Lemma 91.11.3. Let (f, f') be a morphism of first order thickenings of ringed topoi as in Situation 91.9.1. Let \mathcal{F}' be an \mathcal{O}'-module and set \mathcal{F} = i^*\mathcal{F}'. Assume that \mathcal{F}' is flat over \mathcal{O}_{\mathcal{B}'} and that (f, f') is a strict morphism of thickenings. Then the following are equivalent
\mathcal{F}' is an \mathcal{O}'-module of finite presentation, and
\mathcal{F} is an \mathcal{O}-module of finite presentation.
Proof.
The implication (1) \Rightarrow (2) follows from Modules on Sites, Lemma 18.23.4. For the converse, assume \mathcal{F} of finite presentation. We may and do assume that \mathcal{C} = \mathcal{C}'. By Lemma 91.11.2 we have a short exact sequence
0 \to \mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{F} \to \mathcal{F}' \to \mathcal{F} \to 0
Let U be an object of \mathcal{C} such that \mathcal{F}|_ U has a presentation
\mathcal{O}_ U^{\oplus m} \to \mathcal{O}_ U^{\oplus n} \to \mathcal{F}|_ U \to 0
After replacing U by the members of a covering we may assume the map \mathcal{O}_ U^{\oplus n} \to \mathcal{F}|_ U lifts to a map (\mathcal{O}'_ U)^{\oplus n} \to \mathcal{F}'|_ U. The induced map \mathcal{I}^{\oplus n} \to \mathcal{I} \otimes \mathcal{F} is surjective by right exactness of \otimes . Thus after replacing U by the members of a covering we can find a lift (\mathcal{O}'|_ U)^{\oplus m} \to (\mathcal{O}'|_ U)^{\oplus n} of the given map \mathcal{O}_ U^{\oplus m} \to \mathcal{O}_ U^{\oplus n} such that
(\mathcal{O}'_ U)^{\oplus m} \to (\mathcal{O}'_ U)^{\oplus n} \to \mathcal{F}'|_ U \to 0
is a complex. Using right exactness of \otimes once more it is seen that this complex is exact.
\square
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