The Stacks project

65.9 Universally closed morphisms

For a representable morphism of algebraic spaces we have already defined (in Section 65.3) what it means to be universally closed. Hence before we give the natural definition we check that it agrees with this in the representable case.

Lemma 65.9.1. Let $S$ be a scheme. Let $f : X \to Y$ be a representable morphism of algebraic spaces over $S$. The following are equivalent

  1. $f$ is universally closed (in the sense of Section 65.3), and

  2. for every morphism of algebraic spaces $Z \to Y$ the morphism of topological spaces $|Z \times _ Y X| \to |Z|$ is closed.

Proof. Assume (1), and let $Z \to Y$ be as in (2). Choose a scheme $V$ and a surjective étale morphism $V \to Y$. By assumption the morphism of schemes $V \times _ Y X \to V$ is universally closed. By Properties of Spaces, Section 64.4 in the commutative diagram

\[ \xymatrix{ |V \times _ Y X| \ar[r] \ar[d] & |Z \times _ Y X| \ar[d] \\ |V| \ar[r] & |Z| } \]

the horizontal arrows are open and surjective, and moreover

\[ |V \times _ Y X| \longrightarrow |V| \times _{|Z|} |Z \times _ Y X| \]

is surjective. Hence as the left vertical arrow is closed it follows that the right vertical arrow is closed. This proves (2). The implication (2) $\Rightarrow $ (1) is immediate from the definitions. $\square$

Thus we may use the following natural definition.

Definition 65.9.2. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$.

  1. We say $f$ is closed if the map of topological spaces $|X| \to |Y|$ is closed.

  2. We say $f$ is universally closed if for every morphism of algebraic spaces $Z \to Y$ the morphism of topological spaces

    \[ |Z \times _ Y X| \to |Z| \]

    is closed, i.e., the base change $Z \times _ Y X \to Z$ is closed.

Lemma 65.9.3. The base change of a universally closed morphism of algebraic spaces by any morphism of algebraic spaces is universally closed.

Proof. This is immediate from the definition. $\square$

Lemma 65.9.4. The composition of a pair of (universally) closed morphisms of algebraic spaces is (universally) closed.

Proof. Omitted. $\square$

Lemma 65.9.5. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. The following are equivalent

  1. $f$ is universally closed,

  2. for every scheme $Z$ and every morphism $Z \to Y$ the projection $|Z \times _ Y X| \to |Z|$ is closed,

  3. for every affine scheme $Z$ and every morphism $Z \to Y$ the projection $|Z \times _ Y X| \to |Z|$ is closed,

  4. there exists a scheme $V$ and a surjective étale morphism $V \to Y$ such that $V \times _ Y X \to V$ is a universally closed morphism of algebraic spaces, and

  5. there exists a Zariski covering $Y = \bigcup Y_ i$ such that each of the morphisms $f^{-1}(Y_ i) \to Y_ i$ is universally closed.

Proof. We omit the proof that (1) implies (2), and that (2) implies (3).

Assume (3). Choose a surjective étale morphism $V \to Y$. We are going to show that $V \times _ Y X \to V$ is a universally closed morphism of algebraic spaces. Let $Z \to V$ be a morphism from an algebraic space to $V$. Let $W \to Z$ be a surjective étale morphism where $W = \coprod W_ i$ is a disjoint union of affine schemes, see Properties of Spaces, Lemma 64.6.1. Then we have the following commutative diagram

\[ \xymatrix{ \coprod _ i |W_ i \times _ Y X| \ar@{=}[r] \ar[d] & |W \times _ Y X| \ar[r] \ar[d] & |Z \times _ Y X| \ar[d] \ar@{=}[r] & |Z \times _ V (V \times _ Y X)| \ar[ld] \\ \coprod |W_ i| \ar@{=}[r] & |W| \ar[r] & |Z| } \]

We have to show the south-east arrow is closed. The middle horizontal arrows are surjective and open (Properties of Spaces, Lemma 64.16.7). By assumption (3), and the fact that $W_ i$ is affine we see that the left vertical arrows are closed. Hence it follows that the right vertical arrow is closed.

Assume (4). We will show that $f$ is universally closed. Let $Z \to Y$ be a morphism of algebraic spaces. Consider the diagram

\[ \xymatrix{ |(V \times _ Y Z) \times _ V (V \times _ Y X)| \ar@{=}[r] \ar[rd] & |V \times _ Y X| \ar[r] \ar[d] & |Z \times _ Y X| \ar[d] \\ & |V \times _ Y Z| \ar[r] & |Z| } \]

The south-west arrow is closed by assumption. The horizontal arrows are surjective and open because the corresponding morphisms of algebraic spaces are étale (see Properties of Spaces, Lemma 64.16.7). It follows that the right vertical arrow is closed.

Of course (1) implies (5) by taking the covering $Y = Y$. Assume $Y = \bigcup Y_ i$ is as in (5). Then for any $Z \to Y$ we get a corresponding Zariski covering $Z = \bigcup Z_ i$ such that the base change of $f$ to $Z_ i$ is closed. By a simple topological argument this implies that $Z \times _ Y X \to Z$ is closed. Hence (1) holds. $\square$

Example 65.9.6. Strange example of a universally closed morphism. Let $\mathbf{Q} \subset k$ be a field of characteristic zero. Let $X = \mathbf{A}^1_ k/\mathbf{Z}$ as in Spaces, Example 63.14.8. We claim the structure morphism $p : X \to \mathop{\mathrm{Spec}}(k)$ is universally closed. Namely, if $Z/k$ is a scheme, and $T \subset |X \times _ k Z|$ is closed, then $T$ corresponds to a $\mathbf{Z}$-invariant closed subset of $T' \subset |\mathbf{A}^1 \times Z|$. It is easy to see that this implies that $T'$ is the inverse image of a subset $T''$ of $Z$. By Morphisms, Lemma 29.25.12 we have that $T'' \subset Z$ is closed. Of course $T''$ is the image of $T$. Hence $p$ is universally closed by Lemma 65.9.5.

Lemma 65.9.7. Let $S$ be a scheme. A universally closed morphism of algebraic spaces over $S$ is quasi-compact.

Proof. This proof is a repeat of the proof in the case of schemes, see Morphisms, Lemma 29.41.8. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume that $f$ is not quasi-compact. Our goal is to show that $f$ is not universally closed. By Lemma 65.8.8 there exists an affine scheme $Z$ and a morphism $Z \to Y$ such that $Z \times _ Y X \to Z$ is not quasi-compact. To achieve our goal it suffices to show that $Z \times _ Y X \to Z$ is not universally closed, hence we may assume that $Y = \mathop{\mathrm{Spec}}(B)$ for some ring $B$.

Write $X = \bigcup _{i \in I} X_ i$ where the $X_ i$ are quasi-compact open subspaces of $X$. For example, choose a surjective étale morphism $U \to X$ where $U$ is a scheme, choose an affine open covering $U = \bigcup U_ i$ and let $X_ i \subset X$ be the image of $U_ i$. We will use later that the morphisms $X_ i \to Y$ are quasi-compact, see Lemma 65.8.9. Let $T = \mathop{\mathrm{Spec}}(B[a_ i ; i \in I])$. Let $T_ i = D(a_ i) \subset T$. Let $Z \subset T \times _ Y X$ be the reduced closed subspace whose underlying closed set of points is $|T \times _ Y Z| \setminus \bigcup _{i \in I} |T_ i \times _ Y X_ i|$, see Properties of Spaces, Lemma 64.12.3. (Note that $T_ i \times _ Y X_ i$ is an open subspace of $T \times _ Y X$ as $T_ i \to T$ and $X_ i \to X$ are open immersions, see Spaces, Lemmas 63.12.3 and 63.12.2.) Here is a diagram

\[ \xymatrix{ Z \ar[r] \ar[rd] & T \times _ Y X \ar[d]^{f_ T} \ar[r]_ q & X \ar[d]^ f \\ & T \ar[r]^ p & Y } \]

It suffices to prove that the image $f_ T(|Z|)$ is not closed in $|T|$.

We claim there exists a point $y \in Y$ such that there is no affine open neighborhood $V$ of $y$ in $Y$ such that $X_ V$ is quasi-compact. If not then we can cover $Y$ with finitely many such $V$ and for each $V$ the morphism $Y_ V \to V$ is quasi-compact by Lemma 65.8.9 and then Lemma 65.8.8 implies $f$ quasi-compact, a contradiction. Fix a $y \in Y$ as in the claim.

Let $t \in T$ be the point lying over $y$ with $\kappa (t) = \kappa (y)$ such that $a_ i = 1$ in $\kappa (t)$ for all $i$. Suppose $z \in |Z|$ with $f_ T(z) = t$. Then $q(t) \in X_ i$ for some $i$. Hence $f_ T(z) \not\in T_ i$ by construction of $Z$, which contradicts the fact that $t \in T_ i$ by construction. Hence we see that $t \in |T| \setminus f_ T(|Z|)$.

Assume $f_ T(|Z|)$ is closed in $|T|$. Then there exists an element $g \in B[a_ i; i \in I]$ with $f_ T(|Z|) \subset V(g)$ but $t \not\in V(g)$. Hence the image of $g$ in $\kappa (t)$ is nonzero. In particular some coefficient of $g$ has nonzero image in $\kappa (y)$. Hence this coefficient is invertible on some affine open neighborhood $V$ of $y$. Let $J$ be the finite set of $j \in I$ such that the variable $a_ j$ appears in $g$. Since $X_ V$ is not quasi-compact and each $X_{i, V}$ is quasi-compact, we may choose a point $x \in |X_ V| \setminus \bigcup _{j \in J} |X_{j, V}|$. In other words, $x \in |X| \setminus \bigcup _{j \in J} |X_ j|$ and $x$ lies above some $v \in V$. Since $g$ has a coefficient that is invertible on $V$, we can find a point $t' \in T$ lying above $v$ such that $t' \not\in V(g)$ and $t' \in V(a_ i)$ for all $i \notin J$. This is true because $V(a_ i; i \in I \setminus J) = \mathop{\mathrm{Spec}}(B[a_ j; j\in J])$ and the set of points of this scheme lying over $v$ is bijective with $\mathop{\mathrm{Spec}}(\kappa (v)[a_ j; j \in J])$ and $g$ restricts to a nonzero element of this polynomial ring by construction. In other words $t' \not\in T_ i$ for each $i \not\in J$. By Properties of Spaces, Lemma 64.4.3 we can find a point $z$ of $X \times _ Y T$ mapping to $x \in X$ and to $t' \in T$. Since $x \not\in |X_ j|$ for $j \in J$ and $t' \not\in T_ i$ for $i \in I \setminus J$ we see that $z \in |Z|$. On the other hand $f_ T(z) = t' \not\in V(g)$ which contradicts $f_ T(Z) \subset V(g)$. Thus the assumption “$f_ T(|Z|)$ closed” is wrong and we conclude indeed that $f_ T$ is not closed as desired. $\square$

The target of a separated algebraic space under a surjective universally closed morphism is separated.

Lemma 65.9.8. Let $S$ be a scheme. Let $B$ be an algebraic space over $S$. Let $f : X \to Y$ be a surjective universally closed morphism of algebraic spaces over $B$.

  1. If $X$ is quasi-separated, then $Y$ is quasi-separated.

  2. If $X$ is separated, then $Y$ is separated.

  3. If $X$ is quasi-separated over $B$, then $Y$ is quasi-separated over $B$.

  4. If $X$ is separated over $B$, then $Y$ is separated over $B$.

Proof. Parts (1) and (2) are a consequence of (3) and (4) for $S = B = \mathop{\mathrm{Spec}}(\mathbf{Z})$ (see Properties of Spaces, Definition 64.3.1). Consider the commutative diagram

\[ \xymatrix{ X \ar[d] \ar[rr]_{\Delta _{X/B}} & & X \times _ B X \ar[d] \\ Y \ar[rr]^{\Delta _{Y/B}} & & Y \times _ B Y } \]

The left vertical arrow is surjective (i.e., universally surjective). The right vertical arrow is universally closed as a composition of the universally closed morphisms $X \times _ B X \to X \times _ B Y \to Y \times _ B Y$. Hence it is also quasi-compact, see Lemma 65.9.7.

Assume $X$ is quasi-separated over $B$, i.e., $\Delta _{X/B}$ is quasi-compact. Then if $Z$ is quasi-compact and $Z \to Y \times _ B Y$ is a morphism, then $Z \times _{Y \times _ B Y} X \to Z \times _{Y \times _ B Y} Y$ is surjective and $Z \times _{Y \times _ B Y} X$ is quasi-compact by our remarks above. We conclude that $\Delta _{Y/B}$ is quasi-compact, i.e., $Y$ is quasi-separated over $B$.

Assume $X$ is separated over $B$, i.e., $\Delta _{X/B}$ is a closed immersion. Then if $Z$ is affine, and $Z \to Y \times _ B Y$ is a morphism, then $Z \times _{Y \times _ B Y} X \to Z \times _{Y \times _ B Y} Y$ is surjective and $Z \times _{Y \times _ B Y} X \to Z$ is universally closed by our remarks above. We conclude that $\Delta _{Y/B}$ is universally closed. It follows that $\Delta _{Y/B}$ is representable, locally of finite type, a monomorphism (see Lemma 65.4.1) and universally closed, hence a closed immersion, see Étale Morphisms, Lemma 41.7.2 (and also the abstract principle Spaces, Lemma 63.5.8). Thus $Y$ is separated over $B$. $\square$


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 03HH. Beware of the difference between the letter 'O' and the digit '0'.