The Stacks project

Proof. Let $\mathcal{L}$ be an $f$-ample invertible module on $X$. This in particular implies that $f$ is quasi-compact, see Morphisms, Definition 29.37.1. Since $Y$ is quasi-compact by Morphisms, Definition 29.12.1 we see that $X$ is quasi-compact (and hence $X$ itself satisfies the first condition of Morphisms, Definition 29.12.1). Let $x \in X$ with image $y \in Y$. By assumption (2) we can find an invertible $\mathcal{O}_ Y$-module $\mathcal{N}$ and a section $t \in \Gamma (Y, \mathcal{N})$ such that the locus $Y_ t$ where $t$ does not vanish is affine. Then $\mathcal{L}$ is ample over $f^{-1}(Y_ t) = X_{f^*t}$ and hence we can find a section $s \in \Gamma (X_{f^*t}, \mathcal{L})$ such that $(X_{f^*t})_ s$ is affine and contains $x$. By Properties, Lemma 28.17.2 for some $n \geq 0$ the product $(f^*t)^ n s$ extends to a section $s' \in \Gamma (X, f^*\mathcal{N}^{\otimes n} \otimes \mathcal{L})$. Then finally the section $s'' = f^* ts'$ of $f^*\mathcal{N}^{\otimes n + 1} \otimes \mathcal{L}$ vanishes at every point of $X \setminus X_{f^*t}$ hence we see that $X_{s''} = (X_{f^*t})_ s$ is affine as desired. $\square$

Proof. By Morphisms, Lemma 29.37.6 this is a special case of Lemma 37.79.1. $\square$