# Why are morphisms of finite type defined in the usual way?

We say that a morphism of schemes $$f : Y \to X$$ is finite if $$X$$ may be covered by affine open sets $$\text{Spec}(B)$$ such that each $$f^{-1}(\text{Spec}(B))$$ is affine, say of the form $$\text{Spec}(A)$$, where $$A$$ is finitely-generated as a $$B$$-module.

It seems like the natural analogue would then be to say that a morphism of schemes $$f : Y \to X$$ is of finite type if $$X$$ may be covered by affine open sets $$\text{Spec}(B)$$ such that each $$f^{-1}(\text{Spec}(B))$$ is affine, say of the form $$\text{Spec}(A)$$, where $$A$$ is finitely-generated as a $$B$$-algebra.

However, the usual definition is that $$f : Y \to X$$ is locally of finite type if $$X$$ may be covered by affine open sets $$\text{Spec}(B)$$ such that each $$f^{-1}(\text{Spec}(B))$$ is covered by affine open sets $$\text{Spec}(A)$$ where each $$A$$ is finitely-generated as a $$B$$-algebra. Then we say that $$f : Y \to X$$ is of finite type if each $$\text{Spec}(B)$$ in the previous definition can be covered by finitely many $$\text{Spec}(A)$$'s with $$A$$ finitely-generated as a $$B$$-algebra.

I therefore have two questions.

Question 1: Why can't we simply impose the condition that each $$f^{-1}(\text{Spec}(B))$$ is an affine open set in the definition of finite type?

Question 2: Why is it worthwhile to distinguish between morphisms that are "locally of finite type" and "of finite type"? In particular, what is an example of a morphism that is locally of finite type but not of finite type?

For question one, consider the map from $$\mathbb{P}^1$$ to a point. This is locally of finite type (and of finite type) but there's no way to cover the base with affines such that the pre-images are affine.