Sheaf cohomology and singular cohomology Let $X$ be a manifold, $\mathcal{F}$ be an abelian sheaf on $X$. We can consider the etale space $F$ of $\mathcal{F}$, which is the disjoint union of stalks, whose topology is generated by local sections. 
Is there some the relation between sheaf cohomology of $\mathcal{F}$ and cohomology of topological space $F$? 
(Say, is there a relation between $H^i(X,\mathcal{F})$ and singular cohomology of $H^i(F,\mathbb{Z})$? Or in a nicer situation, when $\mathcal{F}$ is a locally constant $\Lambda=\mathbb{Z}/l\mathbb{Z}$ sheaf, is there relation between $H^i(X,\mathcal{F})$ and $H^i(F,\Lambda)$? )
 A: If $\mathcal F$ is a locally constant sheaf of stalk $\Bbb Z/\ell \Bbb Z$, the étalé space $F$ is a degree $\ell$ cover, and the monodromy of this cover $f : F \to X$ is given by $\mathcal F$. What you can say is basically that for any $j$, there is an isomorphism 
$$H^j(F, \underline{\Bbb Z}) \cong H^j(X,f_*\underline{\Bbb Z}) \ \ \ \  (*)$$ 
This is more or less the desired relation, since $f_*\underline{\Bbb Z}$ is "morally" $\mathcal F$. (**)

We will do two examples with $X = S^1$ and $\ell = 2$. We always write $f : F \to X$ for the projection from the étalé space to $X$.
Example 1 :
We take the trivial local system. The étalé space is $F = S^1 \sqcup S^1$, so the cohomology of $F$ is $\Bbb Z^2$ in degree $0$ and $1$. Now, since $f_*\Bbb Z = \Bbb Z \oplus \Bbb Z$, $(*)$ is trivially verified.
Example 2 :
We now assume $\mathcal F$ has non trivial monodromy. We have $F = S^1$ and the projection map is $z \mapsto z^2$. The left hand side is $\Bbb Z$ in degree $0$ and $\Bbb Z$ in degree $1$. 
We now compute the right hand side. The sheaf $\mathcal f_*\Bbb Z$ has stalks $\Bbb Z^2$ and the two components are permuted by the monodromy, say $T$. You can check that for any local system $L$ with monodromy $T : V \to V$ ($V$ is the stalk at $1 \in S^1$ of $L$) the following complex computes the monodromy : $0 \to V \overset{T - id}{\to} V \to 0$. In our case, this is the complex $$0 \to \Bbb Z^2 \overset{\pmatrix{-1 & 1 \\ 1 & -1}}{\to} \Bbb Z^2 \to 0$$
which has cohomology $\Bbb Z$ in degree $0$ and $1$, as expected.

(*) More precisely, they have different coefficients but the same monodromy at a point $x \in X$ : we have $(f_*\underline{\Bbb{Z}})_x \cong \Bbb Z \oplus \dots \oplus \Bbb Z$ ($\ell$ times) canonically once you pick some order on the fiber $f^{-1}(x)$ and $\pi_1(X;x)$ acts by permuting these copy like $\mathcal F$. In a more pedantic way : the natural bijection between the canonical basis of $(f_*\underline{\mathbb Z})_x$ and $\mathcal F_x$ is $\pi_1(X,x)$-equivariant.
