Lemma 81.11.2. In Situation 81.10.6 the functor (81.11.0.1) reflects isomorphisms.

**Proof.**
By a formal argument with base change, this reduces to the following question: A morphism $a : X' \to X$ of algebraic spaces such that $U \times _ X X' \to U$ and $Y \times _ X X' \to Y$ are isomorphisms, is an isomorphism. The family $\{ U \to X, Y \to X\} $ can be refined by an fpqc covering by Lemma 81.10.11. Hence the result follows from Descent on Spaces, Lemma 74.11.15.
$\square$

## 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Your email address will not be published.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)

There are also: