The Stacks project

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 66.4 in the commutative diagram

the horizontal arrows are open and surjective, and moreover

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. $\square$

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

Proof. Omitted. $\square$

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 66.6.1. Then we have the following commutative diagram

We have to show the south-east arrow is closed. The middle horizontal arrows are surjective and open (Properties of Spaces, Lemma 66.16.7). 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

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 66.16.7). 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. $\square$

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 67.8.8 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 67.8.9. 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 66.12.3. (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 65.12.3 and 65.12.2.) Here is a diagram

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 67.8.9 and then Lemma 67.8.8 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 66.4.3 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. $\square$

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 66.3.1). Consider the commutative diagram

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 67.9.7.

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 67.4.1) and universally closed, hence a closed immersion, see Étale Morphisms, Lemma 41.7.2 (and also the abstract principle Spaces, Lemma 65.5.8). Thus $Y$ is separated over $B$. $\square$