Lemma 101.41.3. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. If $f$ is separated, then $f$ satisfies the uniqueness part of the valuative criterion.
Proof. Since $f$ is separated we see that all categories of dotted arrows are setoids by Lemma 101.39.2. Consider a diagram
and a $2$-morphism $\gamma : y \circ j \to f \circ x$ as in Definition 101.39.1. Consider two objects $(a, \alpha , \beta )$ and $(a', \beta ', \alpha ')$ of the category of dotted arrows. To finish the proof we have to show these objects are isomorphic. The isomorphism
means that $(a, a', \beta ' \circ \beta ^{-1})$ is a morphism $\mathop{\mathrm{Spec}}(A) \to \mathcal{X} \times _\mathcal {Y} \mathcal{X}$. On the other hand, $\alpha $ and $\alpha '$ define a $2$-arrow
Here we use that both $(a, \alpha , \beta )$ and $(a', \alpha ', \beta ')$ are dotted arrows with respect to $\gamma $. We obtain a commutative diagram
with $2$-commutativity witnessed by $(\alpha , \alpha ')$. Now $\Delta _ f$ is representable by algebraic spaces (Lemma 101.3.3) and proper as $f$ is separated. Hence by Lemma 101.39.13 and the valuative criterion for properness for algebraic spaces (Morphisms of Spaces, Lemma 67.44.1) we see that there exists a dotted arrow. Unwinding the construction, we see that this means $(a, \alpha , \beta )$ and $(a', \alpha ', \beta ')$ are isomorphic in the category of dotted arrows as desired. $\square$
Comments (0)