A morphism of schemes $f:X\to S$ is said to be quasi-compact if for every OPEN quasi-compact subset $K \subset S$ the subset $f^{-1}(K) \subset X$ is also quasi-compact (and open, of course!). The morphism $f:X\to S$ is said to be universally closed if for every morphism $T\to S$ the resulting base-changed morphism $X\_T \to T$ is closed. The title question (inspired by topology) is then:


**Question 1: If $f:X\to S$ is universally closed, does it follow that $f$ is quasi-compact?**


Here is a variant of this question, asking for a stronger conclusion :


**Question 2: If $f:X\to S$ is universally closed, does it follow that for every quasi-compact subset $K\subset S$, open or not,
 $f^{-1}(K)$ is quasi-compact ?**


**REMARK 1** The converse of Question 1 is false: any morphism between affine schemes is quasi-compact but is not universally closed in general.


**REMARK 2** One might wonder whether $f$ proper implies $f$ quasi-compact. The answer is "yes" but for an irrelevant reason: proper is *defined* as separated, universally closed and of finite type. Since finite type already implies quasi-compact, proper obviously implies quasi-compact.


**REMARK 3** In topology "proper" is (or should be !) defined as universally closed; equivalently, closed with quasi-compact fibres. Topologically proper implies that every quasi-compact subset (open or not) of the codomain has quasi-compact inverse image. The converse is not true in general, but it is for locally compact spaces. 


**REMARK 4** (edited).As BCnrd remarks in his comment below, it is not at all clear that the two questions are equivalent (I had stated they were in the previous version of this post, but I retract that claim ). Also, beware that in topology the notion of quasi-compact continuous map is so weak as to be essentially useless since decent topological spaces, the ones algebraic geometers never use :) , have so few open quasi-compact subsets.



Is a universally closed morphism of schemes quasi-compact ?
Yes, a universally closed morphism is quasi-compact. (I haven't yet checked whether the same approach answers question 2.)


**Proof:** Without loss of generality, we may assume that $S=\operatorname{Spec} A$ for some ring $A$, and that $f$ is surjective. Suppose that $f$ is not quasi-compact. We need to show that $f$ is not universally closed.


Write $X = \bigcup\_{i \in I} X\_i$ where the $X\_i$ are affine open subschemes of $X$. Let $T=\operatorname{Spec} A[\{t\_i:i \in I\}]$, where the $t\_i$ are distinct indeterminates. Let $T\_i=D(t\_i) \subseteq T$. 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 $f\_T \colon X \times\_S T \to T$ is not closed. 


There exists a point $\mathfrak{p} \in \operatorname{Spec} A$ such that there is no neighborhood $U$ of $\mathfrak{p}$ in $S$ such that $X\_U$ is quasi-compact, since otherwise we could cover $S$ with finitely many such $U$ and prove that $X$ itself was quasi-compact. Fix such $\mathfrak{p}$, and let $k$ be its residue field.


First we check that $f\_T(Z\_k) \ne T\_k$. Let $\tau \in T(k)$ be the point such that $t\_i(\tau)=1$ for all $i$. Then $\tau \in T\_i$ for all $i$, and the fiber of $Z\_k \to T\_k$ above $\tau$ is isomorphic to $(X - \bigcup\_{i \in I} X\_i)\_k$, which is empty. Thus $\tau \in T\_k - f\_T(Z\_k)$.


If $f\_T(Z)$ were closed in $T$, there would exist a polynomial $g \in A[\{t\_i:i \in I\}]$ vanishing on $f\_T(Z)$ but not at $\tau$. Since $g(\tau) \ne 0$, some coefficient of $g$ would have nonzero image in $k$, and hence be invertible on some neighborhood $U$ of $\mathfrak{p}$. Let $J$ be the finite set of $j \in I$ such that $t\_j$ 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 $P \in T$ lying above $u$ such that $g(P) \ne 0$ and $t\_i(P)=0$ for all $i \notin J$. Then $P \notin T\_i$ for each $i \notin J$. A point $z$ of $X \times\_S T$ mapping to $x \in X$ and to $P \in T$ then belongs to $Z$. But $g(f\_T(z))=g(P) \ne 0$, so this contradicts the fact that $g$ vanishes on $f\_T(Z)$.


