In every treatment of Grothendieck sites I can find, flasque sheaves are not defined in the way one would naïvely expect from ordinary sheaf cohomology; namely instead of saying that "restriction maps surject," one says something about Čech cohomologies vanishing. The problem is then to set up the theory of flasque sheaves, one needs to already know some rather heavy facts about Čech cohomology that you don't need in the point-set case.


**Question.** Suppose that $\mathcal{F}'$ is an abelian sheaf on a Grothendieck site all of whose restriction maps are surjective. Does there exist a quick proof that if
$$0 \to \mathcal{F}'\to \mathcal{F}\to\mathcal{F}''\to 0$$
is exact, for $\mathcal{F}$, $\mathcal{F}''$ arbitrary abelian sheaves, then $\mathcal{F}(U)\to\mathcal{F}''(U)$ surjects for any $U$?


In the point-set world, I would prove this using Zorn's lemma in a way similar to the Hahn–Banach theorem, and the characterization that "a sheaf morphism is surjective if and only if for each $U$ and each $s \in \mathcal{F}''(U)$, there is an open cover of $U$ by objects $U\_i$ and $t\_i \in \mathcal{F}(U\_i)$ so that $\phi(U\_i)(t\_i) = s\rvert\_{U\_i}$." Unless I have made some mistake, I think this characterization of surjectivity is still true for sites, but I can't make my Zorn's lemma/Hahn–Banach argument work (namely, to construct an upper bound for a chain, I want to ‘take a union’ but I can't figure out any way to make that work here).



Zorn's lemma for Grothendieck sites
The statement you formulate is not true in this generality.


The idea of the following counterexample is to exploit the fact that the assumption "all restriction maps along morphisms of the site are surjective", applied to a thin site (no parallel morphisms), does not ensure that the sheaf is flabby (flasque) when regarded as a sheaf on the locale presented by the thin site.


Consider the topological space / locale $X = \mathbb{C}\setminus \{0\}$.
Let $\mathcal{F} = \mathcal{F}''$ be the abelian sheaf of continuous functions with values in the (multiplicative) group $\mathbb{C}^\times$.
Let $f : \mathcal{F} \to \mathcal{F}''$ be the sheaf homomorphism that takes the pointwise square of any section, that is, $f(s)(x) = (s(x))^2$ for $s \in \mathcal{F}(U)$, $x \in U$.
Then $f\_U : \mathcal{F}(U) \to \mathcal{F}''(U)$ is surjective as long as $U$ is simply connected -- in particular, $f$ is an epimorphism -- but $f\_X : \mathcal{F}(X) \to \mathcal{F}''(X)$ is not surjective. (Roots of nonzero functions exist locally but not globally.) The kernel $\mathcal{F}'$ of $f$ is the abelian sheaf of continuous (= locally constant) functions with values in $\{-1, 1\}$.


Now, $\mathcal{F'}$ is not a flabby (flasque) sheaf. But as a site of definition for $\mathrm{Sh}(X)$ we can take any basis of the topology of $X$, for example the *connected* open subsets of $X$. For connected $U, V$ with $U \subseteq V$, the restriction map $\mathcal{F}'(V) \to \mathcal{F}'(U)$ is surjective (even bijective). At the same time, there is a connected open subset $U$ such that $f\_U$ is not surjective, namely $U = X$. Thus, we have found a Grothendieck site and an exact sequence $0 \to \mathcal{F}' \to \mathcal{F} \to \mathcal{F}'' \to 0$ of abelian sheaves on it that violates the statement in the question.


