I have a question about an argument from Bosch's "Algebraic Geometry and Commutative Algebra" (page 476):
The author reduced in the excerpt above that to verify that (ii) $=>$ (i) (so that $f$ is proper) it suffice to show that $f$ is a closed map since the conditions (ii) are invariant under base change.
So we take a closed $V \subset X$. Futhermore in the next reduction step the author showed that the composition
$$V \hookrightarrow X \xrightarrow{f} Y$$
fulfils lso (ii) so wlog $X=V$.
My question is why after this reduction the resulting morphism
$$f:X \to Y$$
is quasi compact?
My considerations: quasi compact means by definition that if $U \subset Y$ is an open quasi compact set then the preimage $f^{-1}(U)$ is quasi compact.
After reduction step we assume that $X=V$ is reduced so irreducible. But I'm not sure how conclude from this that $f$ is quasi compact.
Quasicompactness is part of the assumption on $f$ here, since $f$ is assumed to be of finite type. Recall that by definition, finite type means locally of finite type and quasicompact.