Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. Let $\mathcal{S} \subset \mathcal{O}$ be a sub-presheaf of sets such that for all $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ the set $\mathcal{S}(U) \subset \mathcal{O}(U)$ is a multiplicative subset, see Algebra, Definition 10.9.1. In this case we can consider the presheaf of rings
The restriction mapping sends the section $f/s$, $f \in \mathcal{O}(U)$, $s \in \mathcal{S}(U)$ to $(f|_ V)/(s|_ V)$ for $V \to U$ in $\mathcal{C}$.
Lemma 18.44.1. In the situation above the map to the sheafification is a homomorphism of sheaves of rings with the following universal property: for any homomorphism of sheaves of rings $\mathcal{O} \to \mathcal{A}$ such that each local section of $\mathcal{S}$ maps to an invertible section of $\mathcal{A}$ there exists a unique factorization $(\mathcal{S}^{-1}\mathcal{O})^\# \to \mathcal{A}$.
\[ \mathcal{O} \longrightarrow (\mathcal{S}^{-1}\mathcal{O})^\# \]
Proof. Omitted. $\square$
Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. Let $\mathcal{S} \subset \mathcal{O}$ be a sub-presheaf of sets such that for all $U \in \mathcal{C}$ the set $\mathcal{S}(U) \subset \mathcal{O}(U)$ is a multiplicative subset. Let $\mathcal{F}$ be a sheaf of $\mathcal{O}$-modules. In this case we can consider the presheaf of $\mathcal{S}^{-1}\mathcal{O}$-modules
The restriction mapping sends the section $t/s$, $t \in \mathcal{F}(U)$, $s \in \mathcal{S}(U)$ to $(t|_ V)/(s|_ V)$ if $V \to U$ is a morphism of $\mathcal{C}$. Then $\mathcal{S}^{-1}\mathcal{F}$ is a presheaf of $\mathcal{S}^{-1}\mathcal{O}$-modules.
Lemma 18.44.2. In the situation above the map to the sheafification has the following universal property: for any homomorphism of $\mathcal{O}$-modules $\mathcal{F} \to \mathcal{G}$ such that each local section of $\mathcal{S}$ acts invertibly on $\mathcal{G}$ there exists a unique factorization $(\mathcal{S}^{-1}\mathcal{F})^\# \to \mathcal{G}$. Moreover we have as sheaves of $(\mathcal{S}^{-1}\mathcal{O})^\# $-modules.
\[ \mathcal{F} \longrightarrow (\mathcal{S}^{-1}\mathcal{F})^\# \]
\[ (\mathcal{S}^{-1}\mathcal{F})^\# = (\mathcal{S}^{-1}\mathcal{O})^\# \otimes _\mathcal {O} \mathcal{F} \]
Proof. Omitted. $\square$
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