The Stacks project

Proof. By Lemma 90.8.11 there exists a morphism $\xi \to \xi '$. By Lemma 90.14.2 the corresponding map $f : R \to R'$ induces a surjection $f|_{R'} : R' \to R'$. This is an isomorphism by Algebra, Lemma 10.31.10. Hence $I = \mathop{\mathrm{Ker}}(f)$ is an ideal of $R$ such that $R = R' \oplus I$. Let $x_1, \ldots , x_ n \in I$ be elements which form a basis for $I/\mathfrak m_ RI$. Consider the map $S = R'[[X_1, \ldots , X_ r]] \to R$ mapping $X_ i$ to $x_ i$. For every $n \geq 1$ we get a surjection of Artinian $R'$-algebras $B = S/\mathfrak m_ S^ n \to R/\mathfrak m_ R^ n = A$. Denote $y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{F}(B)$, resp. $x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{F}(A))$ the pushforward of $\xi '$ along $R' \to S \to B$, resp. $R' \to S \to A$. Note that $x$ is also the pushforward of $\xi $ along $R \to A$ as $\xi $ is the pushforward of $\xi '$ along $R' \to R$. Thus we have a solid diagram

Because $\xi $ is versal, using Remark 90.8.10 we obtain the dotted arrows fitting into these diagrams. In particular, the maps $S/\mathfrak m_ S^ n \to R/\mathfrak m_ R^ n$ have sections $h_ n : R/\mathfrak m_ R^ n \to S/\mathfrak m_ S^ n$. It follows from Lemma 90.4.9 that $S \to R$ is an isomorphism.

As $\xi $ is a pushforward of $\xi '$ along $R' \to R$ we obtain from Remark 90.7.13 a commutative diagram

Since $R' \to R$ has a left inverse (namely $R \to R/I = R'$) we see that $\underline{R}|_{\mathcal{C}_\Lambda } \to \underline{R'}|_{\mathcal{C}_\Lambda }$ is essentially surjective. Hence by Lemma 90.8.7 we see that $\underline{\xi '}$ is smooth, i.e., $\xi '$ is a versal formal object. $\square$