The Stacks project

Lemma 80.6.3. 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 80.6.2). 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 66.20.5. Hence we can apply Lemma 80.6.2 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 80.6.1 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. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 07VY. Beware of the difference between the letter 'O' and the digit '0'.