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.
2026-03-25 16:01:30.1774454490
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Sheaf Hom between locally free sheaves
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Answer: There is a canonical isomorphism (HH.Ex.II.5.1)
$$\phi: E^*\otimes_{\mathcal{O}}F \cong Hom_{\mathcal{O}}(E,F) ,$$
and since $E^*$ is locally free (of the same rank as $E$) it follows $E^*\otimes F$ is locally free of rank $ef$ with $e:=rk(E),f:=rk(F)$
Note: The map $\phi$ is a "vector bundle version" of the map of vector spaces
$$\psi: V^*\otimes_k W \cong Hom_k(V,W)$$
defined by
$$\psi(f\otimes w)(v):=f(v)w.$$
General principle "Most" intrinsic maps of vector spaces globalize to locally free sheaves.
Let $\{U_i\}_{i \in I}, \{V_i\}_{j \in J}$ be covers of $X$ such that $$\mathcal F|_{U_i} \cong \mathcal O_{U_i}^{\oplus m}, \quad \mathcal G|_{V_j} \cong \mathcal O_{V_j}^{\oplus n}.$$
Then $\{U_i \cap V_j\}_{i\in I, j \in J}$ is a cover of $X$ such that $$\mathcal F|_{U_i \cap V_j} \cong \mathcal O_{U_i \cap V_j}^{\oplus m}, \quad \mathcal G|_{U_i \cap V_j} \cong \mathcal O_{U_i \cap V_j}^{\oplus n}.$$
For any open $W \subset U_i \cap V_j$, $$\mathcal{Hom}(\mathcal F, \mathcal G)(W)=\operatorname{Hom}(\mathcal F|_{W}, \mathcal G|_{W})\cong \operatorname{Hom}(\mathcal O_{W}^{\oplus m},\mathcal O_{W}^{\oplus n}) \cong \operatorname{Hom}(\mathcal O_{W},\mathcal O_{W})^{\oplus mn} \cong \mathcal O_W^{\oplus mn}.$$
So, $$\mathcal{Hom}(\mathcal F, \mathcal G)|_{U_i \cap V_j} \cong \mathcal O_{U_i \cap V_j}^{\oplus mn}.$$