Let $f\colon X \to Y$ be a finite morphism of schemes. If $U$ is a quasi-compact open subset of $X$, is $f\colon U \to Y$ a quasi-finite morphism? Obviously it has finite fibers, but I cannot see why it is a quasi-compact morphism. Should we add the assumption that the open immersion $i\colon U \to X$ is quasi-compact? Thanks!
A quasi-finite morphism of schemes is a morphism $f: X \to Y$ such that:
The composite of two quasi-finite morphisms is quasi-finite (this is easy to see because each of the two above properties is preserved under composition). So if the open embedding $U \to X$ is quasi-compact, then it is of finite type, and thus quasi-finite (since the fibers are either a point or empty) and the composite $U \to X \to Y$ is thus quasi-finite.
If the inclusion $U \to X$ is not quasi-compact, then the composite need not be quasi-finite. For instance, take $f$ to be the identity map. Then the open immersion $U \to X$ itself is not quasi-finite.