Lemma 18.31.2. Let $f : \mathop{\mathit{Sh}}\nolimits (\mathcal{C}) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{C}')$ be a morphism of ringed topoi. Then

\[ f^{-1} : \textit{Ab}(\mathcal{C}') \longrightarrow \textit{Ab}(\mathcal{C}), \quad \mathcal{F} \longmapsto f^{-1}\mathcal{F} \]

is exact. If $(f, f^\sharp ) : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O}) \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}'), \mathcal{O}')$ is a flat morphism of ringed topoi then

\[ f^* : \textit{Mod}(\mathcal{O}') \longrightarrow \textit{Mod}(\mathcal{O}), \quad \mathcal{F} \longmapsto f^*\mathcal{F} \]

is exact.

**Proof.**
Given an abelian sheaf $\mathcal{G}$ on $\mathcal{C}'$ the underlying sheaf of sets of $f^{-1}\mathcal{G}$ is the same as $f^{-1}$ of the underlying sheaf of sets of $\mathcal{G}$, see Sites, Section 7.44. Hence the exactness of $f^{-1}$ for sheaves of sets (required in the definition of a morphism of topoi, see Sites, Definition 7.15.1) implies the exactness of $f^{-1}$ as a functor on abelian sheaves.

To see the statement on modules recall that $f^*\mathcal{F}$ is defined as the tensor product $f^{-1}\mathcal{F} \otimes _{f^{-1}\mathcal{O}', f^\sharp } \mathcal{O}$. Hence $f^*$ is a composition of functors both of which are exact.
$\square$

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