Lemma 67.42.2. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$.

1. If $f$ is quasi-separated and universally closed, then $f$ satisfies the existence part of the valuative criterion.

2. If $f$ is quasi-compact and quasi-separated, then $f$ is universally closed if and only if the existence part of the valuative criterion holds.

Proof. If (1) is true then combined with Lemma 67.42.1 we obtain (2). Assume $f$ is quasi-separated and universally closed. Assume given a diagram

$\xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r] \ar[d] & X \ar[d] \\ \mathop{\mathrm{Spec}}(A) \ar[r] & Y }$

as in Definition 67.41.1. A formal argument shows that the existence of the desired diagram

$\xymatrix{ \mathop{\mathrm{Spec}}(K') \ar[r] \ar[d] & \mathop{\mathrm{Spec}}(K) \ar[r] & X \ar[d] \\ \mathop{\mathrm{Spec}}(A') \ar[r] \ar[rru] & \mathop{\mathrm{Spec}}(A) \ar[r] & Y }$

follows from existence in the case of the morphism $X_ A \to \mathop{\mathrm{Spec}}(A)$. Since being quasi-separated and universally closed are preserved by base change, the lemma follows from the result in the next paragraph.

Consider a solid diagram

$\xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r]_-x \ar[d] & X \ar[d]^ f \\ \mathop{\mathrm{Spec}}(A) \ar@{=}[r] \ar@{..>}[ru] & \mathop{\mathrm{Spec}}(A) }$

where $A$ is a valuation ring with field of fractions $K$. By Lemma 67.8.9 and the fact that $f$ is quasi-separated we have that the morphism $x$ is quasi-compact. Since $f$ is universally closed, we have in particular that $|f|(\overline{\{ x\} })$ is closed in $\mathop{\mathrm{Spec}}(A)$. Since this image contains the generic point of $\mathop{\mathrm{Spec}}(A)$ there exists a point $x' \in |X|$ in the closure of $x$ mapping to the closed point of $\mathop{\mathrm{Spec}}(A)$. By Lemma 67.16.5 we can find a commutative diagram

$\xymatrix{ \mathop{\mathrm{Spec}}(K') \ar[r] \ar[d] & \mathop{\mathrm{Spec}}(K) \ar[d] \\ \mathop{\mathrm{Spec}}(A') \ar[r] & X }$

such that the closed point of $\mathop{\mathrm{Spec}}(A')$ maps to $x' \in |X|$. It follows that $\mathop{\mathrm{Spec}}(A') \to \mathop{\mathrm{Spec}}(A)$ maps the closed point to the closed point, i.e., $A'$ dominates $A$ and this finishes the proof. $\square$

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