The Stacks project

Lemma 66.5.1. Let $S$ be a scheme. Let $f : X \to Y$ be a representable morphism of algebraic spaces over $S$. Then $f$ is surjective (in the sense of Section 66.3) if and only if $|f| : |X| \to |Y|$ is surjective.

Proof. Namely, if $f : X \to Y$ is representable, then it is surjective if and only if for every scheme $T$ and every morphism $T \to Y$ the base change $f_ T : T \times _ Y X \to T$ of $f$ is a surjective morphism of schemes, in other words, if and only if $|f_ T|$ is surjective. By Properties of Spaces, Lemma 65.4.3 the map $|T \times _ Y X| \to |T| \times _{|Y|} |X|$ is always surjective. Hence $|f_ T| : |T \times _ Y X| \to |T|$ is surjective if $|f| : |X| \to |Y|$ is surjective. Conversely, if $|f_ T|$ is surjective for every $T \to Y$ as above, then by taking $T$ to be the spectrum of a field we conclude that $|X| \to |Y|$ is surjective. $\square$


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