Let $f:X \to Y$ be a faithfully flat morphism of schemes. If necessary, one can put moderate mild assumptions on these as they are locally noetherian and the map is locally of finite type.
It occurred to me the necessity of a statement as follows: there exists AFFINE open coverings $Y_i$ of $Y$ and $X_i$ of $X$ such that $f(X_i) \subseteq Y_i$ and $f|_{X_i}: X_i \to Y_i$ is faithfully flat. To obtain that, one should prove in particular that the maps $ f|_{X_i}$ are surjective. I wasn't able to ind such a statement in the literature neither to prove this. Is this even true?
Best Answer
The general answer is "no". Depending on what you are trying to do exactly, you may want to require $f$ to be affine, or quasi-compact. If $f$ is affine, then for any affine open $Y'\subset Y$, the preimage $X'$ of $Y'$ in $X$ is an affine open subscheme of $X$ and the canonical map $X'\to Y'$ is faithfully flat. If $f$ is quasi-compact, then $X'$ can be covered by finitely many affine open subschemes of $X$, and the map from their disjoint union to $Y'$ is a faithfully flat morphism of affine schemes.
For a counterexample to the general statement, consider $Y=\operatorname{Spec} \mathbb Z$. For each closed point $y$ of $Y$, let $X_y$ be the localization of $Y$ at $y$. Call $X$ the disjoint union of the $X_y$ and $f$ the natural map $X \to Y$. Then, $f$ is faithfully flat, but there exist no nonempty affine open subschemes $Y'\subset Y$ and $X'\subset X$ such that $f(X')=Y'$ set-theoretically.
To see this, suppose given such $X',Y'$. Since $Y'$ contains infinitely many closed points of $Y$, we know that $X'$ must contain infinitely many of the $X_y$, so $X'$ is not quasi-compact and in particular not affine: we have a contradiction.