Lemma 29.41.2 (02K7)—The Stacks project

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Table of contents
Part 2: Schemes
Chapter 29: Morphisms of Schemes
Section 29.41: Proper morphisms
Lemma 29.41.2 (cite)

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Lemma 29.41.2. Let $f : X \to S$ be a morphism of schemes. The following are equivalent:

The morphism $f$ is universally closed.
There exists an open covering $S = \bigcup V_ j$ such that $f^{-1}(V_ j) \to V_ j$ is universally closed for all indices $j$ .

Proof. This is clear from the definition. $\square$

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