Lemma 81.6.3. 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 81.6.2). Base change gives a functor

$F : (\textit{Spaces}/Y') \longrightarrow (\textit{Spaces}/Y) \times _{(\textit{Spaces}/Y')} (\textit{Spaces}/X')$

given by $V' \longmapsto (V' \times _{Y'} Y, V' \times _{Y'} X', 1)$ which sends $(\mathit{Sch}/Y')$ into $(\mathit{Sch}/Y) \times _{(\mathit{Sch}/Y')} (\mathit{Sch}/X')$. The functor $F$ has a left adjoint

$G : (\textit{Spaces}/Y) \times _{(\textit{Spaces}/Y')} (\textit{Spaces}/X') \longrightarrow (\textit{Spaces}/Y')$

which sends the triple $(V, U', \varphi )$ to the pushout $V \amalg _{(V \times _ Y X)} U'$ in the category of algebraic spaces over $S$. The functor $G$ sends $(\mathit{Sch}/Y) \times _{(\mathit{Sch}/Y')} (\mathit{Sch}/X')$ into $(\mathit{Sch}/Y')$.

Proof. The proof is completely formal. Since the morphisms $X \to X'$ and $X \to Y$ are representable it is clear that $F$ sends $(\mathit{Sch}/Y')$ into $(\mathit{Sch}/Y) \times _{(\mathit{Sch}/Y')} (\mathit{Sch}/X')$.

Let us construct $G$. Let $(V, U', \varphi )$ be an object of the fibre product category. Set $U = U' \times _{X'} X$. Note that $U \to U'$ is a thickening. Since $\varphi : V \times _ Y X \to U' \times _{X'} X = U$ is an isomorphism we have a morphism $U \to V$ over $X \to Y$ which identifies $U$ with the fibre product $X \times _ Y V$. In particular $U \to V$ is affine, see Morphisms of Spaces, Lemma 67.20.5. Hence we can apply Lemma 81.6.2 to get a pushout $V' = V \amalg _ U U'$. Denote $V' \to Y'$ the morphism we obtain in virtue of the fact that $V'$ is a pushout and because we are given morphisms $V \to Y$ and $U' \to X'$ agreeing on $U$ as morphisms into $Y'$. Setting $G(V, U', \varphi ) = V'$ gives the functor $G$.

If $(V, U', \varphi )$ is an object of $(\mathit{Sch}/Y) \times _{(\mathit{Sch}/Y')} (\mathit{Sch}/X')$ then $U = U' \times _{X'} X$ is a scheme too and we can form the pushout $V' = V \amalg _ U U'$ in the category of schemes by More on Morphisms, Lemma 37.14.3. By Lemma 81.6.1 this is also a pushout in the category of schemes, hence $G$ sends $(\mathit{Sch}/Y) \times _{(\mathit{Sch}/Y')} (\mathit{Sch}/X')$ into $(\mathit{Sch}/Y')$.

Let us prove that $G$ is a left adjoint to $F$. Let $Z$ be an algebraic space over $Y'$. We have to show that

$\mathop{\mathrm{Mor}}\nolimits (V', Z) = \mathop{\mathrm{Mor}}\nolimits ((V, U', \varphi ), F(Z))$

where the morphism sets are taking in their respective categories. Let $g' : V' \to Z$ be a morphism. Denote $\tilde g$, resp. $\tilde f'$ the composition of $g'$ with the morphism $V \to V'$, resp. $U' \to V'$. Base change $\tilde g$, resp. $\tilde f'$ by $Y \to Y'$, resp. $X' \to Y'$ to get a morphism $g : V \to Z \times _{Y'} Y$, resp. $f' : U' \to Z \times _{Y'} X'$. Then $(g, f')$ is an element of the right hand side of the equation above (details omitted). Conversely, suppose that $(g, f') : (V, U', \varphi ) \to F(Z)$ is an element of the right hand side. We may consider the composition $\tilde g : V \to Z$, resp. $\tilde f' : U' \to Z$ of $g$, resp. $f$ by $Z \times _{Y'} X' \to Z$, resp. $Z \times _{Y'} Y \to Z$. Then $\tilde g$ and $\tilde f'$ agree as morphism from $U$ to $Z$. By the universal property of pushout, we obtain a morphism $g' : V' \to Z$, i.e., an element of the left hand side. We omit the verification that these constructions are mutually inverse. $\square$

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