Lemma 80.6.2. 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 75.9.6 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 64.10.5.

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 75.9.6. 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 80.6.1 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. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).