Lemma 65.45.7. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. The following are equivalent

$f$ is integral, and

$f$ is affine and universally closed.

Lemma 65.45.7. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. The following are equivalent

$f$ is integral, and

$f$ is affine and universally closed.

**Proof.**
In both cases the morphism is representable, and you can check the condition after a base change by an affine scheme mapping into $Y$, see Lemmas 65.45.3, 65.20.3, and 65.9.5. Hence the result follows from Morphisms, Lemma 29.44.7.
$\square$

Your email address will not be published.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)