Inverse of invertible sheaf

Question

Let $\mathcal{L}$ be sheaf on scheme $X$ such that it's locally is $\mathcal{O}$ i want to prove that $$\mathrm{Hom}(\mathcal{L},\mathcal{O})\otimes \mathcal{L} \simeq \mathcal{O}$$ it's quite obvious intuitively or if we considering line bundles. Can someone explain it in details?

Answer 1

We have a natural map $\mathscr{Hom}(\mathscr L, \mathscr O) \otimes \mathscr L \to \mathscr O$ given by $\varphi \otimes g \mapsto \varphi(f)$. Indeed, cover $X$ by affines such that the formula makes sense, and note that the same formula holds on the intersections, so we have a well-defined global map.

Now we can consider the stalks of these two sheaves at a closed point $P \in X$. But since $\mathscr L$ is locally free, $\mathscr L_P \simeq \mathscr O_P$, so we get $\mathscr{Hom}(\mathscr O_P, \mathscr O_P) \otimes \mathscr O_P \to \mathscr O_P$, which is an isomorphism.

Hence the map is locally an isomorphism, hence it is an isomorphism.