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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!

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    $\begingroup$ This is indeed standard and should probably be asked at the sister site math.stackexchange.com Anyway it follows because the stalk of F at x is just the inverse image sheaf under the inclusion ${x}\to X$, and the inverse image is an exact functor. $\endgroup$
    – Dan Petersen
    Commented Sep 12, 2010 at 18:45
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    $\begingroup$ That seems a little unfair, Dan. The questions on math.stackexchange.com seem mostly undergraduate and high school level. This one is at least is at the graduate level. Also, at least as I understand the foundations, your answer is a little circular. In all the sources I'm familiar with, the exactness of the inverse image functor is proven by studying what it does on stalks (eg this is how it is done in Hartschorne, which AFAIK never talks about the categorical version of a monomorphism). $\endgroup$
    – Nikita
    Commented Sep 12, 2010 at 18:56
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    $\begingroup$ Let me be a little more precise. In Hartschorne, a morphism of sheaves is called injective if its kernel is 0 (by the way, of course that only makes sense for sheaves of abelian groups, not sets as I have in my question). Thus when it is proven that the inverse image functor is exact, it is proven that if a map of sheaves has trivial kernel, so does the image of the map under the inverse image. But for this to give what I want, you would have to first prove that if a morphism of sheaves is a categorical monomorphism, then it has trivial kernel, which is exactly what I asked about! $\endgroup$
    – Nikita
    Commented Sep 12, 2010 at 19:02
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    $\begingroup$ An open U in X can be viewed as a sheaf on X: sections over V are inclusions of V in U (so either empty or one-element). We then have Hom(U,F) = F(U) for any sheaf F. Thus if F --> G is a monomorphism, it is injective on sections over any open, hence injective on stalks. $\endgroup$
    – Dustin Clausen
    Commented Sep 12, 2010 at 19:37
  • $\begingroup$ Thanks Dustin (you wrote that while I composed the edit above)! $\endgroup$
    – Nikita
    Commented Sep 12, 2010 at 19:46

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