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The Stacks project

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


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