The Stacks project

Proof. The final statement is clear from the equivalence of (1) and (2). It is also clear that (2) implies (1). Assume (1) holds. Let $S = \bigcup V_ i$ and $f^{-1}(V_ i) = \bigcup U_{ij}$ be affine open coverings as in Definition 37.58.1. Let $U \subset X$ and $V \subset S$ be as in (2). By More on Algebra, Lemma 15.80.8 it suffices to find a standard open covering $U = \bigcup U_ k$ of $U$ such that $\mathcal{F}(U_ k)$ is finitely presented relative to $\mathcal{O}_ S(V)$. In other words, for every $u \in U$ it suffices to find a standard affine open $u \in U' \subset U$ such that $\mathcal{F}(U')$ is finitely presented relative to $\mathcal{O}_ S(V)$. Pick $i$ such that $f(u) \in V_ i$ and then pick $j$ such that $u \in U_{ij}$. By Schemes, Lemma 26.11.5 we can find $v \in V' \subset V \cap V_ i$ which is standard affine open in $V'$ and $V_ i$. Then $f^{-1}V' \cap U$, resp. $f^{-1}V' \cap U_{ij}$ are standard affine opens of $U$, resp. $U_{ij}$. Applying the lemma again we can find $u \in U' \subset f^{-1}V' \cap U \cap U_{ij}$ which is standard affine open in both $f^{-1}V' \cap U$ and $f^{-1}V' \cap U_{ij}$. Thus $U'$ is also a standard affine open of $U$ and $U_{ij}$. By More on Algebra, Lemma 15.80.4 the assumption that $\mathcal{F}(U_{ij})$ is finitely presented relative to $\mathcal{O}_ S(V_ i)$ implies that $\mathcal{F}(U')$ is finitely presented relative to $\mathcal{O}_ S(V_ i)$. Since $\mathcal{O}_ X(U') = \mathcal{O}_ X(U') \otimes _{\mathcal{O}_ S(V_ i)} \mathcal{O}_ S(V')$ we see from More on Algebra, Lemma 15.80.5 that $\mathcal{F}(U')$ is finitely presented relative to $\mathcal{O}_ S(V')$. Applying More on Algebra, Lemma 15.80.4 again we conclude that $\mathcal{F}(U')$ is finitely presented relative to $\mathcal{O}_ S(V)$. This finishes the proof. $\square$