Sheaf cohomology of a space and cohomology of the espace etale

Question

I have a very basic question about cohomology of sheaves. Suppose $\mathcal{F}$ is a sheaf of abelian groups over a topological space $X$. Then $\mathcal{F}$ itself is a topological space with a continuous map $\sigma : \mathcal{F} \rightarrow X$.

How are the sheaf cohomology groups $H^i(X,\mathcal{F})$ of $X$ related to $H^i(\mathcal{F},\mathbb{Z})$, the singular cohomology groups of the space $\mathcal{F}$?

Can we also consider "relative" cohomology groups $H^i(\mathcal{F}, X)$ using the zero section and relate it to the other cohomology groups, perhaps with a spectral sequence?

Answer 1

They have nothing much to do with each other. Take the special case that $X$ is discrete, so a sheaf $F$ of abelian groups is just a collection $F_x, x \in X$ of abelian groups indexed by $X$. The étale space, which I'll denote $Y$, is the disjoint union $\bigsqcup_{x \in X} F_x$, so it is again a discrete space, so it satisfies $H^0(Y, \mathbb{Z}) \cong \mathbb{Z}^Y$ and higher cohomology vanishes. On the other hand the sheaf cohomology is $H^0(X, F) \cong \prod_{x \in X} F_x$ and higher cohomology vanishes.

So the cohomology of $Y$ is much bigger and not particularly interesting; in particular it is completely insensitive to the group structure on each stalk $F_x$ (and this generalizes to the general case, since the étale space is only sensitive to the underlying sheaf of sets).

Answer 2

I had some fun playing with examples of the espace étalé in different settings. I'll write $X_\mathcal{F}$ for the espace étalé of $\mathcal{F}$ on $X$. Summary:

1. In differential/topological settings, the cohomology of the espace étalé will almost always be extremely pathological.
2. The singular cohomology of $X_{\mathcal{F}}$ is more closely related to the singular cohomology of the base $X$, often a very different beast than the $\mathcal{F}$-sheaf cohomology. So the comparison is a bit apples-to-oranges.

Differential-type Settings

Let's first look at sheaves that are fine (roughly, where bump functions exist -- think topological and differential-geometric settings). Take for example $X = S^1$ and $\mathcal{F} = C^\infty(X)$ to be smooth real-valued functions. Since the sheaf is fine, $H^1(X, \mathcal{F}) = 0$ (this is a standard Cech cohomology fact, similar to the proof that flabby sheaves are acyclic).

To get some intuition for what $X_{\mathcal{F}}$ looks like, let's think about its fundamental group. Pick two distinct points $p, q$ on $X$, and fix a germ $f_p$ at $p$ (hence, a point in $X_\mathcal{F}$). Given any $g_q$ germ at $q$, we can smoothly interpolate from $f_p$ to $g_q$ and then around the other side of the circle to complete a loop in the espace étalé. None of these loops for distinct $g_q$ can be deformed into each other: roughly, this is because the espace étalé equips the fiber over $q$ with the discrete topology: any deformation of our loop has to give rise to a continuously moving point in that fiber, which therefore has to be constant. All of these nonequivalent loops live over a generator for $\pi_1(S^1)$. In general, we can specify arbitrary germs at an arbitrary finite sequence of points along with signs specifying which way around to get there, and there will be uncountably many nonequivalent loops satisfying that data. Yikes! Note also that this space is definitely NOT locally simply connected.

Okay, but all that was about the fundamental group. There is some hope that $H_1 = \pi_1^{ab}$ might be much smaller, since the group described above is highly noncommutative, but I am not optimistic. In particular, since any commutator of loops lives in the kernel of $\pi_1(X_{\mathcal{F}})$, we at least have a surjective map $H_1(X_{\mathcal{F}}) \to \pi_1(S^1) = H_1(S^1) = \mathbb{Z}$, so we

Moral: the singular (co)homology of $X_{\mathcal{F}}$ will always be able to see the singular (co)homology of the base space, while the sheaf cohomology can't, in general.

Of course, if we take $\mathcal{F} = \underline{\mathbb{Z}}$ the constant sheaf on $X = S^1$, then $H^i(X_{\mathcal{F}}, \mathbb{Z}) = \bigoplus_{i \in \mathbb{Z}} H^i(X, \mathbb{Z})$, and so the two are not so far apart. But you can see this is a uniquely convenient choice: we are still comparing singular cohomology of the base to its sheaf cohomology, and for $\underline{\mathbb{Z}}$ these happen to coincide.

Rigid Settings

Things are less huge in rigid (e.g. algebraic or complex-analytic settings), because a germ at a point determines the function uniquely on its full domain, but here the espace étalé isn't very useful or interesting for the same reason. Given $X$ an (integral) algebraic variety or complex manifold over $\mathbb{C}$ and $\mathcal{F} = \mathcal{O}$ (algebraic or holomorphic functions), each function $f$ determines a connected component of $X_{\mathcal{F}}$ which is just homeomorphic to $U$, the maximal domain of $f$. So the espace étalé ends up just being a bunch of disjoint copies of open subsets of $X$, indexed by all the rational/meromorphic functions on $X$. Not very interesting, and in particular its singular cohomology can't see anything except topological properties of the variety, while $H^i(X, \mathcal{O})$ sees more geometric/analytic information.