Lemma 81.6.2. Let $S$ be a 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 $S$. Then there exists a pushout
\[ \xymatrix{ X \ar[r] \ar[d]_ f & X' \ar[d]^{f'} \\ Y \ar[r] & Y \amalg _ X X' } \]
in the category of algebraic spaces over $S$. Moreover $Y' = Y \amalg _ X X'$ is a thickening of $Y$ and
\[ \mathcal{O}_{Y'} = \mathcal{O}_ Y \times _{f_*\mathcal{O}_ X} f'_*\mathcal{O}_{X'} \]
as sheaves on $Y_{\acute{e}tale}= (Y')_{\acute{e}tale}$.
Proof.
Choose a scheme $V$ and a surjective étale morphism $V \to Y$. Set $U = V \times _ Y X$. This is a scheme affine over $V$ with a surjective étale morphism $U \to X$. By More on Morphisms of Spaces, Lemma 76.9.6 there exists a $U' \to X'$ surjective étale with $U = U' \times _{X'} X$. In particular the morphism of schemes $U \to U'$ is a thickening too. Apply More on Morphisms, Lemma 37.14.3 to obtain a pushout $V' = V \amalg _ U U'$ in the category of schemes.
We repeat this procedure to construct a pushout
\[ \xymatrix{ U \times _ X U \ar[d] \ar[r] & U' \times _{X'} U' \ar[d] \\ V \times _ Y V \ar[r] & R' } \]
in the category of schemes. Consider the morphisms
\[ U \times _ X U \to U \to V',\quad U' \times _{X'} U' \to U' \to V',\quad V \times _ Y V \to V \to V' \]
where we use the first projection in each case. Clearly these glue to give a morphism $t' : R' \to V'$ which is étale by More on Morphisms, Lemma 37.14.6. Similarly, we obtain $s' : R' \to V'$ étale. The morphism $j' = (t', s') : R' \to V' \times _ S V'$ is unramified (as $t'$ is étale) and a monomorphism when restricted to the closed subscheme $V \times _ Y V \subset R'$. As $V \times _ Y V \subset R'$ is a thickening it follows that $j'$ is a monomorphism too. Finally, $j'$ is an equivalence relation as we can use the functoriality of pushouts of schemes to construct a morphism $c' : R' \times _{s', V', t'} R' \to R'$ (details omitted). At this point we set $Y' = U'/R'$, see Spaces, Theorem 65.10.5.
We have morphisms $X' = U'/U' \times _{X'} U' \to V'/R' = Y'$ and $Y = V/V \times _ Y V \to V'/R' = Y'$. By construction these fit into the commutative diagram
\[ \xymatrix{ X \ar[r] \ar[d]_ f & X' \ar[d]^{f'} \\ Y \ar[r] & Y' } \]
Since $Y \to Y'$ is a thickening we have $Y_{\acute{e}tale}= (Y')_{\acute{e}tale}$, see More on Morphisms of Spaces, Lemma 76.9.6. The commutativity of the diagram gives a map of sheaves
\[ \mathcal{O}_{Y'} \longrightarrow \mathcal{O}_ Y \times _{f_*\mathcal{O}_ X} f'_*\mathcal{O}_{X'} \]
on this set. By More on Morphisms, Lemma 37.14.3 this map is an isomorphism when we restrict to the scheme $V'$, hence it is an isomorphism.
To finish the proof we show that the diagram above is a pushout in the category of algebraic spaces. To see this, let $Z$ be an algebraic space and let $a' : X' \to Z$ and $b : Y \to Z$ be morphisms of algebraic spaces. By Lemma 81.6.1 we obtain a unique morphism $h : V' \to Z$ fitting into the commutative diagrams
\[ \vcenter { \xymatrix{ U' \ar[d] \ar[r] & V' \ar[d]^ h \\ X' \ar[r]^{a'} & Z } } \quad \text{and}\quad \vcenter { \xymatrix{ V \ar[r] \ar[d] & V' \ar[d]^ h \\ Y \ar[r]^ b & Z } } \]
The uniqueness shows that $h \circ t' = h \circ s'$. Hence $h$ factors uniquely as $V' \to Y' \to Z$ and we win.
$\square$
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