Lemma 70.15.1. Let $S$ be a scheme. Let $f : X \to Y$ be an integral morphism of algebraic spaces. Then $\dim (X) \leq \dim (Y)$. If $f$ is surjective then $\dim (X) = \dim (Y)$.
Proof. Choose $V \to Y$ surjective étale with $V$ a scheme. Then $U = X \times _ Y V$ is a scheme and $U \to V$ is integral (and surjective if $f$ is surjective). By Properties of Spaces, Lemma 64.22.5 we have $\dim (X) = \dim (U)$ and $\dim (Y) = \dim (V)$. Thus the result follows from the case of schemes which is Morphisms, Lemma 29.43.9. $\square$
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