Hartshorne exercise II.3.3(c) asks the reader to prove that if $f:X\to Y$ is finite type, then for any $\mathrm{Spec}A\subset Y$ and $\mathrm{Spec}B\subset X$ with $\mathrm{Spec}B\subset f^{-1} \mathrm{Spec}A$, we have $A\to B$ is of finite type.
However, the proof I produced seems to work in the case that $f$ is only locally of finite type, but exercise 2 of that section of Hartshorne asks to prove a strictly weaker condition in that case. (That is, for any affine of $Y$, the preimage can be covered by affines giving us finite-type homomorphisms.)
Is the property of local finite-typeness affine-local on the source? My proof seemed very straightforward:
Take any affine open $\mathrm{Spec}\ B$ of $X$, and also choose a cover $\mathrm{Spec}\ B_i$ of $X$ by affine opens so that each $A\to B_i$ is finite type. (This cover is guaranteed to exist by part II.3.2.(b).) Then by the "Affine Communication Lemma" (5.3.2 of Vakil's FOAG - page 155 in the Feb 25, 2013 version) it suffices to show that if $A\to B$ is finite type and $f\in B$, so is $A\to B_f$, and if $f_1,\dots,f_n\in B$ generate the unit and each $A\to B_{f_i}$ is finite type, then so is $A\to B$.
The first claim is trivial. As for the second, take $b\in B$, and choose $b_{i1},\dots,b_{ik}\in B$ which generate $B_{f_i}$ over $A$. Then we can choose $n$ and $g_{ij}\in A$ for each $i$ we have $f_i^nb=\sum_j g_{ij}b_{ij}$. Some linear combination of the $f_i^n$ is equal to one, and so we can write $b$ as an $A$-linear combination of all the $g_{ij}$.
Does this reasoning make sense?
