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A finite morphism $f:X\rightarrow Y$ firstly requires an affine cover $V_i\subset Y$, such that $f^{-1}(V_i)$ are affine open sets. However a morphism of (locally) finite type $f:X\rightarrow Y$ involves a cover $V_i\subset Y$ such that $f^{-1}(V_i)$ can be covered by affine open sets.

The latter sense of local must be broader than the first one. Why are the first part of their definition so different? What if I define a finite morphism by requiring the preimage covered by affine open sets, such that the induced $f^*$ are integral, will undesirable things happen? (BTW, is it right that there's actually no difference when $X$ is affine?)

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up vote 2 down vote accepted

A finite morphism $f : X\to Y$ has a strong topological property: it is proper ($f$ is a closed map, and even universally closed : for any $Z\to Y$, the projection $X\times_Y Z\to Z$ is closed). If you only ask $f^{-1}(V_i)$ to be "locally finite", then you only get so called quasi-finite morphisms. For example, any open immersion is quasi-finite, but not finite.

If $X$ is affine and $Y$ is separated, then it is true that $f^{-1}(V)$ is affine for any affine open subset $V$ of $Y$.

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Dear QiL, If you actually ask that $f^{-1}(V)$ be covered by finitely many affine open sets that are finite over $V$, for an open affine cover $\{V\}$ of $Y$, then $f$ is still universally closed, as well as quasi-finite. So if $f$ is also separated, then $f$ is actually automatically finite. See this answer. Best wishes, – Matt E May 10 '13 at 6:46
Dear @MattE: what you say is certainly correct, but to be honest, I no longer understand the original question.Best wishes, – user18119 May 26 '13 at 19:44

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