The Stacks project

Lemma 99.13.7. Let

\[ \xymatrix{ T \ar[r] \ar[d] & T' \ar[d] \\ S \ar[r] & S' } \]

be a pushout in the category of schemes where $T \to T'$ is a thickening and $T \to S$ is affine, see More on Morphisms, Lemma 37.14.3. Then the functor on fibre categories

\[ \begin{matrix} \mathcal{S}\! \mathit{paces}'_{fp, flat, proper, S'} \\ \downarrow \\ \mathcal{S}\! \mathit{paces}'_{fp, flat, proper, S} \times _{\mathcal{S}\! \mathit{paces}'_{fp, flat, proper, T}} \mathcal{S}\! \mathit{paces}'_{fp, flat, proper, T'} \end{matrix} \]

is an equivalence.

Proof. The functor is an equivalence if we drop “proper” from the list of conditions and replace “of finite presentation” by “locally of finite presentation”, see Pushouts of Spaces, Lemma 81.6.7. Thus it suffices to show that given a morphism $X' \to S'$ of an algebraic space to $S'$ which is flat and locally of finite presentation, then $X' \to S'$ is proper if and only if $S \times _{S'} X' \to S$ and $T' \times _{S'} X' \to T'$ are proper. One implication follows from the fact that properness is preserved under base change (Morphisms of Spaces, Lemma 67.40.3) and the other from the fact that properness of $S \times _{S'} X' \to S$ implies properness of $X' \to S'$ by More on Morphisms of Spaces, Lemma 76.10.2. $\square$


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Your email address will not be published.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 0D1J. Beware of the difference between the letter 'O' and the digit '0'.