Lemma 67.42.1. Let S be a scheme. Let f : X \to Y be a morphism of algebraic spaces over S. Assume
f is quasi-compact, and
f satisfies the existence part of the valuative criterion.
Then f is universally closed.
Lemma 67.42.1. Let S be a scheme. Let f : X \to Y be a morphism of algebraic spaces over S. Assume
f is quasi-compact, and
f satisfies the existence part of the valuative criterion.
Then f is universally closed.
Proof. By Lemmas 67.8.4 and 67.41.7 properties (1) and (2) are preserved under any base change. By Lemma 67.9.5 we only have to show that |T \times _ Y X| \to |T| is closed, whenever T is an affine scheme over S mapping into Y. Hence it suffices to prove: If Y is an affine scheme, f : X \to Y is quasi-compact and satisfies the existence part of the valuative criterion, then f : |X| \to |Y| is closed. In this situation X is a quasi-compact algebraic space. By Properties of Spaces, Lemma 66.6.3 there exists an affine scheme U and a surjective étale morphism \varphi : U \to X. Let T \subset |X| closed. The inverse image \varphi ^{-1}(T) \subset U is closed, and hence is the set of points of an affine closed subscheme Z \subset U. Thus, by Algebra, Lemma 10.41.5 we see that f(T) = f(\varphi (|Z|)) \subset |Y| is closed if it is closed under specialization.
Let y' \leadsto y be a specialization in Y with y' \in f(T). Choose a point x' \in T \subset |X| mapping to y' under f. We may represent x' by a morphism \mathop{\mathrm{Spec}}(K) \to X for some field K. Thus we have the following diagram
see Schemes, Section 26.13 for the existence of the left vertical map. Choose a valuation ring A \subset K dominating the image of the ring map \mathcal{O}_{Y, y} \to K (this is possible since the image is a local ring and not a field as y' \not= y, see Algebra, Lemma 10.50.2). By assumption there exists a field extension K'/K and a valuation ring A' \subset K' dominating A, and a morphism \mathop{\mathrm{Spec}}(A') \to X fitting into the commutative diagram. Since A' dominates A, and A dominates \mathcal{O}_{Y, y} we see that the closed point of \mathop{\mathrm{Spec}}(A') maps to a point x \in X with f(x) = y which is a specialization of x'. Hence x \in T as T is closed, and hence y \in f(T) as desired. \square
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