Lemma 101.41.1. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Consider a $2$-commutative solid diagram

\[ \xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[rr]_-x \ar[d]_ j & & \mathcal{X} \ar[d]^{\Delta _ f} \\ \mathop{\mathrm{Spec}}(A) \ar[rr]^{(a_1, a_2, \varphi )} \ar@{..>}[rru] & & \mathcal{X} \times _\mathcal {Y} \mathcal{X} } \]

where $A$ is a valuation ring with field of fractions $K$. Let $\gamma : (a_1, a_2, \varphi ) \circ j \longrightarrow \Delta _ f \circ x$ be a $2$-morphism witnessing the $2$-commutativity of the diagram. Then

Writing $\gamma = (\alpha _1, \alpha _2)$ with $\alpha _ i : a_ i \circ j \to x$ we obtain two dotted arrows $(a_1, \alpha _1, \text{id})$ and $(a_2, \alpha _2, \varphi )$ in the diagram

\[ \xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r]_-x \ar[d]_ j & \mathcal{X} \ar[d]^ f \\ \mathop{\mathrm{Spec}}(A) \ar[r]^-{f \circ a_1} \ar@{..>}[ru] & \mathcal{Y} } \]

The category of dotted arrows for the original diagram and $\gamma $ is a setoid whose set of isomorphism classes of objects equal to the set of morphisms $(a_1, \alpha _1, \text{id}) \to (a_2, \alpha _2, \varphi )$ in the category of dotted arrows.

**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.
$\square$

Here is a converse.

**Proof.**
Since $f$ is separated we see that all categories of dotted arrows are setoids by Lemma 101.39.2. Consider a diagram

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

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

\[ f \circ a \xrightarrow {\beta ^{-1}} y \xrightarrow {\beta '} f \circ a' \]

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

\[ (a, a', \beta ' \circ \beta ^{-1}) \circ j = (a \circ j, a' \circ j, (\beta ' \star \text{id}_ j) \circ (\beta \star \text{id}_ j)^{-1}) \xrightarrow {(\alpha , \alpha ')} (x, x, \text{id}) = \Delta _ f \circ x \]

Here we use that both $(a, \alpha , \beta )$ and $(a', \alpha ', \beta ')$ are dotted arrows with respect to $\gamma $. We obtain a commutative diagram

\[ \xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[d]_ j \ar[rr]_ x & & \mathcal{X} \ar[d]^{\Delta _ f} \\ \mathop{\mathrm{Spec}}(A) \ar[rr]^{(a, a', \beta ' \circ \beta ^{-1})} & & \mathcal{X} \times _\mathcal {Y} \mathcal{X} } \]

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.
$\square$

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