Lemma 80.6.6. Let $S$ be a base scheme. Let $X \to X'$ be a thickening of algebraic spaces over $S$ and let $X \to Y$ be an affine morphism of algebraic spaces over $S$. Let $Y' = Y \amalg _ X X'$ be the pushout (see Lemma 80.6.2). Let $V' \to Y'$ be a morphism of algebraic spaces over $S$. Set $V = Y \times _{Y'} V'$, $U' = X' \times _{Y'} V'$, and $U = X \times _{Y'} V'$. There is an equivalence of categories between

1. quasi-coherent $\mathcal{O}_{V'}$-modules flat over $Y'$, and

2. the category of triples $(\mathcal{G}, \mathcal{F}', \varphi )$ where

1. $\mathcal{G}$ is a quasi-coherent $\mathcal{O}_ V$-module flat over $Y$,

2. $\mathcal{F}'$ is a quasi-coherent $\mathcal{O}_{U'}$-module flat over $X$, and

3. $\varphi : (U \to V)^*\mathcal{G} \to (U \to U')^*\mathcal{F}'$ is an isomorphism of $\mathcal{O}_ U$-modules.

The equivalence maps $\mathcal{G}'$ to $((V \to V')^*\mathcal{G}', (U' \to V')^*\mathcal{G}', can)$. Suppose $\mathcal{G}'$ corresponds to the triple $(\mathcal{G}, \mathcal{F}', \varphi )$. Then

1. $\mathcal{G}'$ is a finite type $\mathcal{O}_{V'}$-module if and only if $\mathcal{G}$ and $\mathcal{F}'$ are finite type $\mathcal{O}_ Y$ and $\mathcal{O}_{U'}$-modules.

2. if $V' \to Y'$ is locally of finite presentation, then $\mathcal{G}'$ is an $\mathcal{O}_{V'}$-module of finite presentation if and only if $\mathcal{G}$ and $\mathcal{F}'$ are $\mathcal{O}_ Y$ and $\mathcal{O}_{U'}$-modules of finite presentation.

Proof. A quasi-inverse functor assigns to the triple $(\mathcal{G}, \mathcal{F}', \varphi )$ the fibre product

$(V \to V')_*\mathcal{G} \times _{(U \to V')_*\mathcal{F}} (U' \to V')_*\mathcal{F}'$

where $\mathcal{F} = (U \to U')^*\mathcal{F}'$. This works, because on affines étale over $V'$ and $Y'$ we recover the equivalence of More on Algebra, Lemma 15.7.5. Details omitted.

Parts (a) and (b) reduce by étale localization (Properties of Spaces, Section 65.30) to the case where $V'$ and $Y'$ are affine in which case the result follows from More on Algebra, Lemmas 15.7.4 and 15.7.6. $\square$

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