Lemma 65.45.9. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. The following are equivalent

1. $f$ is finite, and

2. $f$ is affine and proper.

Proof. In both cases the morphism is representable, and you can check the condition after base change to an affine scheme mapping into $Y$, see Lemmas 65.45.3, 65.20.3, and 65.40.2. Hence the result follows from Morphisms, Lemma 29.43.11. $\square$

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