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I stumbled upon a question in Vakil's notes which asks us to prove that for two locally free sheaves $\mathcal{F}$ and $\mathcal{G}$ each of finite rank, the sheaf $\mathcal{Hom_O(F,G)}$ is locally free with rank equal to the product of the ranks of $\mathcal{F}$ and $\mathcal{G}$. I reduced the question to $\mathcal{F}$ and $\mathcal{G}$ being free of rank 1. Now, my plan was to give a map $$\mathcal{Hom_O(O,O)}\to \mathcal{O}$$ and show that the stalks are isomorphic. The map is over a section U is $$\phi \mapsto \phi(U)(1)$$ However, I am not sure if this is correct since this proves that dual of a locally free sheaf is isomorphic to itself which is not true. Also, proving the stalk part does not seem to be easy. Any other ideas of proving this lemma would be very helpful.

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  • $\begingroup$ Reduce to the affine case; then you're just proving the Yoneda lemma for commutative rings. $\endgroup$ – Qiaochu Yuan May 25 '18 at 5:29
  • $\begingroup$ Vakil made a comment in his text saying the lemma holds for ringed spaces in general. $\endgroup$ – Samiron May 25 '18 at 5:31
  • $\begingroup$ You can show that the map $\mathcal{H}om_{\mathcal{O}}(\mathcal{O},\mathcal{O})\to \mathcal{O}$ you wrote is an isomorphism directly, without using stalks (in the other direction, the map is $s\mapsto m_s$ where $m_s$ is the multiplication by $s$). It does not prove that the dual of a locally free sheaf is isomorphic to itself, just that the dual of a locally free sheaf is locally isomorphic to itself (which is obvious). But these local isomorphisms usually don't glue. $\endgroup$ – Roland May 25 '18 at 9:42

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