**Proof.**
Write $\xi = (R, \xi _ n, f_ n)$. Set $R'_1 = k$ and $\xi '_1 = \xi _1$. Suppose that we have constructed minimal objects $\xi '_ m \to \xi _ m$ of $\mathcal{S}_{\xi _ m}$ lying over $R'_ m \subset R/\mathfrak m_ R^ m$ for $m \leq n$ and morphisms $f'_ m : \xi '_{m + 1} \to \xi '_ m$ compatible with $f_ m$ for $m \leq n - 1$. By Lemma 90.14.1 (2) there exists a minimal object $\xi '_{n + 1} \to \xi _{n + 1}$ lying over $R'_{n + 1} \subset R/\mathfrak m_ R^{n + 1}$ whose image is $\xi '_ n \to \xi _ n$ over $R'_ n \subset R/\mathfrak m_ R^ n$. This produces the commutative diagram

\[ \xymatrix{ \xi '_{n + 1} \ar[r]_{f'_ n} \ar[d] & \xi '_ n \ar[d] \\ \xi _{n + 1} \ar[r]^{f_ n} & \xi _ n } \]

by construction. Moreover the ring map $R'_{n + 1} \to R'_ n$ is surjective. Set $R' = \mathop{\mathrm{lim}}\nolimits _ n R'_ n$. Then $R' \to R$ is injective.

However, it isn't a priori clear that $R'$ is Noetherian. To prove this we use that $\xi $ is versal. Namely, versality implies that there exists a morphism $\xi \to \xi '_ n$ in $\widehat{\mathcal{F}}$, see Lemma 90.8.11. The corresponding map $R \to R'_ n$ has to be surjective (as $\xi '_ n \to \xi _ n$ is minimal in $\mathcal{S}_{\xi _ n}$). Thus the dimensions of the cotangent spaces are bounded and Lemma 90.4.8 implies $R'$ is Noetherian, i.e., an object of $\widehat{\mathcal{C}}_\Lambda $. By Lemma 90.7.4 (plus the result on filtrations of Lemma 90.4.8) the sequence of elements $\xi '_ n$ defines a formal object $\xi '$ over $R'$ and we have a map $\xi ' \to \xi $.

By construction (1) holds for $R \to R/\mathfrak m_ R^ n$ for each $n$. Since each $R \to A$ as in (1) factors through $R \to R/\mathfrak m_ R^ n \to A$ we see that (1) for $x' \to x$ over $f(R) \subset A$ follows from the minimality of $\xi '_ n \to \xi _ n$ over $R'_ n \to R/\mathfrak m_ R^ n$ by Lemma 90.14.1 (1).

If $R'' \to R'$ as in (2) is not surjective, then $R'' \to R' \to R'_ n$ would not be surjective for some $n$ and $\xi '_ n \to \xi _ n$ wouldn't be minimal, a contradiction. This contradiction proves (2).
$\square$

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