Let $\phi: \mathscr{F} \to \mathscr{G}$ be a morphism of sheaves. Recall that $\phi$ is an epimorphism if given $\psi_1, \psi_2: \mathscr{G} \to \mathscr{H}$, then $\psi_1 \circ \phi = \psi_2 \circ \phi$ implies $\psi_1 = \psi_2$.
One direction of this is clear: let $p \in X$ (assuming $\mathscr{F}, \mathscr{G},$ and $\mathscr{H}$ are sheaves on $X$); if $\phi_p: \mathscr{F}_p \to \mathscr{G}$ is surjective for all $p$, then if $\psi_1\circ\phi = \psi_2\circ\phi$, they induce the same map on stalks. $\phi_p$ is surjective on stalks, so is an epimorphism in $\mathsf{Set}$ or $\mathsf{Grp}$ or whatever category. Thus $\psi_{1,p} = \psi_{2,p}$. Maps that are equal on stalks are equal, so $\psi_1 = \psi_2$.
It is the other direction that I'm failing to see clearly. I understand that we must exhibit a sheaf $\mathscr{H}$, along with maps $\psi_1, \psi_2: \mathscr{G} \to \mathscr{H}$ such that the following conditions hold:
- $\psi_1 \circ \phi = \psi_2 \circ \phi$, and
- $\psi_1 = \psi_2$ implies $\phi_p$ is surjective.
Any hints into what the right sheaf or sheaf morphisms to consider are would be greatly appreciated. For reference, this comes from Exercise 2.4.O in Vakil's Algebraic Geometry notes.
Best Answer
One way to get out of this may be as follows: Let $\phi_1, \phi_2$ be such that $\phi_1\not= \phi_2, \phi_1\circ \phi=\phi_2\circ \phi$. Since $\phi_1\not=\phi_2$, there exist some point $x\in X$ such that the morphisms on the stalks are different. But by definition after compose it with $\phi_{x}$ we get the same map on the stalks. So we reached a contradiction.
The forward proof you claimed seems to be fishy. For a concise proof, see:
An 'easy' way to prove that epimorphism of sheaves implies surjectivity on stalks