The Stacks project

Lemma 35.23.3. The property $\mathcal{P}(f) =$“$f$ is universally closed” is fpqc local on the base.

Proof. A base change of a universally closed morphism is universally closed by definition. Being universally closed is Zariski local on the base (from the definition or by Morphisms, Lemma 29.41.2). Finally, let $S' \to S$ be a flat surjective morphism of affine schemes, and let $f : X \to S$ be a morphism. Assume that the base change $f' : X' \to S'$ is universally closed. Let $T \to S$ be any morphism. Consider the diagram

\[ \xymatrix{ X' \ar[d] & S' \times _ S T \times _ S X \ar[d] \ar[r] \ar[l] & T \times _ S X \ar[d] \\ S' & S' \times _ S T \ar[r] \ar[l] & T } \]

in which both squares are cartesian. Thus the assumption implies that the middle vertical arrow is closed. The right horizontal arrows are flat, quasi-compact and surjective (as base changes of $S' \to S$). Hence a subset of $T$ is closed if and only if its inverse image in $S' \times _ S T$ is closed, see Morphisms, Lemma 29.25.12. An easy diagram chase shows that the right vertical arrow is closed too, and we conclude $X \to S$ is universally closed. Therefore Lemma 35.22.4 applies and we win. $\square$

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