Lemma 101.41.2. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Assume $f$ is quasi-separated. If $f$ satisfies the uniqueness part of the valuative criterion, then $f$ is separated.
Proof. The assumption on $f$ means that $\Delta _ f$ is quasi-compact and quasi-separated (Definition 101.4.1). We have to show that $\Delta _ f$ is proper. Lemma 101.40.1 says that $\Delta _ f$ is separated. By Lemma 101.3.3 we know that $\Delta _ f$ is locally of finite type. To finish the proof we have to show that $\Delta _ f$ is universally closed. A formal argument (see Lemma 101.41.1) shows that the uniqueness part of the valuative criterion implies that we have the existence of a dotted arrow in any solid diagram like so:
\[ \xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[d] \ar[r] & \mathcal{X} \ar[d]^{\Delta _ f} \\ \mathop{\mathrm{Spec}}(A) \ar[r] \ar@{..>}[ru] & \mathcal{X} \times _\mathcal {Y} \mathcal{X} } \]
Using that this property is preserved by any base change we conclude that any base change by $\Delta _ f$ by an algebraic space mapping into $\mathcal{X} \times _\mathcal {Y} \mathcal{X}$ has the existence part of the valuative criterion and we conclude is universally closed by the valuative criterion for morphisms of algebraic spaces, see Morphisms of Spaces, Lemma 67.42.1. $\square$
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