Lemma 21.49.1. Let $(\mathcal{C}, \mathcal{O})$ be a ringed space. Set $R = \Gamma (\mathcal{C}, \mathcal{O})$. The category of $\mathcal{O}$-modules which are summands of finite free $\mathcal{O}$-modules is equivalent to the category of finite projective $R$-modules.
Proof. Observe that a finite projective $R$-module is the same thing as a summand of a finite free $R$-module. The equivalence is given by the functor $\mathcal{E} \mapsto \Gamma (\mathcal{C}, \mathcal{E})$. The inverse functor is given by the following construction. Consider the morphism of topoi $f : \mathop{\mathit{Sh}}\nolimits (\mathcal{C}) \to \mathop{\mathit{Sh}}\nolimits (\text{pt})$ with $f_*$ given by taking global sections and $f^{-1}$ by sending a set $S$, i.e., an object of $\mathop{\mathit{Sh}}\nolimits (\text{pt})$, to the constant sheaf with value $S$. We obtain a morphism $(f, f^\sharp ) : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O}) \to (\mathop{\mathit{Sh}}\nolimits (\text{pt}), R)$ of ringed topoi by using the identity map $R \to f_*\mathcal{O}$. Then the inverse functor is given by $f^*$. $\square$
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