Lemma 66.45.7. 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 integral, and

2. $f$ is affine and universally closed.

Proof. In both cases the morphism is representable, and you can check the condition after a base change by an affine scheme mapping into $Y$, see Lemmas 66.45.3, 66.20.3, and 66.9.5. Hence the result follows from Morphisms, Lemma 29.44.7. $\square$

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