# $Hom$ Sheaf invertible

Let $(X, \mathcal{O}_X$ ringed space, $\mathcal{F}, \mathcal{G}$ two invertible sheaves on $X$ and $\underline{Hom}_{\mathcal{O}_X}(\mathcal{F},\mathcal{G})$ the $Hom$ - sheaf. Why is then $\underline{Hom}_{\mathcal{O}_X}(\mathcal{F},\mathcal{G})$ also invertible?

My ideas: Let considering the canonical evaluation map $\mathcal{F} \otimes \underline{Hom}_{\mathcal{O}_X}(\mathcal{F},\mathcal{G}) \to \mathcal{G}$. I know that for invertible $\mathcal{F}, \mathcal{G}$ it is an isomorphism. Futhermore, there exist a $\mathcal{O}_X$-module $\mathcal{N}$ such that $\mathcal{N} \otimes \mathcal{F} \cong \mathcal{O}_X$ (because $\mathcal{F}$invertible).

By tesor product properties tensoring the evaluation map with $\mathcal{N}$ provides the isomorphism $\underline{Hom}_{\mathcal{O}_X}(\mathcal{F}) \cong \mathcal{N} \otimes \mathcal{F} \otimes \underline{Hom}_{\mathcal{O}_X}(\mathcal{F},\mathcal{G}) \to \mathcal{N} \otimes \mathcal{G}$. If $\mathcal{N}$ is invertible then I have done, but I don't why the $\mathcal{O}_X$-module $\mathcal{N}$ as taken above is invertible.

Let $F,G$ be invertible sheaves, $Hom(F,G)=F\otimes G^*$ where $G^*$ is the dual of $G$. Let$F_1,G_1$ the inverse of respectively $F$ and $G$. $(F\otimes G^*)\otimes( G_1^*\otimes F_1)=F\otimes (G^*\otimes G_1^*)\otimes F_1^*=F\otimes F_1=O_X$.
Since $\mathcal{N}\otimes\mathcal{F}\cong\mathcal{O}_X$, the sheaf $\mathcal{N}$ is invertible with inverse $\mathcal{F}$.
• And this holds because locally we have $\mathcal{F}|_U \cong \mathcal{O}_X|_U \cong \mathcal{O}_U$, so $\mathcal{O}_U \cong \mathcal{N} \otimes \mathcal{F} |_U \cong \mathcal{N}|_U \otimes \mathcal{F}|_U \cong \mathcal{N}|_U$? – KarlPeter Oct 29 '17 at 12:31
• If your definition of "invertible" is "locally isomorphic to $\mathcal{O}_X$, sure. I was assuming it was "has an inverse". – Eric Wofsey Oct 29 '17 at 17:41