Let $X$ and $Y$ be schemes and $f:X\to Y$ a morphism between them.
We say that $f$ is of finite type if for every affine open $spec(A)=U\subset Y$ there exists a finite affine (open) cover of $f^{-1}(U)$ by $\{ V_i\}_{i=1}^n$, where $V_i=spec(B_i)$, such that $B_i$ is a finitely generated $A$-algbera.
On the other hand $f$ is finite if $f^{-1}(U)$ is affine, say $f^{-1}(U)=spec(B)$, and $A$ is finite $B$-module.
I was wondering why don't we have the intermediate notion as following.
The map $f$ is such that for every affine $U=spec(A)$ we have $f^{-1}(U)=\bigcup_{i=1}^n V_i$ where $V_i=spec(B_i)$ and $B_i$ is finite $A$-module. Let's call the this property quasi finite type.
So the question is as follows: Suppose $f$ a morphism of schemes was of quasi finite type then would it imply that it is actually finite.
If yes, then there is no need of such a definition.
If no, is it just because we don't see this kind of property in "nature" and hence don't need to define it.
The notion of a quasi-finite morphism already exists: it is a morphism of finite type with finite fibers.
Zariski's main theorem, in its Grothendieck's version, asserts that a quasi-finite morphism decomposes as an open immersion into a finite scheme.