Lemma 101.42.2. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Assume
$f$ is quasi-separated, and
$f$ is universally closed.
Then $f$ satisfies the existence part of the valuative criterion.
Proof. Consider a solid diagram
\[ \xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r]_-x \ar[d]_ j & \mathcal{X} \ar[d]^ f \\ \mathop{\mathrm{Spec}}(A) \ar[r]^-y \ar@{..>}[ru] & \mathcal{Y} } \]
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
\[ \xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r]_-x \ar[d]_ j & \mathcal{X} \ar[d]^ f \\ \mathop{\mathrm{Spec}}(A) \ar@{=}[r] \ar@{..>}[ru] & \mathop{\mathrm{Spec}}(A) } \]
where $A$ is a valuation ring with field of fractions $K$ 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
\[ \xymatrix{ \mathop{\mathrm{Spec}}(K') \ar[r] \ar[d] & \mathop{\mathrm{Spec}}(K) \ar[d] \\ \mathop{\mathrm{Spec}}(A') \ar[r] & \mathcal{X} } \]
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. $\square$
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