This seems ridiculously simple, but it's eluding me.
Suppose $f:X\rightarrow Y$ is a morphism of affine schemes. Let $V$ be an open affine subscheme of $Y$. Why is $f^{-1}(V)$ affine?
I noted that $V$ is quasi-compact and wrote it as a finite union of principal open sets. Because the pullbacks of principal open sets are principal open sets, we can write $f^{-1}(V)$ as the union of such sets. But I'm not sure how to show this union is affine. I don't think this is the right way to go, because such unions are not affine in general.
I'm particularly perplexed because this occurs as an exercise in the chapter on separated morphisms and base change in Liu. Of course, all affine schemes are separated, but I don't see the relevance of that here.
Best Answer
The slogan I remember is, "open immersions are stable under base change". Suppose you have a diagram \[ \begin{matrix} & & X' \\ & & \downarrow {\scriptstyle f} \\ U & \xrightarrow{i} & X \end{matrix} \] and $i$ is an open immersion, in which case we might as well identify $U$ with an open subset of $X$. The fibered product exists and one realization of it is the open subscheme $f^{-1}(U)$ of $X'$. To show this you would just verify that it has the universal property. [You don't need to know that fibered products exist in general to do this.]
In the setting of the problem $X, X'$, and $U$ are affine, so you have another realization of $U \times_X X'$ as the spectrum of a ring. Hence there is an isomorphism between it and $f^{-1}(U)$.