102.7 Universally closed morphisms
This section is the analogue of Limits of Spaces, Section 70.20.
Lemma 102.7.1. Let $g : Z \to Y$ be a morphism of affine schemes. Let $f : \mathcal{X} \to Y$ be a quasi-compact morphism of algebraic stacks. Let $z \in Z$ and let $T \subset |\mathcal{X} \times _ Y Z|$ be a closed subset with $z \not\in \mathop{\mathrm{Im}}(T \to |Z|)$. If $\mathcal{X}$ is quasi-compact, then there exist an open neighbourhood $V \subset Z$ of $z$, 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
$Z'$ is an affine scheme of finite presentation over $Y$,
with $z' = a(z)$ we have $z' \not\in \mathop{\mathrm{Im}}(T' \to |Z'|)$, and
the inverse image of $T$ in $|\mathcal{X} \times _ Y V|$ maps into $T'$ via $|\mathcal{X} \times _ Y V| \to |\mathcal{X} \times _ Y Z'|$.
Proof.
We will deduce this from the corresponding result for morphisms of schemes. Since $\mathcal{X}$ is quasi-compact, we may choose an affine scheme $W$ and a surjective smooth morphism $W \to \mathcal{X}$. 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
$Z'$ is an affine scheme of finite presentation over $Y$,
with $z' = a(z)$ we have $z' \not\in \mathop{\mathrm{Im}}(T' \to |Z'|)$, and
$T_1 = T_ W \cap |W \times _ Y V|$ maps into $T'$ via $|W \times _ Y V| \to |W \times _ Y Z'|$.
The commutative diagram
\[ \xymatrix{ W \times _ Y Z \ar[d] & W \times _ Y V \ar[l] \ar[rr]_{a_1} \ar[d]_ c & & W \times _ Y Z' \ar[d]^ q \\ \mathcal{X} \times _ Y Z & \mathcal{X} \times _ Y V \ar[l] \ar[rr]^{a_2} & & \mathcal{X} \times _ Y Z' } \]
has cartesian squares and the vertical maps are surjective, smooth, 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 $T_2 = T \cap |\mathcal{X} \times _ Y V|$ by $c$. By Properties of Stacks, Lemma 100.4.3 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 $\overline{a_2(T_2)} \subset |\mathcal{X} \times _ Y Z'|$ as a solution to the problem posed by the lemma.
$\square$
Lemma 102.7.2. Let $f : \mathcal{X} \to \mathcal{Y}$ be a quasi-compact morphism of algebraic stacks. The following are equivalent
$f$ is universally closed,
for every morphism $Z \to \mathcal{Y}$ which is locally of finite presentation and where $Z$ is an affine scheme the map $|\mathcal{X} \times _ Y Z| \to |Z|$ is closed, and
there exists a scheme $V$ and a surjective smooth morphism $V \to \mathcal{Y}$ such that $|\mathbf{A}^ n \times (\mathcal{X} \times _\mathcal {Y} V)| \to |\mathbf{A}^ n \times V|$ is closed for all $n \geq 0$.
Proof.
It is clear that (1) implies (2).
Assume (2). Choose a scheme $V$ which is the disjoint union of affine schemes and a surjective smooth morphism $V \to \mathcal{Y}$. In order to show that $f$ is universally closed, it suffices to show that the base change $\mathcal{X} \times _\mathcal {Y} V \to V$ of $f$ is universally closed, see Morphisms of Stacks, Lemma 101.13.5. Note that property (2) holds for this base change. Hence in order to prove that (2) implies (1) we may assume $Y = \mathcal{Y}$ is an affine scheme.
Assume (2) and assume $\mathcal{Y} = Y$ is an affine scheme. If $f$ is not universally closed, then there exists an affine scheme $Z$ over $Y$ such that $|\mathcal{X} \times _ Y Z| \to |Z|$ is not closed, see Morphisms of Stacks, Lemma 101.13.5. This means that there exists some closed subset $T \subset |\mathcal{X} \times _ Y Z|$ 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 102.7.1. We find an open neighbourhood $V \subset Z$, a commutative diagram
\[ \xymatrix{ V \ar[d] \ar[r]_ a & Z' \ar[d]^ b \\ Z \ar[r]^ g & Y, } \]
and a closed subset $T' \subset |\mathcal{X} \times _ Y Z'|$ such that
$Z'$ is an affine scheme of finite presentation over $Y$,
with $z' = a(z)$ we have $z' \not\in \mathop{\mathrm{Im}}(T' \to |Z'|)$, and
the inverse image of $T$ in $|\mathcal{X} \times _ Y V|$ maps into $T'$ via $|\mathcal{X} \times _ Y V| \to |\mathcal{X} \times _ Y Z'|$.
We claim that $z'$ is in the closure of $\mathop{\mathrm{Im}}(T' \to |Z'|)$. This implies that $|\mathcal{X} \times _ Y Z'| \to |Z'|$ is not closed and this is absurd as we assumed (2), in other words, the claim shows that (2) implies (1). To see the claim is true we contemplate the following commutative diagram
\[ \xymatrix{ \mathcal{X} \times _ Y Z \ar[d] & \mathcal{X} \times _ Y V \ar[l] \ar[d] \ar[r] & \mathcal{X} \times _ Y Z' \ar[d] \\ Z & V \ar[l] \ar[r]^ a & Z' } \]
Let $T_ V \subset |\mathcal{X} \times _ Y V|$ be the inverse image of $T$. By Properties of Stacks, Lemma 100.4.3 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 $z$ is in the closure of the image of $T_ V \to |V|$. Since the image of $T_ V$ in $|\mathcal{X} \times _ Y Z'|$ is contained in $|T'|$ it follows immediately that $z' = a(z)$ is in the closure of the image of $T'$.
It is clear that (1) implies (3). Let $V \to \mathcal{Y}$ be as in (3). If we can show that $\mathcal{X} \times _ Y V \to V$ is universally closed, then $f$ is universally closed by Morphisms of Stacks, Lemma 101.13.5. Thus it suffices to show that $f : \mathcal{X} \to \mathcal{Y}$ satisfies (2) if $f$ is a quasi-compact morphism of algebraic stacks, $\mathcal{Y} = Y$ is a scheme, and $|\mathbf{A}^ n \times \mathcal{X}| \to |\mathbf{A}^ n \times Y|$ is closed for all $n$. Let $Z \to Y$ be locally of finite presentation where $Z$ is an affine scheme. We have to show the map $|\mathcal{X} \times _ Y Z| \to |Z|$ is closed. 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
\[ \vcenter { \xymatrix{ \mathcal{X} \times _ Y Z \ar[d] \ar[r] & \mathbf{A}^ n \times \mathcal{X} \ar[d] \\ Z \ar[r] & \mathbf{A}^ n \times Y } } \quad \begin{matrix} \text{inducing the}
\\ \text{cartesian square}
\end{matrix} \quad \vcenter { \xymatrix{ |\mathcal{X} \times _ Y Z| \ar[d] \ar[r] & |\mathbf{A}^ n \times \mathcal{X}| \ar[d] \\ |Z| \ar[r] & |\mathbf{A}^ n \times Y| } } \]
of topological spaces whose horizontal arrows are homeomorphisms onto locally closed subsets (Properties of Stacks, Lemma 100.9.6). 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)