# 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 $$H^i(F,\Lambda)$$? )

• The space etale gives you a fibration, so I'd expect you want to use the whole structure of that to say anything about the cohomology of the sheaf. – Pedro Tamaroff Nov 24 '18 at 15:22

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.