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This is related to the answer to Does Hom commute with stalks for locally free sheaves?.

Like Enrique, I also understand how to do this problem, as long as I can exhibit an isomorphism

$$\mathcal{Hom}_{\mathcal{O}_X}(\mathcal{F},\mathcal{G})_x \cong \mathcal{Hom}_{\mathcal{O}_{X,x}} (\mathcal{F}_x,\mathcal{G}_x) ,$$ (this is done, for example, in Serre's FAC). However, in the answer to this post, it is claimed that we don't have to go down to the level of stalks, because we can just check that the evaluation morphism we defined is an isomorphism on some open (trivialising) cover.

My question is the following: Given that sheaf $\mathcal{Hom}$ is defined as the sheaf associated to the presheaf $$U\mapsto \text{Hom}_{\mathcal{O}_X|_U}(\mathcal{F}|_U,\mathcal{G}|_U), $$ how would we actually check that the sheafification of the evaluation map

$$ \mathcal{E}\to ((\mathcal{E})^{\vee})^{\vee} $$

is an isomorphism without going down to the level of stalks?

user041193
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  • You don't need sheafification ! $U\mapsto\operatorname{Hom}(\mathcal{F}{|U},\mathcal{G}{|U})$ is always a sheaf. Now if $\mathcal{E}$, to check that $\mathcal{E}\rightarrow\mathcal{E}^{\vee\vee}$ is an isomorphism, it is enough to prove it on an open cover. You can then choose a cover where $\mathcal{E}$ is trivial. – Roland Apr 21 '17 at 07:44
  • Oh yes, I thought that was only true in the category of sheaves of abelian groups, but this post http://stacks.math.columbia.edu/tag/01CM, explains why it is also true in the category of sheaves of $\mathcal{O}_X$-modules.

    Thanks!

     – user041193 Apr 21 '17 at 13:56

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