Suppose $\mathcal{F}$,$\mathcal{G}$ are 2 sheaves valued in an arbitrary category(such that all limits and colimits(hence the stalks of sheaves) exist) on a topological space $ X$, and $f\colon \mathcal{F}\rightarrow\mathcal{G} $ a morphism between sheaves, what is the condition that $f $ is an epimorphism.
It is clear when $\mathcal{F} $ and $\mathcal{G} $ are sheaves of sets, which is, for every open subset $U$, there exists an open cover $\{U_i,i\in I \} $, such that the restriction of $s\in \mathcal{g}\left(U\right)$ on each $U_i$ comes from the image of $f_{U_i}\colon \mathcal{F}\left(U_i\right)\rightarrow\mathcal{G}\left(U_i\right) $. For sheaves in arbitrary category, what I had guessed is that$f $ is epimorphic if the restriction morphism $\mathcal{G}\left(U\right)\rightarrow\mathcal{G}\left(U_i\right) $ can be factored through $f_{U_i} $, but I failed to prove that this gives an epimorphism of sheaves, so does the construction given above gives an epimorphism, and does every epimorphism satisfies the condition shown above? If not, what is the condition for a sheaf morphism being epimorphic?