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Let $F$ and $G$ be sheaves on $X$. Under what conditions is the natural map from the stalk at $p$ of $SheafHom(F,G)$ to $Hom(F_p, G_p)$ an isomorphism?

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It's true when X is a locally Noetherian scheme, F is a coherent sheaf and G is any O_X module. This is Chapter III, Prop 6.8 of Hartshorne's Algebraic Geometry, so hopefully I'm not just telling you something you knew.

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By the way, here's a counterexample for the most sweeping generalization ("it's always an isomorphism"), which I found online in a book called "Topological Invariants of Stratified Spaces". Let X = [0,1] and F be the skyscraper sheaf Z at 0. Let G be the constant sheaf Z. If U contains 0 then Hom(F|U,G|U)=0, so Hom(F,G)_0 = 0, but of course Hom(Z,Z)=Z.

What about for coherent O_X modules on a ringed space X that need not be a scheme? Say, a complex manifold? Just idle curiosity...

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The result in Hartshorne if I recall correctly only really uses the fact that affine locally a coherent sheaf on a scheme has a locally free resolution by finite rank projectives and that one can compute stalks affine locally. In particular, as pointed out by David in the comments we only ready need the first two steps so that one can use the exactness properties of Hom/SheafHom.

So the right condition on F as in the comments is that it be finitely presented.

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Nitpick: you want to remove "finite" from "finite locally free resolution": the finiteness is only guaranteed around nonsingular points. The finiteness isn't used in the proof, so it's irrelevant. – Steven Sam Oct 15 '09 at 21:58
Fair point, but I actually meant resolutions by finite rank locally free modules. Looking at what I wrote this isn't clear, but it isn't irrelevant. – Greg Stevenson Oct 15 '09 at 22:26
To clarify Stevens comment: A full free resolution by finite rank locally free modules is indeed not needed, only the "first two steps", i.e., it is enough that F is of finite presentation (=coherent if X is loc. noetherian). So the general statement is: If X is a ringed space and F, G are O_X-modules such that F is of finite presentation, then SheafHom(F,G) commutes with talking stalks. (this can be found in EGA I) – David Rydh Oct 16 '09 at 0:45
Yeh... I must be tired, that is obviously the right statement... I have no idea why I was thinking of actual resolutions. I'll edit the post so the answer isn't so bad. – Greg Stevenson Oct 16 '09 at 1:12

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