The Stacks project

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

Proof. Omitted. $\square$

Proof. A composition of closed morphisms is closed. If $X \to Y \to Z$ are universally closed morphisms and $Z' \to Z$ is any morphism, then we see that $Z' \times _ Z X = (Z' \times _ Z Y) \times _ Y X \to Z' \times _ Z Y$ is closed and $Z' \times _ Z Y \to Z'$ is closed. Hence the result for universally closed morphisms. We have seen that “separated” and “finite type” are preserved under compositions (Schemes, Lemma 26.21.12 and Lemma 29.15.3). Hence the result for proper morphisms. $\square$

Proof. This is true by definition for universally closed morphisms. It is true for separated morphisms (Schemes, Lemma 26.21.12). It is true for morphisms of finite type (Lemma 29.15.4). Hence it is true for proper morphisms. $\square$

Proof. The base change of a closed immersion is a closed immersion (Schemes, Lemma 26.18.2). Hence it is universally closed. A closed immersion is separated (Schemes, Lemma 26.23.8). A closed immersion is of finite type (Lemma 29.15.5). Hence a closed immersion is proper. $\square$

Proof. Assume that $X \to S$ is universally closed (resp. proper). We factor the morphism as $X \to X \times _ S Y \to Y$. The first morphism is a closed immersion, see Schemes, Lemma 26.21.10. Hence the first morphism is proper (Lemma 29.41.6). The projection $X \times _ S Y \to Y$ is the base change of a universally closed (resp. proper) morphism and hence universally closed (resp. proper), see Lemma 29.41.5. Thus $X \to Y$ is universally closed (resp. proper) as the composition of universally closed (resp. proper) morphisms (Lemma 29.41.4). $\square$

Proof. Let $f : X \to S$ be a morphism. Assume that $f$ is not quasi-compact. Our goal is to show that $f$ is not universally closed. By Schemes, Lemma 26.19.2 there exists an affine open $V \subset S$ such that $f^{-1}(V)$ is not quasi-compact. To achieve our goal it suffices to show that $f^{-1}(V) \to V$ is not universally closed, hence we may assume that $S = \mathop{\mathrm{Spec}}(A)$ for some ring $A$.

Write $X = \bigcup _{i \in I} X_ i$ where the $X_ i$ are affine open subschemes of $X$. Let $T = \mathop{\mathrm{Spec}}(A[y_ i ; i \in I])$. Let $T_ i = D(y_ i) \subset T$. Let $Z$ be the closed set $(X \times _ S T) - \bigcup _{i \in I} (X_ i \times _ S T_ i)$. It suffices to prove that the image $f_ T(Z)$ of $Z$ under $f_ T : X \times _ S T \to T$ is not closed.

There exists a point $s \in S$ such that there is no neighborhood $U$ of $s$ in $S$ such that $X_ U$ is quasi-compact. Otherwise we could cover $S$ with finitely many such $U$ and Schemes, Lemma 26.19.2 would imply $f$ quasi-compact. Fix such an $s \in S$.

First we check that $f_ T(Z_ s) \ne T_ s$. Let $t \in T$ be the point lying over $s$ with $\kappa (t) = \kappa (s)$ such that $y_ i = 1$ in $\kappa (t)$ for all $i$. Then $t \in T_ i$ for all $i$, and the fiber of $Z_ s \to T_ s$ above $t$ is isomorphic to $(X - \bigcup _{i \in I} X_ i)_ s$, which is empty. Thus $t \in T_ s - f_ T(Z_ s)$.

Assume $f_ T(Z)$ is closed in $T$. Then there exists an element $g \in A[y_ i; i \in I]$ 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 $\kappa (s)$. Hence this coefficient is invertible on some neighborhood $U$ of $s$. Let $J$ be the finite set of $j \in I$ such that $y_ j$ appears in $g$. Since $X_ U$ is not quasi-compact, we may choose a point $x \in X - \bigcup _{j \in J} X_ j$ lying above some $u \in U$. Since $g$ has a coefficient that is invertible on $U$, we can find a point $t' \in T$ lying above $u$ such that $t' \not\in V(g)$ and $t' \in V(y_ i)$ for all $i \notin J$. This is true because $V(y_ i; i \in I, i \not\in J) = \mathop{\mathrm{Spec}}(A[t_ j; j\in J])$ and the set of points of this scheme lying over $u$ is bijective with $\mathop{\mathrm{Spec}}(\kappa (u)[t_ j; j \in J])$. In other words $t' \notin T_ i$ for each $i \notin J$. By Schemes, Lemma 26.17.5 we can find a point $z$ of $X \times _ S T$ 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. Assume $X$ is universally closed and $f$ surjective. Denote $p : X \to S$, $q : Y \to S$ the structure morphisms. Let $S' \to S$ be a morphism of schemes. The base change $f' : X_{S'} \to Y_{S'}$ is surjective (Lemma 29.9.4), and the base change $p' : X_{S'} \to S'$ is closed. If $T \subset Y_{S'}$ is closed, then $(f')^{-1}(T) \subset X_{S'}$ is closed, hence $p'((f')^{-1}(T)) = q'(T)$ is closed. So $q'$ is closed. This proves the first statement. Thus $Y \to S$ is quasi-compact by Lemma 29.41.8 and hence $Y \to S$ is proper by definition if in addition $Y \to S$ is locally of finite type and separated. $\square$

Proof. The scheme theoretic image of $h$ is constructed in Section 29.6. Since $f$ is quasi-compact (Lemma 29.41.8) we find that $h$ is quasi-compact (Schemes, Lemma 26.21.14). Hence $h(X) \subset Z$ is dense (Lemma 29.6.3). On the other hand $h(X)$ is closed in $Y$ (Lemma 29.41.7) hence $X \to Z$ is surjective. Thus $Z \to S$ is a proper (Lemma 29.41.9). $\square$

Proof. Parts (1) and (2) are a consequence of (3) and (4) for $S = \mathop{\mathrm{Spec}}(\mathbf{Z})$ (see Schemes, Definition 26.21.3). 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 $X \times _ S X \to X \times _ S Y \to Y \times _ S Y$. Hence it is also quasi-compact, see Lemma 29.41.8.

Assume $X$ is quasi-separated over $S$, i.e., $\Delta _{X/S}$ is quasi-compact. If $V \subset Y \times _ S Y$ is a quasi-compact open, then $V \times _{Y \times _ S Y} X \to \Delta _{Y/S}^{-1}(V)$ is surjective and $V \times _{Y \times _ S Y} X$ is quasi-compact by our remarks above. We conclude that $\Delta _{Y/S}$ is quasi-compact, i.e., $Y$ is quasi-separated over $S$.

Assume $X$ is separated over $S$, i.e., $\Delta _{X/S}$ is a closed immersion. Then $X \to Y \times _ S Y$ is closed as a composition of closed morphisms. Since $X \to Y$ is surjective, it follows that $\Delta _{Y/S}(Y)$ is closed in $Y \times _ S Y$. Hence $Y$ is separated over $S$ by the discussion following Schemes, Definition 26.21.3. $\square$