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.