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It is well-known that if $\mathscr{F},\mathscr{G}$ are sheaves on $X$, then $\mathscr{H}om(\mathscr{F},\mathscr{G})$ is a sheaf. I came up with two proofs; one is a direct proof, and the other, more succinct one uses Mayer-Vitories argument. What I find strange is that the direct proof doesn't seem to require that $\mathscr{F}$ is a sheaf, while Mayer-Vitories argument uses the fact that both $\mathscr{F},\mathscr{G}$ are sheaves. Am I missing something, or is it true in general that only $\mathscr{G}$ needs to be a sheaf?

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Yes! I have not enough time for an answer but it is true. – Simon Markett Jun 7 '12 at 15:40
This is true. If you're satisfied with an appeal to authority: see Exercise 3.3.C of Vakil. – Dylan Moreland Jun 7 '12 at 15:56
Thank you both! (It is Ex 3.3.B in Vakil's note that I found) I'm now thoroughly satisfied. – ashpool Jun 7 '12 at 16:05
It's also true in the context of sheaves of sets. Morally, it is because any presheaf morphism $\mathscr{F} \to \mathscr{G}$, where $\mathscr{F}$ is a presheaf and $\mathscr{G}$ is a sheaf, must factor through the associated sheaf $\mathscr{F}^+$. So $\mathscr{Hom}(\mathscr{F}, \mathscr{G})$, whatever it is, ought to be isomorphic to $\mathscr{Hom}(\mathscr{F}^+, \mathscr{G})$. – Zhen Lin Jun 7 '12 at 16:46
@Zhen Lin: A good point! One should note, of course, that $\mathscr{F}^+|_U\simeq(\mathscr{F}|_U)^+$. – ashpool Jun 7 '12 at 21:59

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