Lemma 67.45.9. Let S be a scheme. Let f : X \to Y be a morphism of algebraic spaces over S. The following are equivalent
f is finite, and
f is affine and proper.
Lemma 67.45.9. Let S be a scheme. Let f : X \to Y be a morphism of algebraic spaces over S. The following are equivalent
f is finite, and
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 67.45.3, 67.20.3, and 67.40.2. Hence the result follows from Morphisms, Lemma 29.44.11. \square
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