The Stacks project

Proof. Consider a solid diagram

where $A$ is a valuation ring with field of fractions $K$ and $\gamma : y \circ j \longrightarrow f \circ x$ as in Definition 101.39.1. By Lemma 101.39.4 in order to find a dotted arrow (after possibly replacing $K$ by an extension and $A$ by a valuation ring dominating it) we may replace $\mathcal{Y}$ by $\mathop{\mathrm{Spec}}(A)$ and $\mathcal{X}$ by $\mathop{\mathrm{Spec}}(A) \times _\mathcal {Y} \mathcal{X}$. Of course we use here that being quasi-separated and universally closed are preserved under base change. Thus we reduce to the case discussed in the next paragraph.

Consider a solid diagram

where $A$ is a valuation ring with field of fractions $K$ as in Definition 101.39.1. By Lemma 101.7.7 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 $x' \in |\mathcal{X}|$ in the closure of $x$ mapping to the closed point of $\mathop{\mathrm{Spec}}(A)$. By Lemma 101.7.9 we can find a commutative diagram

such that the closed point of $\mathop{\mathrm{Spec}}(A')$ maps to $x' \in |\mathcal{X}|$. 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. $\square$