The Stacks project

Proof. Choose an affine scheme $V$, a point $v \in V$ and a smooth morphism $V \to \mathcal{Y}$ mapping $v$ to $y$. The map $|V| \to |\mathcal{Y}|$ is open and by Properties of Stacks, Lemma 100.4.3 the image of $|\mathcal{X} \times _\mathcal {Y} V| \to |V|$ is the inverse image of the image of $|f|$. We conclude that the point $v$ is in the closure of the image of $|\mathcal{X} \times _\mathcal {Y} V| \to |V|$. If we prove the lemma for $\mathcal{X} \times _\mathcal {Y} V \to V$ and the point $v$, then the lemma follows for $f$ and $y$. In this way we reduce to the situation described in the next paragraph.

Assume we have $f : \mathcal{X} \to Y$ and $y \in |Y|$ as in the lemma where $Y$ is a Noetherian affine scheme. Since $f$ is quasi-compact, we conclude that $\mathcal{X}$ is quasi-compact. Hence we can choose an affine scheme $W$ and a surjective smooth morphism $W \to \mathcal{X}$. Then the image of $|f|$ is the same as the image of $|W| \to |Y|$. In this way we reduce to the case of schemes which is Limits, Lemma 32.15.1. $\square$

Proof. To prove that $f$ is separated we have to show that $\Delta : \mathcal{X} \to \mathcal{X} \times _\mathcal {Y} \mathcal{X}$ is proper. We already know that $\Delta $ is representable by algebraic spaces, locally of finite type (Morphisms of Stacks, Lemma 101.3.3) and quasi-compact and quasi-separated (by definition of $f$ being quasi-separated). Choose a scheme $U$ and a surjective smooth morphism $U \to \mathcal{X} \times _\mathcal {Y} \mathcal{X}$. Set

It suffices to show that the morphism of algebraic spaces $V \to U$ is proper (Properties of Stacks, Lemma 100.3.3). Observe that $U$ is locally Noetherian (use Morphisms of Stacks, Lemma 101.17.5 and the fact that $U \to \mathcal{Y}$ is locally of finite type) and $V \to U$ is of finite type and quasi-separated (as the base change of $\Delta $ and properties of $\Delta $ listed above). Applying Cohomology of Spaces, Lemma 69.19.2 it suffices to show: Given a commutative diagram

where $A$ is a discrete valuation ring and $K$ its fraction field, there is a unique dashed arrow making the diagram commute. By Morphisms of Stacks, Lemma 101.39.4 the categories of dashed and dotted arrows are equivalent. Assumption (3) implies there is a unique dotted arrow up to isomorphism, see Morphisms of Stacks, Lemma 101.41.1. We conclude there is a unique dashed arrow as desired. $\square$

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

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. $\square$

Proof. We have to prove $\Delta $ is a proper morphism. Assume first that $\Delta $ is separated. Then we may apply Lemma 106.11.3 to the morphisms $\mathcal{U} \to \mathcal{X}$ and $\Delta : \mathcal{X} \to \mathcal{X} \times _\mathcal {Y} \mathcal{X}$. Observe that $\Delta $ is quasi-compact as $f$ is quasi-separated. Of course $\Delta $ is locally of finite type (true for any diagonal morphism, see Morphisms of Stacks, Lemma 101.3.3). Finally, suppose given a $2$-commutative diagram

where $A$ is a discrete valuation ring with field of fractions $K$ and $\gamma : y \circ j \to \Delta \circ h \circ u$. By Morphisms of Stacks, Lemma 101.41.1 and the assumption in the lemma we find there exists a unique dotted arrow. This proves the last assumption of Lemma 106.11.3 holds and the result follows.

In the general case, it suffices to prove $\Delta $ is separated since then we'll be back in the previous case. In fact, we claim that the assumptions of the lemma hold for

Namely, since $\Delta $ is representable by algebraic spaces, the category of dotted arrows for a diagram as in the previous paragraph is a setoid (see for example Morphisms of Stacks, Lemma 101.39.2). The argument in the preceding paragraph shows these categories are either empty or have one isomorphism class. Thus $\Delta $ is separated. $\square$

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 101.39.4). 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 101.39.3.) 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 102.7.2. 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 106.11.1 we may choose a commutative diagram

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. $\square$