The Stacks project

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. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0CM9. Beware of the difference between the letter 'O' and the digit '0'.