Lemma 32.8.15. Notation and assumptions as in Situation 32.8.1. If

1. $f$ is surjective, and

2. $f_0$ is locally of finite presentation,

then there exists an $i \geq 0$ such that $f_ i$ is surjective.

Proof. The morphism $f_0$ is of finite presentation. Hence $E = f_0(X_0)$ is a constructible subset of $Y_0$, see Morphisms, Lemma 29.22.2. Since $f_ i$ is the base change of $f_0$ by $Y_ i \to Y_0$ we see that the image of $f_ i$ is the inverse image of $E$ in $Y_ i$. Moreover, we know that $Y \to Y_0$ maps into $E$. Hence we win by Lemma 32.4.10. $\square$

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