The Stacks project

Proof. The proof is exactly the same as the proof of the implication “(2) implies (1)” in Proposition 32.6.1. Choose any affine opens $U \subset X$ and $V \subset S$ such that $f(U) \subset V$. We have to show that $\mathcal{O}_ S(V) \to \mathcal{O}_ X(U)$ is of finite presentation. Let $(A_ i, \varphi _{ii'})$ be a directed system of $\mathcal{O}_ S(V)$-algebras. Set $A = \mathop{\mathrm{colim}}\nolimits _ i A_ i$. According to Algebra, Lemma 10.127.3 it suffices to show that

is surjective. Consider the schemes $T_ i = \mathop{\mathrm{Spec}}(A_ i)$. They form an inverse system of $V$-schemes over $I$ with transition morphisms $f_{ii'} : T_ i \to T_{i'}$ induced by the $\mathcal{O}_ S(V)$-algebra maps $\varphi _{i'i}$. Set $T := \mathop{\mathrm{Spec}}(A) = \mathop{\mathrm{lim}}\nolimits _ i T_ i$. The formula above becomes in terms of morphism sets of schemes

We first observe that $\mathop{\mathrm{Mor}}\nolimits _ V(T_ i, U) = \mathop{\mathrm{Mor}}\nolimits _ S(T_ i, U)$ and $\mathop{\mathrm{Mor}}\nolimits _ V(T, U) = \mathop{\mathrm{Mor}}\nolimits _ S(T, U)$. Hence we have to show that

is surjective and we are given that

is surjective. Hence it suffices to prove that given a morphism $g_ i : T_ i \to X$ over $S$ such that the composition $T \to T_ i \to X$ ends up in $U$ there exists some $i' \geq i$ such that the composition $g_{i'} : T_{i'} \to T_ i \to X$ ends up in $U$. Denote $Z_{i'} = g_{i'}^{-1}(X \setminus U)$. Assume each $Z_{i'}$ is nonempty to get a contradiction. By Lemma 32.4.8 there exists a point $t$ of $T$ which is mapped into $Z_{i'}$ for all $i' \geq i$. Such a point is not mapped into $U$. A contradiction. $\square$