About stalk of sheaves in an unbalanced category


Let $\phi:\fs\to\gs$ be an epimorphism in ${\bf Sh}(X,{\bf Ring})$. Is it true that $\phi_x:\fs_x\to\gs_x$ is an epimorphism in ${\bf Ring}$ for every $x\in X$ ?

Note that epimorphism in ${\bf Ring}$ is not necessarily surjective.

If instead of ${\bf Ring}$, we use the category of abelian groups, then the statement is a familiar result (true). The main trouble seems to be that, for ${\bf Ring}$, there is no zero object, and we can't define things like kernel and cokernel.

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I have one tiny nagging issue: the fact that colimits preserve epimorphisms implies that if $\phi$ is an epimorphism in the category of presheaves, then $\phi_x$ is an epimorphism for every $x\in X$. $\phi$ being an epimorphism in the category of sheaves does not necessarily imply that $\phi(U)$ is an epimorphism for every open subset $U\subset X$, does it? – ashpool Jan 20 '13 at 2:05
But then your argument doesn't seem to apply to sheaves? To use the fact that colimits preserve epimorphisms seems to require that each $\phi(U)$ is an epimorphism. – ashpool Jan 20 '13 at 13:48
It's left adjoint to the "Skyscraper functor"(I don't know of any real name for the general functor). If $G$ is a group, let's call by $\mathscr{G}$ the skyscraper sheaf at a fixed point $x \in X$. Then $\mathrm{Hom}_{\mathsf{Sh(X;Ab)}}(\mathscr{F},\mathscr{G}) = \mathrm{Hom}_{Ab}(\mathscr{F}_x,G)$. I think this generalizes easily to other categories besides $\mathsf{Ab}$, but I haven't really thought carefully about it. – John Stalfos Jan 21 '13 at 19:42