**Proof.**
Assume (1), (2), (3), and (4) hold. Choose an object $R \in \mathop{\mathrm{Ob}}\nolimits (\widehat{\mathcal{C}}_\Lambda )$ such that $\underline{R}|_{\mathcal{C}_\Lambda }$ is smooth. See Lemma 90.9.5. Let $r = \dim _ k T\mathcal{F}$ and put $S = R[[X_1, \ldots , X_ r]]$.

We are going to inductively construct for $n \geq 2$ pairs $(J_ n, f_{n - 1} : \xi _ n \to \xi _{n - 1})$ where $J_ n \subset S$ is an decreasing sequence of ideals and $f_{n - 1} : \xi _ n \to \xi _{n - 1}$ is a morphism of $\mathcal{F}$ lying over the projection $S/J_ n \to S/J_{n - 1}$.

Step 1. Let $J_1 = \mathfrak m_ S$. Let $\xi _1$ be the unique (up to unique isomorphism) object of $\mathcal{F}$ over $k = S/J_1 = S/\mathfrak m_ S$

Step 2. Let $J_2 = \mathfrak m_ S^2 + \mathfrak {m}_ R S$. Then $S/J_2 = k[V]$ with $V = kX_1 \oplus \ldots \oplus kX_ r$ By (S2) for $\overline{\mathcal{F}}$ we get a bijection

\[ \overline{\mathcal{F}}(S/J_2) \longrightarrow T\mathcal{F} \otimes _ k V, \]

see Lemmas 90.10.5 and 90.12.2. Choose a basis $\theta _1, \ldots , \theta _ r$ for $T\mathcal{F}$ and set $\xi _2 = \sum \theta _ i \otimes X_ i \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{F}(S/J_2))$. The point of this choice is that

\[ d\xi _2 : \mathop{\mathrm{Mor}}\nolimits _{\mathcal{C}_\Lambda }(S/J_2, k[\epsilon ]) \longrightarrow T\mathcal{F} \]

is surjective. Let $f_1 : \xi _2 \to \xi _1$ be the unique morphism.

Induction step. Assume $(J_ n, f_{n - 1} : \xi _ n \to \xi _{n - 1})$ has been constructed for some $n \geq 2$. There is a minimal element $J_{n + 1}$ of the set of ideals $J \subset S$ satisfying: (a) $\mathfrak m_ S J_ n \subset J \subset J_ n$ and (b) there exists a morphism $\xi _{n + 1} \to \xi _ n$ lying over $S/J \to S/J_ n$, see Lemma 90.13.3. Let $f_ n : \xi _{n + 1} \to \xi _ n$ be any morphism of $\mathcal{F}$ lying over $S/J_{n + 1} \to S/J_ n$.

Set $J = \bigcap J_ n$. Set $\overline{S} = S/J$. Set $\overline{J}_ n = J_ n/J$. By Lemma 90.4.7 the sequence of ideals $(\overline{J}_ n)$ induces the $\mathfrak m_{\overline{S}}$-adic topology on $\overline{S}$. Since $(\xi _ n, f_ n)$ is an object of $\widehat{\mathcal{F}}_\mathcal {I}(\overline{S})$, where $\mathcal{I}$ is the filtration $(\overline{J}_ n)$ of $\overline{S}$, we see that $(\xi _ n, f_ n)$ induces an object $\xi $ of $\widehat{\mathcal{F}}(\overline{S})$. see Lemma 90.7.4.

We prove $\xi $ is versal. For versality it suffices to check conditions (1) and (2) of Lemma 90.13.2. Condition (1) follows from our choice of $\xi _2$ in Step 2 above. Suppose given a diagram in $\widehat{\mathcal{F}}$

\[ \vcenter { \xymatrix{ & y \ar[d] \\ \xi \ar[r] & x } } \quad \text{lying over}\quad \vcenter { \xymatrix{ & B \ar[d]^{f} \\ \overline{S} \ar[r] & A } } \]

in $\widehat{\mathcal{C}}_\Lambda $ with $f: B \to A$ a small extension of Artinian rings. We have to show there is a map $\overline{S} \to B$ fitting into the diagram on the right. Choose $n$ such that $\overline{S} \to A$ factors through $\overline{S} \to S/J_ n$. This is possible as the sequence $(\overline{J}_ n)$ induces the $\mathfrak m_{\overline{S}}$-adic topology as we saw above. The pushforward of $\xi $ along $\overline{S} \to S/J_ n$ is $\xi _ n$. We may factor $\xi \to x$ as $\xi \to \xi _ n \to x$ hence we get a diagram in $\mathcal{F}$

\[ \vcenter { \xymatrix{ & y \ar[d] \\ \xi _ n \ar[r] & x } } \quad \text{lying over}\quad \vcenter { \xymatrix{ & B \ar[d]^{f} \\ S/J_ n \ar[r] & A . } } \]

To check condition (2) of Lemma 90.13.2 it suffices to complete the diagram

\[ \xymatrix{ S/J_{n + 1} \ar[d] \ar@{-->}[r] & B \ar[d]^{f} \\ S/J_ n \ar[r] & A } \]

or equivalently, to complete the diagram

\[ \xymatrix{ & S/J_ n \times _ A B \ar[d]^{p_1} \\ S/J_{n + 1} \ar@{-->}[ur] \ar[r] & S/J_ n. } \]

If $p_1$ has a section we are done. If not, by Lemma 90.3.8 (2) $p_1$ is a small extension, so by Lemma 90.3.12 (4) $p_1$ is an essential surjection. Recall that $S = R[[X_1, \ldots , X_ r]]$ and that we chose $R$ such that $\underline{R}|_{\mathcal{C}_\Lambda }$ is smooth. Hence there exists a map $h : R \to B$ lifting the map $R \to S \to S/J_ n \to A$. By the universal property of a power series ring there is an $R$-algebra map $h : S = R[[X_1, \ldots , X_ r]] \to B$ lifting the given map $S \to S/J_ n \to A$. This induces a map $g: S \to S/J_ n \times _ A B$ making the solid square in the diagram

\[ \xymatrix{ S \ar[d] \ar[r]_-g & S/J_ n \times _ A B \ar[d]^{p_1} \\ S/J_{n + 1} \ar@{-->}[ur] \ar[r] & S/J_ n } \]

commute. Then $g$ is a surjection since $p_1$ is an essential surjection. We claim the ideal $K = \mathop{\mathrm{Ker}}(g)$ of $S$ satisfies conditions (a) and (b) of the construction of $J_{n + 1}$ in the induction step above. Namely, $K \subset J_ n$ is clear and $\mathfrak m_ SJ_ n \subset K$ as $p_1$ is a small extension; this proves (a). By (S1) applied to

\[ \xymatrix{ & y \ar[d] \\ \xi _ n \ar[r] & x, } \]

there exists a lifting of $\xi _ n$ to $S/K \cong S/J_ n \times _ A B$, so (b) holds. Since $J_{n + 1}$ was the minimal ideal with properties (a) and (b) this implies $J_{n + 1} \subset K$. Thus the desired map $S/J_{n+1} \to S/K \cong S/J_ n \times _ A B$ exists.
$\square$

## Comments (3)

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