Lemma 99.7.8. In Situation 99.7.1. Let
\[ \xymatrix{ Z \ar[r] \ar[d] & Z' \ar[d] \\ Y \ar[r] & Y' } \]
be a pushout in the category of schemes over $B$ where $Z \to Z'$ is a thickening and $Z \to Y$ is affine, see More on Morphisms, Lemma 37.14.3. Then the natural map
\[ Q_{\mathcal{F}/X/B}(Y') \longrightarrow Q_{\mathcal{F}/X/B}(Y) \times _{Q_{\mathcal{F}/X/B}(Z)} Q_{\mathcal{F}/X/B}(Z') \]
is bijective. If $X \to B$ is locally of finite presentation, then the same thing is true for $Q^{fp}_{\mathcal{F}/X/B}$.
Proof.
Let us construct an inverse map. Namely, suppose we have $\mathcal{F}_ Y \to \mathcal{A}$, $\mathcal{F}_{Z'} \to \mathcal{B}'$, and an isomorphism $\mathcal{A}|_{X_ Z} \to \mathcal{B}'|_{X_ Z}$ compatible with the given surjections. Then we apply Pushouts of Spaces, Lemma 81.6.6 to get a quasi-coherent module $\mathcal{A}'$ on $X_{Y'}$ flat over $Y'$. Since this sheaf is constructed as a fibre product (see proof of cited lemma) there is a canonical map $\mathcal{F}_{Y'} \to \mathcal{A}'$. That this map is surjective can be seen because it factors as
\[ \begin{matrix} \mathcal{F}_{Y'}
\\ \downarrow
\\ (X_ Y \to X_{Y'})_*\mathcal{F}_ Y \times _{(X_ Z \to X_{Y'})_*\mathcal{F}_ Z} (X_{Z'} \to X_{Y'})_*\mathcal{F}_{Z'}
\\ \downarrow
\\ \mathcal{A}' = (X_ Y \to X_{Y'})_*\mathcal{A} \times _{(X_ Z \to X_{Y'})_*\mathcal{A}|_{X_ Z}} (X_{Z'} \to X_{Y'})_*\mathcal{B}'
\end{matrix} \]
and the first arrow is surjective by More on Algebra, Lemma 15.6.5 and the second by More on Algebra, Lemma 15.6.6.
In the case of $Q^{fp}_{\mathcal{F}/X/B}$ all we have to show is that the construction above produces a finitely presented module. This is explained in More on Algebra, Remark 15.7.8 in the commutative algebra setting. The current case of modules over algebraic spaces follows from this by étale localization.
$\square$
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