Lemma 58.18.1. In More on Morphisms, Situation 37.64.1, for example if $Z \to Y$ and $Z \to X$ are closed immersions of schemes, there is an equivalence of categories

## 58.18 Pushouts and fundamental groups

Here is the main result.

**Proof.**
The pushout exists by More on Morphisms, Proposition 37.64.3. The functor is given by sending a scheme $U$ finite étale over the pushout to the base changes $Y' = U \times _{Y \amalg _ Z X} Y$ and $X' = U \times _{Y \amalg _ Z X} X$ and the natural isomorphism $Y' \times _ Y Z \to X' \times _ X Z$ over $Z$. To prove this functor is an equivalence we use More on Morphisms, Lemma 37.64.7 to construct a quasi-inverse functor. The only thing left to prove is to show that given a morphism $U \to Y \amalg _ Z X$ which is separated, quasi-finite and étale such that $X' \to X$ and $Y' \to Y$ are finite, then $U \to Y \amalg _ Z X$ is finite. This can either be deduced from the corresponding algebra fact (More on Algebra, Lemma 15.6.7) or it can be seen because

is surjective and $X'$ and $Y'$ are proper over $Y \amalg _ Z X$ (this uses the description of the pushout in More on Morphisms, Proposition 37.64.3) and then we can apply Morphisms, Lemma 29.41.10 to conclude that $U$ is proper over $Y \amalg _ Z X$. Since a quasi-finite and proper morphism is finite (More on Morphisms, Lemma 37.43.1) we win. $\square$

## 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Your email address will not be published.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.

## Comments (0)