Let $f:X \to Y$ be a morphism of separated schemes and $V,U$ be affine open subschemes of $X$ and $Y$ respectively. Why is the intersection $U \cap f^{-1}(V)$ affine? It is easy to prove that intersection of 2 affine schemes on separated scheme is affine, but $f^{-1}(V)$ could be not affine.
[Math] The intersection of affine subscheme and preimage of affine on separated scheme is affine
affine-schemesalgebraic-geometryschemes
Best Answer
First things first: since $X$ and $Y$ are separated, the product $X \times Y$ is also separated. I claim the morphism $\langle \mathrm{id}, f \rangle : X \to X \times Y$ is a closed immersion. Indeed, it even has a retraction (namely the projection $X \times Y \to X$), and any morphism into a separated scheme with a retraction is always a closed immersion. (Exercise.)
Now, consider the following diagram: $$\require{AMScd} \begin{CD} U \cap f^{-1} V @>>> X \\ @VVV @VV{\langle \mathrm{id}, f \rangle}V \\ U \times V @>>> X \times Y \end{CD}$$ It is a pullback diagram, so $U \cap f^{-1} V \to U \times V$ is also a closed immersion. Since $U$ and $V$ are affine, so too are $U \times V$ and $U \cap f^{-1} V$.
We could do all this over an arbitrary base scheme $S$, so long as we remember that "affine $S$-scheme" refers to an $S$-scheme whose structure morphism is affine, not an affine scheme with a morphism to $S$.