**Proof.**
Assume (1). Then the functor $- \otimes _\mathcal {O} \mathcal{L}$ is essentially surjective, hence there exists an $\mathcal{O}$-module $\mathcal{N}$ as in (2). If (2) holds, then the functor $- \otimes _\mathcal {O} \mathcal{N}$ is a quasi-inverse to the functor $- \otimes _\mathcal {O} \mathcal{L}$ and we see that (1) holds.

Assume (1) and (2) hold. Since $- \otimes _\mathcal {O} \mathcal{L}$ is an equivalence, it is exact, and hence $\mathcal{L}$ is flat. Denote $\psi : \mathcal{L} \otimes _\mathcal {O} \mathcal{N} \to \mathcal{O}$ the given isomorphism. Let $U$ be an object of $\mathcal{C}$. We will show that the restriction $\mathcal{L}$ to the members of a covering of $U$ is a direct summand of a free module, which will certainly imply that $\mathcal{L}$ is of finite presentation. By construction of $\otimes $ we may assume (after replacing $U$ by the members of a covering) that there exists an integer $n \geq 1$ and sections $x_ i \in \mathcal{L}(U)$, $y_ i \in \mathcal{N}(U)$ such that $\psi (\sum x_ i \otimes y_ i) = 1$. Consider the isomorphisms

\[ \mathcal{L}|_ U \to \mathcal{L}|_ U \otimes _{\mathcal{O}_ U} \mathcal{L}|_ U \otimes _{\mathcal{O}_ U} \mathcal{N}|_ U \to \mathcal{L}|_ U \]

where the first arrow sends $x$ to $\sum x_ i \otimes x \otimes y_ i$ and the second arrow sends $x \otimes x' \otimes y$ to $\psi (x' \otimes y)x$. We conclude that $x \mapsto \sum \psi (x \otimes y_ i)x_ i$ is an automorphism of $\mathcal{L}|_ U$. This automorphism factors as

\[ \mathcal{L}|_ U \to \mathcal{O}_ U^{\oplus n} \to \mathcal{L}|_ U \]

where the first arrow is given by $x \mapsto (\psi (x \otimes y_1), \ldots , \psi (x \otimes y_ n))$ and the second arrow by $(a_1, \ldots , a_ n) \mapsto \sum a_ i x_ i$. In this way we conclude that $\mathcal{L}|_ U$ is a direct summand of a finite free $\mathcal{O}_ U$-module.

Assume (1) and (2) hold. Consider the evaluation map

\[ \mathcal{L} \otimes _\mathcal {O} \mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(\mathcal{L}, \mathcal{O}_ X) \longrightarrow \mathcal{O}_ X \]

To finish the proof of the lemma we will show this is an isomorphism. By Lemma 18.27.6 we have

\[ \mathop{\mathrm{Hom}}\nolimits _\mathcal {O}(\mathcal{O}, \mathcal{O}) = \mathop{\mathrm{Hom}}\nolimits _\mathcal {O} (\mathcal{N} \otimes _\mathcal {O} \mathcal{L}, \mathcal{O}) \longrightarrow \mathop{\mathrm{Hom}}\nolimits _\mathcal {O} (\mathcal{N}, \mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(\mathcal{L}, \mathcal{O})) \]

The image of $1$ gives a morphism $\mathcal{N} \to \mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(\mathcal{L}, \mathcal{O})$. Tensoring with $\mathcal{L}$ we obtain

\[ \mathcal{O} = \mathcal{L} \otimes _\mathcal {O} \mathcal{N} \longrightarrow \mathcal{L} \otimes _\mathcal {O} \mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(\mathcal{L}, \mathcal{O}) \]

This map is the inverse to the evaluation map; computation omitted.
$\square$

## Comments (1)

Comment #8242 by Fan on

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