Quasi-compact morphism

Remarks

It is enough to check this for the sets $V_i$ of one cover of $Y$ by open affine subschemes. Every affine scheme is quasi-compact, by quasi-compactness of the spectrum. Example: every closed immersion $f \colon Z \to Y$ is quasi-compact. For $V$ open affine, $f^{-1}(V)$ is homeomorphic to a closed subset of the quasi-compact space $V$, and closed subsets of quasi-compact spaces are quasi-compact. Example: if $X$ is a Noetherian scheme, every morphism $X \to Y$ is quasi-compact, since every subset of a Noetherian space is quasi-compact. Non-example: let $X$ be the disjoint union of infinitely many copies of $\operatorname{Spec} k$ for a field $k$, and $f \colon X \to \operatorname{Spec} k$ the obvious map. Then $f^{-1}(\operatorname{Spec} k) = X$ is covered by the copies, with no finite subcover, so $f$ is not quasi-compact.