Lemma 67.53.1. Let $S$ be a scheme. Let $f : X \to Y$ be a representable morphism of algebraic spaces over $S$. Then $f$ is a universal homeomorphism (in the sense of Section 67.3) if and only if for every morphism of algebraic spaces $Z \to Y$ the base change map $Z \times _ Y X \to Z$ induces a homeomorphism $|Z \times _ Y X| \to |Z|$.

Proof. If for every morphism of algebraic spaces $Z \to Y$ the base change map $Z \times _ Y X \to Z$ induces a homeomorphism $|Z \times _ Y X| \to |Z|$, then the same is true whenever $Z$ is a scheme, which formally implies that $f$ is a universal homeomorphism in the sense of Section 67.3. Conversely, if $f$ is a universal homeomorphism in the sense of Section 67.3 then $X \to Y$ is integral, universally injective and surjective (by Spaces, Lemma 65.5.8 and Morphisms, Lemma 29.45.5). Hence $f$ is universally closed, see Lemma 67.45.7 and universally injective and (universally) surjective, i.e., $f$ is a universal homeomorphism. $\square$

There are also:

• 1 comment(s) on Section 67.53: Universal homeomorphisms

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).