Lemma 18.24.2. Let $\mathcal{C}$ be a category viewed as a site with the chaotic topology, see Sites, Example 7.6.6. Let $\mathcal{O}$ be a sheaf of rings on $\mathcal{C}$ and let $\mathcal{F}$ be a sheaf of $\mathcal{O}$-modules. Then $\mathcal{F}$ is quasi-coherent if and only if for all $U \to V$ in $\mathcal{C}$ the canonical map

\[ \mathcal{F}(V) \otimes _{\mathcal{O}(V)} \mathcal{O}(U) \longrightarrow \mathcal{F}(U) \]

is an isomorphism.

**Proof.**
Assume $\mathcal{F}$ is quasi-coherent and let $U \to V$ be a morphism of $\mathcal{C}$. Since every covering of $V$ is given by an isomorphism we conclude from Definition 18.23.1 that there exists a presentation

\[ \bigoplus \nolimits _{j \in J} \mathcal{O}_ V \longrightarrow \bigoplus \nolimits _{i \in I} \mathcal{O}_ V \longrightarrow \mathcal{F}|_{\mathcal{C}/V} \longrightarrow 0 \]

Since the topology on $\mathcal{C}$ is chaotic, taking sections over any object of $\mathcal{C}$ is exact. We conclude that we obtain a presentation

\[ \bigoplus \nolimits _{j \in J} \mathcal{O}(V) \longrightarrow \bigoplus \nolimits _{i \in I} \mathcal{O}(V) \longrightarrow \mathcal{F}(V) \longrightarrow 0 \]

of $\mathcal{F}(V)$ as an $\mathcal{O}(V)$-module and similarly for $\mathcal{F}(U)$. This easily shows that the displayed map in the statement of the lemma is an isomorphism.

Assume the displayed map in the statement of the lemma is an isomorphism for every morphism $U \to V$ in $\mathcal{C}$. Fix $V$ and choose a presentation

\[ \bigoplus \nolimits _{j \in J} \mathcal{O}(V) \longrightarrow \bigoplus \nolimits _{i \in I} \mathcal{O}(V) \longrightarrow \mathcal{F}(V) \longrightarrow 0 \]

of $\mathcal{F}(V)$ as an $\mathcal{O}(V)$-module. Then the assumption on $\mathcal{F}$ exactly means that the corresponding sequence

\[ \bigoplus \nolimits _{j \in J} \mathcal{O}_ V \longrightarrow \bigoplus \nolimits _{i \in I} \mathcal{O}_ V \longrightarrow \mathcal{F}|_{\mathcal{C}/V} \longrightarrow 0 \]

is exact and we conclude that $\mathcal{F}$ is quasi-coherent.
$\square$

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