The Stacks project

Proof. Since $Y_0$ is quasi-compact and quasi-separated, the scheme $Y$, which is affine over $Y_0$, is quasi-compact and quasi-separated. Choose $n \geq 0$ and $Z_ i \to Y$, $i = 1, \ldots , n$ as in Lemma 37.78.4. Denote $\sigma _ i : Z_ i \to X$ morphisms over $Y$ which exist by our choice of $Z_ i$. After increasing $0 \in \Lambda $ we may assume there exist morphisms $Z_{i, 0} \to Y_0$ of finite presentation whose base changes to $S$ are the morphisms $Z_ i \to Y$, see Limits, Lemma 32.10.1. By Limits, Lemma 32.8.13 we may assume, after possibly increasing $0$, that $Z_{i, 0} \to Y_0$ is an immersion. Since $\coprod Z_ i \to Y$ is surjective, we may assume, after possibly increasing $0$, that $\coprod Z_{i, 0} \to Y_0$ is surjective, see Limits, Lemma 32.8.15. Observe that $Z_ i = \mathop{\mathrm{lim}}\nolimits _{\lambda \geq 0} Z_{i, \lambda }$ where $Z_{i, \lambda } = Y_\lambda \times _{Y_0} Z_{i, 0}$. Let us view the compositions

as morphisms over $Y_0$. Since $f_0$ is locally of finite presentation by Limits, Proposition 32.6.1 we can find a $\lambda \geq 0$ such that there exist morphisms $\sigma '_{i, \lambda } : Z_{i, \lambda } \to X_0$ over $Y_0$ whose precomposition with $Z_ i \to Z_{i, \lambda }$ are the displayed arrows. Of course, then $\sigma '_{i, \lambda }$ determines a morphism $\sigma _{i, \lambda } : Z_{i, \lambda } \to X_\lambda = X_0 \times _{Y_0} Y_\lambda $ over $Y_\lambda $. Since $\coprod Z_{i, \lambda } \to Y_\lambda $ is surjective we conclude that $X_\lambda \to Y_\lambda $ is completely decomposed. $\square$