Lemma 105.11.5. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Assume that $\mathcal{Y}$ is locally Noetherian and that $f$ is of finite type. If given any $2$-commutative diagram

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

where $A$ is a discrete valuation ring with field of fractions $K$ and $\gamma : y \circ j \to f \circ x$ there exist an extension $K'/K$ of fields, a valuation ring $A' \subset K'$ dominating $A$ such that the category of dotted arrows for the induced diagram

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

with induced $2$-arrow $\gamma ' : y' \circ j' \to f \circ x'$ is nonempty (Morphisms of Stacks, Definition 100.39.1), then $f$ is universally closed.

Proof. Let $V \to \mathcal{Y}$ be a smooth morphism where $V$ is an affine scheme. The category of dotted arrows behaves well with respect to base change (Morphisms of Stacks, Lemma 100.39.4). Hence the assumption on existence of dotted arrows (after extension) is inherited by the morphism $\mathcal{X} \times _\mathcal {Y} V \to V$. Therefore the assumptions of the lemma are satisfied for the morphism $\mathcal{X} \times _\mathcal {Y} V \to V$. Hence we may assume $\mathcal{Y}$ is an affine scheme.

Assume $\mathcal{Y} = Y$ is a Noetherian affine scheme. (From now on we no longer have to worry about the $2$-arrows $\gamma$ and $\gamma '$, see Morphisms of Stacks, Lemma 100.39.3.) To prove that $f$ is universally closed it suffices to show that $|\mathcal{X} \times \mathbf{A}^ n| \to |Y \times \mathbf{A}^ n|$ is closed for all $n$ by Limits of Stacks, Lemma 101.7.2. Since the assumption in the lemma is inherited by the product morphism $\mathcal{X} \times \mathbf{A}^ n \to Y \times \mathbf{A}^ n$ (details omitted) we reduce to proving that $|\mathcal{X}| \to |Y|$ is closed.

Assume $Y$ is a Noetherian affine scheme. Let $T \subset |\mathcal{X}|$ be a closed subset. We have to show that the image of $T$ in $|Y|$ is closed. We may replace $\mathcal{X}$ by the reduced induced closed subspace structure on $T$; we omit the verification that property on the existence of dotted arrows is preserved by this replacement. Thus we reduce to proving that the image of $|\mathcal{X}| \to |Y|$ is closed.

Let $y \in |Y|$ be a point in the closure of the image of $|\mathcal{X}| \to |Y|$. By Lemma 105.11.1 we may choose a commutative diagram

$\xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r] \ar[d] & \mathcal{X} \ar[d]^ f \\ \mathop{\mathrm{Spec}}(A) \ar[r] & Y }$

where $A$ is a discrete valuation ring and $K$ is its field of fractions mapping the closed point of $\mathop{\mathrm{Spec}}(A)$ to $y$. It follows immediately from the assumption in the lemma that $y$ is in the image of $|\mathcal{X}| \to |Y|$ and the proof is complete. $\square$

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