The Stacks project

Proof. Let $B \to A$ be a surjective ring map in $\mathcal{C}_\Lambda $. Let $y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{G}(B))$, $x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{F}(A))$, and $y \to \varphi (x)$ be a morphism lying over $B \to A$. By Lemma 90.3.3 we can factor $B \to A$ into small extensions $B = B_ n \to B_{n-1} \to \ldots \to B_0 = A$. We argue by induction on $n$. If $n = 1$ the result is true by assumption. If $n > 1$, then denote $f : B = B_ n \to B_{n - 1}$ and denote $g : B_{n - 1} \to B_0 = A$. Choose a pushforward $y \to f_* y$ of $y$ along $f$, so that the morphism $y \to \varphi (x)$ factors as $y \to f_* y \to \varphi (x)$. By the induction hypothesis we can find $x_{n - 1} \to x$ lying over $g : B_{n - 1} \to A$ and $a : \varphi (x_{n - 1}) \to f_*y$ lying over $\text{id} : B_{n - 1} \to B_{n - 1}$ such that

commutes. We can apply the assumption to the composition $y \to \varphi (x_{n - 1})$ of $y \to f_*y$ with $a^{-1} : f_*y \to \varphi (x_{n - 1})$. We obtain $x_ n \to x_{n - 1}$ lying over $B_ n \to B_{n - 1}$ and $\varphi (x_ n) \to y$ lying over $\text{id} : B_ n \to B_ n$ so that the diagram

commutes. Then the composition $x_ n \to x_{n - 1} \to x$ and $\varphi (x_ n) \to y$ are the morphisms required by the definition of smoothness. $\square$