Lemma 106.11.3. Let $f : \mathcal{X} \to \mathcal{Y}$ and $h : \mathcal{U} \to \mathcal{X}$ be morphisms of algebraic stacks. Assume that $\mathcal{Y}$ is locally Noetherian, that $f$ and $h$ are of finite type, that $f$ is separated, and that the image of $|h| : |\mathcal{U}| \to |\mathcal{X}|$ is dense in $|\mathcal{X}|$. If given any $2$-commutative diagram

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

where $A$ is a discrete valuation ring with field of fractions $K$ and $\gamma : y \circ j \to f \circ h \circ u$ 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 101.39.1), then $f$ is proper.

**Proof.**
It suffices to prove that $f$ is universally closed. Let $V \to \mathcal{Y}$ be a smooth morphism where $V$ is an affine scheme. By Properties of Stacks, Lemma 100.4.3 the image $I$ of $|\mathcal{U} \times _\mathcal {Y} V| \to |\mathcal{X} \times _\mathcal {Y} V|$ is the inverse image of the image of $|h|$. Since $|\mathcal{X} \times _\mathcal {Y} V| \to |\mathcal{X}|$ is open (Morphisms of Stacks, Lemma 101.27.15) we conclude that $I$ is dense in $|\mathcal{X} \times _\mathcal {Y} V|$. Also since the category of dotted arrows behaves well with respect to base change (Morphisms of Stacks, Lemma 101.39.4) the assumption on existence of dotted arrows (after extension) is inherited by the morphisms $\mathcal{U} \times _\mathcal {Y} V \to \mathcal{X} \times _\mathcal {Y} V \to V$. Therefore the assumptions of the lemma are satisfied for the morphisms $\mathcal{U} \times _\mathcal {Y} V \to \mathcal{X} \times _\mathcal {Y} V \to V$. Hence we may assume $\mathcal{Y}$ is an affine scheme.

Assume $\mathcal{Y} = Y$ is an affine scheme. (From now on we no longer have to worry about the $2$-arrows $\gamma $ and $\gamma '$, see Morphisms of Stacks, Lemma 101.39.3.) Then $\mathcal{U}$ is quasi-compact. Choose an affine scheme $U$ and a surjective smooth morphism $U \to \mathcal{U}$. Then we may and do replace $\mathcal{U}$ by $U$. Thus we may assume that $\mathcal{U}$ is an affine scheme.

Assume $\mathcal{Y} = Y$ and $\mathcal{U} = U$ are affine schemes. By Chow's lemma (Theorem 106.10.3) we can choose a surjective proper morphism $X \to \mathcal{X}$ where $X$ is an algebraic space. We will use below that $X \to Y$ is separated as a composition of separated morphisms. Consider the algebraic space $W = X \times _\mathcal {X} U$. The projection morphism $W \to X$ is of finite type. We may replace $X$ by the scheme theoretic image of $W \to X$ and hence we may assume that the image of $|W|$ in $|X|$ is dense in $|X|$ (here we use that the image of $|h|$ is dense in $|\mathcal{X}|$, so after this replacement, the morphism $X \to \mathcal{X}$ is still surjective). We claim that for every solid commutative diagram

\[ \xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r] \ar[d] & W \ar[r] & X \ar[d] \\ \mathop{\mathrm{Spec}}(A) \ar[rr] \ar@{..>}[rru] & & Y } \]

where $A$ is a discrete valuation ring with field of fractions $K$, there exists a dotted arrow making the diagram commute. First, it is enough to prove there exists a dotted arrow after replacing $K$ by an extension and $A$ by a valuation ring in this extension dominating $A$, see Morphisms of Spaces, Lemma 67.41.4. By the assumption of the lemma we get an extension $K'/K$ and a valuation ring $A' \subset K'$ dominating $A$ and an arrow $\mathop{\mathrm{Spec}}(A') \to \mathcal{X}$ lifting the composition $\mathop{\mathrm{Spec}}(A') \to \mathop{\mathrm{Spec}}(A) \to Y$ and compatible with the composition $\mathop{\mathrm{Spec}}(K') \to \mathop{\mathrm{Spec}}(K) \to W \to X$. Because $X \to \mathcal{X}$ is proper, we can use the valuative criterion of properness (Morphisms of Stacks, Lemma 101.43.1) to find an extension $K''/K'$ and a valuation ring $A'' \subset K''$ dominating $A'$ and a morphism $\mathop{\mathrm{Spec}}(A'') \to X$ lifting the composition $\mathop{\mathrm{Spec}}(A'') \to \mathop{\mathrm{Spec}}(A') \to \mathcal{X}$ and compatible with the composition $\mathop{\mathrm{Spec}}(K'') \to \mathop{\mathrm{Spec}}(K') \to \mathop{\mathrm{Spec}}(K) \to X$. Then $K''/K$ and $A'' \subset K''$ and the morphism $\mathop{\mathrm{Spec}}(A'') \to X$ is a solution to the problem posed above and the claim holds.

The claim implies the morphism $X \to Y$ is proper by the case of the lemma for algebraic spaces (Limits of Spaces, Lemma 70.22.1). By Morphisms of Stacks, Lemma 101.37.6 we conclude that $\mathcal{X} \to Y$ is proper as desired.
$\square$

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