I have a question concerning part (c) of Exercise 5.1 in Chapter II of Hartshorne: He asks to prove that if $\mathcal{E}$ is a locally free $\mathcal{O}_X$-module of finite rank and $\mathcal{F},\mathcal{G}$ are arbitrary $\mathcal{O}_X$-modules, then $$ \operatorname{Hom}_{\mathcal{O}_X}(\mathcal{E}\otimes_{\mathcal{O}_X}\mathcal{F},\mathcal{G})\cong \operatorname{Hom}_{\mathcal{O}_X}(\mathcal{F},\mathcal{Hom}_{\mathcal{O}_X}(\mathcal{E},\mathcal{F})). $$ My question is: is it really necessary to suppose that $\mathcal{E}$ is locally free of finite rank for this to hold? Because I seem to be able to construct canonical maps in both directions which are mutually inverse, without using that $\mathcal{E}$ is locally free. Namely $$ F:\operatorname{Hom}_{\mathcal{O}_X}(\mathcal{E}\otimes_{\mathcal{O}_X}\mathcal{F},\mathcal{G}) \to \operatorname{Hom}_{\mathcal{O}_X}(\mathcal{F},\mathcal{Hom}_{\mathcal{O}_X}(\mathcal{E},\mathcal{F}))\\ \varphi\mapsto \left(\substack{F(\varphi):\mathcal{F}\to\mathcal{Hom}_{\mathcal{O}_X}(\mathcal{E},\mathcal{F})\\ b\in\mathcal{F}(U)\mapsto F(\varphi)(U)(b)\in\mathcal{Hom}_{\mathcal{O}_X|_U}(\mathcal{E}|_U,\mathcal{F}|_U)}\right) $$ where we define for open sets $V\subseteq U\subseteq X$ and sections $a\in\mathcal{E}(V)$ and $b\in\mathcal{F}(U)$ $$ [F(\varphi)(U)(b)](V)(a):=\varphi^{\text{pre}}(V)(a\otimes b|_V). $$ Here $\varphi^{\text{pre}}$ is the precomposition of $\varphi$ with the canonical map from the presheaf tensorproduct to the sheaf tensor product.
On the other hand, we also have the map $$ G:\operatorname{Hom}_{\mathcal{O}_X}(\mathcal{F},\mathcal{Hom}_{\mathcal{O}_X}(\mathcal{E},\mathcal{F}))\to\operatorname{Hom}_{\mathcal{O}_X}(\mathcal{E}\otimes_{\mathcal{O}_X}\mathcal{F},\mathcal{G})\\ \psi\mapsto G(\psi) $$ where on simple tensors $a\otimes b$ with $a\in\mathcal{E}(U)$ and $b\in\mathcal{F}(U)$ (and I believe the values on simple tensors suffice to specify all of $G(\psi)$) we define $$ G(\psi)(U)(a\otimes b):=[\psi(U)(b)](U)(a). $$ It seems to me that these two constructions are mutually inverse and don't use any assumption on $\mathcal{E},\mathcal{F},\mathcal{G}$. Am I mistaking somewhere?
