaf on $U$ with the required properties. To see what kind of bad things can happen, let us try to understand why this works for $X$ an elliptic curve and the sheaf $\cal{O}^{\times}$ on it.

We have the derived global sections functor from the $D^b$ of sheaves on $X$ to the $D^b$ of sheaves on a point, i.e. graded vector spaces. But given a sheaf $F$ on $X$ we can compute its global sections in a roundabout way: we can first take the pullback to $U$, then take global sections and then take the $G$-invariants where $G=\pi_1(X)$. Passing to the derived categories we get $$R\Gamma (F)=R(R\Gamma f^{-1}(F))^G$$ where $F\in D^b(X)$, $f:U\to X$ is the projection, $R\Gamma f^{-1}$ is the right derived functor of the left exact functor $\Gamma f^{-1}$ and $R(\cdot)^G$ is the right derived functor of the functor of $G$-invariants (this functor goes from the $D^b$ of $G$-modules to graded vector spaces).

We have the Grothendieck spectral sequence that converges to $H^\ast(X,F)$ with the $E_2$ sheet given by $$E_2^{p,q}=H^p(G,H^q(U,f^{-1}(F)).$$

Now if $X$ is an elliptic curve, $F=\cal{O}^{\times}$ and $U=\mathbf{C}$, then it follows from the exponential exact sequence that $H^q(U,f^{-1}(F))=0$ for $q\neq 0$, so the above spectral sequence collapses and we get the required isomorphism $H^\ast (X,F)=H^\ast(G,H^0(U,f^{-1}(F))$. This also happens when say $F$ is locally constant and $U$ is contractible. But in general there seems no reason to expect the spectral sequence to collapse, let alone to be concentrated in one row only.

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