## 69.20 Universally closed morphisms

In this section we discuss when a quasi-compact (but not necessarily separated) morphism is universally closed. We first prove a lemma which will allow us to check universal closedness after a base change which is locally of finite presentation.

Lemma 69.20.1. Let $S$ be a scheme. Let $f : X \to Y$ and $g : Z \to Y$ be morphisms of algebraic spaces over $S$. Let $z \in |Z|$ and let $T \subset |X \times _ Y Z|$ be a closed subset with $z \not\in \mathop{\mathrm{Im}}(T \to |Z|)$. If $f$ is quasi-compact, then there exists an étale neighbourhood $(V, v) \to (Z, z)$, a commutative diagram

$\xymatrix{ V \ar[d] \ar[r]_ a & Z' \ar[d]^ b \\ Z \ar[r]^ g & Y, }$

and a closed subset $T' \subset |X \times _ Y Z'|$ such that

1. the morphism $b : Z' \to Y$ is locally of finite presentation,

2. with $z' = a(v)$ we have $z' \not\in \mathop{\mathrm{Im}}(T' \to |Z'|)$, and

3. the inverse image of $T$ in $|X \times _ Y V|$ maps into $T'$ via $|X \times _ Y V| \to |X \times _ Y Z'|$.

Moreover, we may assume $V$ and $Z'$ are affine schemes and if $Z$ is a scheme we may assume $V$ is an affine open neighbourhood of $z$.

Proof. We will deduce this from the corresponding result for morphisms of schemes. Let $y \in |Y|$ be the image of $z$. First we choose an affine étale neighbourhood $(U, u) \to (Y, y)$ and then we choose an affine étale neighbourhood $(V, v) \to (Z, z)$ such that the morphism $V \to Y$ factors through $U$. Then we may replace

1. $X \to Y$ by $X \times _ Y U \to U$,

2. $Z \to Y$ by $V \to U$,

3. $z$ by $v$, and

4. $T$ by its inverse image in $|(X \times _ Y U) \times _ U V| = |X \times _ Y V|$.

In fact, below we will show that after replacing $V$ by an affine open neighbourhood of $v$ there will be a morphism $a : V \to Z'$ for some $Z' \to U$ of finite presentation and a closed subset $T'$ of $|(X \times _ Y U) \times _ U Z'| = |X \times _ Y Z'|$ such that $T$ maps into $T'$ and $a(v) \not\in \mathop{\mathrm{Im}}(T' \to |Z'|)$. Thus we may and do assume that $Z$ and $Y$ are affine schemes with the proviso that we need to find a solution where $V$ is an open neighbourhood of $z$.

Since $f$ is quasi-compact and $Y$ is affine, the algebraic space $X$ is quasi-compact. Choose an affine scheme $W$ and a surjective étale morphism $W \to X$. Let $T_ W \subset |W \times _ Y Z|$ be the inverse image of $T$. Then $z$ is not in the image of $T_ W$. By the schemes case (Limits, Lemma 32.14.1) we can find an open neighbourhood $V \subset Z$ of $z$ a commutative diagram of schemes

$\xymatrix{ V \ar[d] \ar[r]_ a & Z' \ar[d]^ b \\ Z \ar[r]^ g & Y, }$

and a closed subset $T' \subset |W \times _ Y Z'|$ such that

1. the morphism $b : Z' \to Y$ is locally of finite presentation,

2. with $z' = a(z)$ we have $z' \not\in \mathop{\mathrm{Im}}(T' \to Z')$, and

3. $T_1 = T_ W \cap |W \times _ Y V|$ maps into $T'$ via $|W \times _ Y V| \to |W \times _ Y Z'|$.

The commutative diagram

$\xymatrix{ W \times _ Y Z \ar[d] & W \times _ Y V \ar[l] \ar[rr]_{a_1} \ar[d]_ c & & W \times _ Y Z' \ar[d]^ q \\ X \times _ Y Z & X \times _ Y V \ar[l] \ar[rr]^{a_2} & & X \times _ Y Z' }$

has cartesian squares and the vertical maps are, surjective, étale and a fortiori open. Looking at the left hand square we see that $T_1 = T_ W \cap |W \times _ Y V|$ is the inverse image of $T_2 = T \cap |X \times _ Y V|$ by $c$. By Properties of Spaces, Lemma 65.4.3 we get $a_1(T_1) = q^{-1}(a_2(T_2))$. By Topology, Lemma 5.6.4 we get

$q^{-1}\left(\overline{a_2(T_2)}\right) = \overline{q^{-1}(a_2(T_2))} = \overline{a_1(T_1)} \subset T'$

As $q$ is surjective the image of $\overline{a_2(T_2)} \to |Z'|$ does not contain $z'$ since the same is true for $T'$. Thus we can take the diagram with $Z', V, a, b$ above and the closed subset $\overline{a_2(T_2)} \subset |X \times _ Y Z'|$ as a solution to the problem posed by the lemma. $\square$

Lemma 69.20.2. Let $S$ be a scheme. Let $f : X \to Y$ be a quasi-compact morphism of algebraic spaces over $S$. The following are equivalent

1. $f$ is universally closed,

2. for every morphism $Z \to Y$ which is locally of finite presentation the map $|X \times _ Y Z| \to |Z|$ is closed, and

3. there exists a scheme $V$ and a surjective étale morphism $V \to Y$ such that $|\mathbf{A}^ n \times (X \times _ Y V)| \to |\mathbf{A}^ n \times V|$ is closed for all $n \geq 0$.

Proof. It is clear that (1) implies (2). Suppose that $|X \times _ Y Z| \to |Z|$ is not closed for some morphism of algebraic spaces $Z \to Y$ over $S$. This means that there exists some closed subset $T \subset |X \times _ Y Z|$ such that $\mathop{\mathrm{Im}}(T \to |Z|)$ is not closed. Pick $z \in |Z|$ in the closure of the image of $T$ but not in the image. Apply Lemma 69.20.1. We find an étale neighbourhood $(V, v) \to (Z, z)$, a commutative diagram

$\xymatrix{ V \ar[d] \ar[r]_ a & Z' \ar[d]^ b \\ Z \ar[r]^ g & Y, }$

and a closed subset $T' \subset |X \times _ Y Z'|$ such that

1. the morphism $b : Z' \to Y$ is locally of finite presentation,

2. with $z' = a(v)$ we have $z' \not\in \mathop{\mathrm{Im}}(T' \to |Z'|)$, and

3. the inverse image of $T$ in $|X \times _ Y V|$ maps into $T'$ via $|X \times _ Y V| \to |X \times _ Y Z'|$.

We claim that $z'$ is in the closure of $\mathop{\mathrm{Im}}(T' \to |Z'|)$ which implies that $|X \times _ Y Z'| \to |Z'|$ is not closed. The claim shows that (2) implies (1). To see the claim is true we contemplate following commutative diagram

$\xymatrix{ X \times _ Y Z \ar[d] & X \times _ Y V \ar[l] \ar[d] \ar[r] & X \times _ Y Z' \ar[d] \\ Z & V \ar[l] \ar[r]^ a & Z' }$

Let $T_ V \subset |X \times _ Y V|$ be the inverse image of $T$. By Properties of Spaces, Lemma 65.4.3 the image of $T_ V$ in $|V|$ is the inverse image of the image of $T$ in $|Z|$. Then since $z$ is in the closure of the image of $T \to |Z|$ and since $|V| \to |Z|$ is open, we see that $v$ is in the closure of the image of $T_ V \to |V|$. Since the image of $T_ V$ in $|X \times _ Y Z'|$ is contained in $|T'|$ it follows immediately that $z' = a(v)$ is in the closure of the image of $T'$.

It is clear that (1) implies (3). Let $V \to Y$ be as in (3). If we can show that $X \times _ Y V \to V$ is universally closed, then $f$ is universally closed by Morphisms of Spaces, Lemma 66.9.5. Thus it suffices to show that $f : X \to Y$ satisfies (2) if $f$ is a quasi-compact morphism of algebraic spaces, $Y$ is a scheme, and $|\mathbf{A}^ n \times X| \to |\mathbf{A}^ n \times Y|$ is closed for all $n$. Let $Z \to Y$ be locally of finite presentation. We have to show the map $|X \times _ Y Z| \to |Z|$ is closed. This question is étale local on $Z$ hence we may assume $Z$ is affine (some details omitted). Since $Y$ is a scheme, $Z$ is affine, and $Z \to Y$ is locally of finite presentation we can find an immersion $Z \to \mathbf{A}^ n \times Y$, see Morphisms, Lemma 29.39.2. Consider the cartesian diagram

$\vcenter { \xymatrix{ X \times _ Y Z \ar[d] \ar[r] & \mathbf{A}^ n \times X \ar[d] \\ Z \ar[r] & \mathbf{A}^ n \times Y } } \quad \begin{matrix} \text{inducing the} \\ \text{cartesian square} \end{matrix} \quad \vcenter { \xymatrix{ |X \times _ Y Z| \ar[d] \ar[r] & |\mathbf{A}^ n \times X| \ar[d] \\ |Z| \ar[r] & |\mathbf{A}^ n \times Y| } }$

of topological spaces whose horizontal arrows are homeomorphisms onto locally closed subsets (Properties of Spaces, Lemma 65.12.1). Thus every closed subset $T$ of $|X \times _ Y Z|$ is the pullback of a closed subset $T'$ of $|\mathbf{A}^ n \times Y|$. Since the assumption is that the image of $T'$ in $|\mathbf{A}^ n \times X|$ is closed we conclude that the image of $T$ in $|Z|$ is closed as desired. $\square$

Lemma 69.20.3. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume $f$ separated and of finite type. The following are equivalent

1. The morphism $f$ is proper.

2. For any morphism $Y \to Z$ which is locally of finite presentation the map $|X \times _ Y Z| \to |Z|$ is closed, and

3. there exists a scheme $V$ and a surjective étale morphism $V \to Y$ such that $|\mathbf{A}^ n \times (X \times _ Y V)| \to |\mathbf{A}^ n \times V|$ is closed for all $n \geq 0$.

Proof. In view of the fact that a proper morphism is the same thing as a separated, finite type, and universally closed morphism, this lemma is a special case of Lemma 69.20.2. $\square$

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