The following is surely pretty standard, but I have been unable to prove it or find a proof in the literature (many sources assert it without proof).

Let $\phi : \mathcal{F} \rightarrow \mathcal{G}$ be a monomorphism in the category of sheaves of sets on some space $X$ (by this, I mean a monomorphism in the categorical sense : if $\psi_1,\psi_2 : \mathcal{H} \rightarrow \mathcal{F}$ are morphisms of sheaves on $X$ such that $\phi \circ \psi_1 = \phi \circ \psi_2$, then $\psi_1=\psi_2$). Question : must $\phi$ induce an injection on all stalks?

The converse is pretty trivial, but this seems harder in the sense that without knowing much about $\mathcal{F}$ and $\mathcal{G}$, I don't know how to construct many interesting morphisms into $\mathcal{F}$ to test injectivity on stalks.

Thanks for any help!

EDIT : Let me make a few comments up here in reply to Dan's comment. He points out that the result I want is a trivial consequence of the exactness of the inverse image functor. However, I want to emphasize that I am working merely with sheaves of sets! These do not form an abelian category, so it doesn't make sense for a functor to be exact.

Of course, I expect that the inverse image functor still takes monomorphisms and epimorphisms to monomorphisms and epimorphisms, which is a weak for of exactness. But the only way I know to prove this (at least for monomorphisms) is to use the result I ask about above!

Fatxis just the inverse image sheaf under the inclusion ${x}\to X$, and the inverse image is an exact functor. $\endgroup$ – Dan Petersen Sep 12 '10 at 18:45