The Stacks project

Lemma 66.53.1. 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 66.3) 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 66.3. Conversely, if $f$ is a universal homeomorphism in the sense of Section 66.3 then $X \to Y$ is integral, universally injective and surjective (by Spaces, Lemma 64.5.8 and Morphisms, Lemma 29.45.5). Hence $f$ is universally closed, see Lemma 66.45.7 and universally injective and (universally) surjective, i.e., $f$ is a universal homeomorphism. $\square$


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