Consider sheaves of sets on a topological space $X$. A standard fact (and exercise) is the following equivalence for a morphism $\phi\colon \mathscr{F}\rightarrow \mathscr{G}$ of such sheaves:
(a) $\phi$ is an epimorphism in this category of sheaves
(b) The induced morphisms $\phi_x$ on stalks are surjective for all $x\in X$
I found it surprisingly hard to come up with an idea for a proof of (a)=>(b), though I think I managed to do it using an appropriate skyscraper sheaf and thus showing that the $\phi_x$ are epimorphisms too.
Since I found shorter proofs for the analogous statements for monomorphisms and injectivity, I was wondering if there was some elegant and at the same time 'elementary' way to do it. That is, I'd like to see where the surjectivity comes from.
I know there are short proofs using the fact that colimits commute with left adjoints, but I don't want to use that.
Examples for what I would like are usage of the product of all stalks, the fact that morphisms are equal if they agree on all stalks or something alike. I didn't have any idea up to now, though.
Thanks for any insight.
TL;DR: Is there an elegant proof for the equivalence not using adjoints?
Best Answer
Eric's answer is very nice, but addresses the case when $\mathcal{F}$ and $\mathcal{G}$ are sheaves of abelian groups, whereas the original question asks about sheaves of sets. I thought I might show how to modify things in this case. The idea is quite similar; one again wants to use a skyscraper sheaf (as noted in the OP) to make an ''indicator function''-like argument. I'll use the notation of Vakil's ''Foundations of Algebraic Geometry''.
Let $\phi \colon \cal{F} \to \cal{G}$ be our epimorphism of sheaves. Define $\rho \colon \cal{G} \to i_{p, \ast}\{0, 1\}$ by
$$\rho(U)(g) = \begin{cases} 0 \text{ if } (g, U) \in Im(\phi_{p}) \\ 1 \text{ otherwise } \end{cases}$$
for open sets $U$ containing $p$, and trivial maps otherwise. It is straightforward to see that $\rho$ is a morphism of sheaves. Now define $\alpha \colon \cal{G} \to i_{p, \ast}\{0, 1\}$ by
$$\alpha(U)(g) = 0$$
for open sets $U$ containing $p$, and trivial maps otherwise. Likewise, it is obvious that $\alpha$ is a morphism of sheaves. Furthermore, it is clear that $\rho \circ \phi = \alpha \circ \phi$, so $\rho = \alpha$, since $\phi$ is an epimorphism. Surjectivity of the stalk maps is is now easily deduced.