hcju/stacksqa huggingface dataset:
Assume $S$ is a power series ring over $R$. Say $S = R[[x_1, \ldots, x_n]]$. Smoothness of $\underline{S}|_{\mathcal{C}_\Lambda} \to \underline{R}|_{\mathcal{C}_\Lambda}$ means the following (see Remark \ref{remark-compare-smooth-schlessinger}): Given a surjective ring map $B \to A$ in $\mathcal{C}_\Lambda$, a ring map $R \to B$, a ring map $S \to A$ such that the solid diagram $$ \xymatrix{ S \ar[r] \ar@{..>}[rd] & A \\ R \ar[u] \ar[r] & B \ar[u] } $$ is commutative then a dotted arrow exists making the diagram commute. (Note the similarity with Algebra, Definition \ref{algebra-definition-formally-smooth}.) To construct the dotted arrow choose elements $b_i \in B$ whose images in $A$ are equal to the images of $x_i$ in $A$. Note that $b_i \in \mathfrak m_B$ as $x_i$ maps to an element of $\mathfrak m_A$. Hence there is a unique $R$-algebra map $R[[x_1, \ldots, x_n]] \to B$ which maps $x_i$ to $b_i$ and which can serve as our dotted arrow. \medskip\noindent Conversely, assume $\underline{S}|_{\mathcal{C}_\Lambda} \to \underline{R}|_{\mathcal{C}_\Lambda}$ is smooth. Let $x_1, \ldots, x_n \in S$ be elements whose images form a basis in the relative cotangent space $\mathfrak m_S/(\mathfrak m_R S + \mathfrak m_S^2)$ of $S$ over $R$. Set $T = R[[X_1, \ldots, X_n]]$. Note that both $$ S/(\mathfrak m_R S + \mathfrak m_S^2) \cong R/\mathfrak m_R[x_1, \ldots, x_n]/(x_ix_j) $$ and $$ T/(\mathfrak m_R T + \mathfrak m_T^2) \cong R/\mathfrak m_R[X_1, \ldots, X_n]/(X_iX_j). $$ Let $S/(\mathfrak m_R S + \mathfrak m_S^2) \to T/(\mathfrak m_R T + \mathfrak m_T^2)$ be the local $R$-algebra isomorphism given by mapping the class of $x_i$ to the class of $X_i$. Let $f_1 : S \to T/(\mathfrak m_R T + \mathfrak m_T^2)$ be the composition $S \to S/(\mathfrak m_R S + \mathfrak m_S^2) \to T/(\mathfrak m_R T + \mathfrak m_T^2)$. The assumption that $\underline{S}|_{\mathcal{C}_\Lambda} \to \underline{R}|_{\mathcal{C}_\Lambda}$ is smooth means we can lift $f_1$ to a map $f_2 : S \to T/\mathfrak{m}_T^2$, then to a map $f_3 : S \to T/\mathfrak{m}_T^3$, and so on, for all $n \geq 1$. Thus we get an induced map $f : S \to T = \lim T/\mathfrak m_T^n$ of local $R$-algebras. By our choice of $f_1$, the map $f$ induces an isomorphism $\mathfrak m_S/(\mathfrak m_R S + \mathfrak m_S^2) \to \mathfrak m_T/(\mathfrak m_R T + \mathfrak m_T^2)$ of relative cotangent spaces. Hence $f$ is surjective by Lemma \ref{lemma-surjective-cotangent-space} (where we think of $f$ as a map in $\widehat{\mathcal{C}}_R$). Choose preimages $y_i \in S$ of $X_i \in T$ under $f$. As $T$ is a power series ring over $R$ there exists a local $R$-algebra homomorphism $s : T \to S$ mapping $X_i$ to $y_i$. By construction $f \circ s = \text{id}$. Then $s$ is injective. But $s$ induces an isomorphism on relative cotangent spaces since $f$ does, so it is also surjective by Lemma \ref{lemma-surjective-cotangent-space} again. Hence $s$ and $f$ are isomorphisms.
Statements (1) and (2) follow immediately from the definitions. Proof of (3) omitted. Hints: use the formulation of smoothness given in Remark \ref{remark-smoothness-2-categorical} and use that $\mathcal{F} \times_\mathcal{G} \mathcal{G}'$ is the $2$-fibre product, see Remarks \ref{remarks-cofibered-groupoids} (\ref{item-fibre-product}).
By assumption the morphism $\underline{\xi} : \underline{R}|_{\mathcal{C}_\Lambda} \to \mathcal{F}$ is smooth. Then $\iota(\xi) : \underline{R} \to \widehat{\mathcal{F}}$ is the completion of $\underline{\xi}$, see Remark \ref{remark-formal-objects-yoneda}. By Lemma \ref{lemma-smooth-morphism-essentially-surjective} there exists an object $f$ of $\underline{R}$ such that $\iota(\xi)(f) = \eta$. Then $f$ is a ring map $f : R \to S$ in $\widehat{\mathcal{C}}_\Lambda$. And $\iota(\xi)(f) = \eta$ means that $f_*\xi \cong \eta$ which means exactly that there is a morphism $\xi \to \eta$ lying over $f$.
Smoothness of $p : \underline{R}|_{\mathcal{C}_\Lambda} \to \mathcal{C}_\Lambda$ means that given $B \to A$ surjective in $\mathcal{C}_\Lambda$ and given $R \to A$ we can find the dotted arrow in the diagram $$ \xymatrix{ R \ar[r] \ar@{-->}[rd] & A \\ \Lambda \ar[r] \ar[u] & B \ar[u] } $$ This is certainly true if $\Lambda \to R$ is formally smooth in the $\mathfrak m_R$-adic topology, see More on Algebra, Definitions \ref{more-algebra-definition-formally-smooth-adic} and \ref{more-algebra-definition-formally-smooth}. Conversely, if this holds, then we see that $\Lambda \to R$ is formally smooth in the $\mathfrak m_R$-adic topology by More on Algebra, Lemma \ref{more-algebra-lemma-fs-local}. Thus (1) and (2) are equivalent. \medskip\noindent The equivalence of (2), (3), and (4) is More on Algebra, Proposition \ref{more-algebra-proposition-fs-flat-fibre-fs}. The equivalence with (5) follows for example from Lemma \ref{lemma-smooth-morphism-power-series} and the fact that $\mathcal{C}_\Lambda$ is the same as $\underline{\Lambda}|_{\mathcal{C}_\Lambda}$ in the classical case.
If (1) holds, i.e., if $\mathcal{F}$ is unobstructed, then the composition $$ \underline{R}|_{\mathcal{C}_\Lambda} \xrightarrow{\underline{\xi}} \mathcal{F} \to \mathcal{C}_\Lambda $$ is smooth, see Lemma \ref{lemma-smooth-properties}. Hence we see that (2) holds by Lemma \ref{lemma-smooth}. Conversely, if (2) holds, then the composition is smooth and moreover the first arrow is essentially surjective by Lemma \ref{lemma-versal-object-quasi-initial}. Hence we find that the second arrow is smooth by Lemma \ref{lemma-smooth-properties} which means that $\mathcal{F}$ is unobstructed by definition. The equivalence with (3) in the classical case follows from Lemma \ref{lemma-smooth}.
Note that $\text{id}_{A_1} \times g : A_1 \times_A A_2 \to A_1 \times_k k[I]$ is an isomorphism and that $k[I] \cong k[\epsilon]$. Hence we have a diagram $$ \vcenter{ \xymatrix{ y \ar[r] \ar[d] & g_*y \ar[d] \\ x_1 \ar[r]        & x_0 } } \quad\text{lying over}\quad \vcenter{ \xymatrix{ A_1 \times_k k[\epsilon] \ar[r] \ar[d] & k[\epsilon] \ar[d] \\ A_1 \ar[r] & k. } } $$ where $x_0$ is an object of $\mathcal{F}$ lying over $k$ (every object of $\mathcal{F}$ has a unique morphism to $x_0$, see discussion following Definition \ref{definition-predeformation-category}). If we have a morphism $x_1 \to g_*y$ then Lemma \ref{lemma-lifting-section} provides us with a section $s : x_1 \to y$ of the map $y \to x_1$. Composing this with the map $y \to x_2$ we obtain $b : x_1 \to x_2$ which has the property that $a_1 =  a_2 \circ b$ because the diagram of the lemma commutes and because $s$ is a section.
For $V \in \Ob(\text{Mod}^{fg}_K)$, the isomorphism $L(K) \otimes_K V \to L(V)$ is given on pure tensors by $x \otimes v \mapsto L(f_v)(x)$, where $f_v: K \to V$ is the $K$-linear map sending $1 \mapsto v$.  When $V = K$, this is the isomorphism $L(K) \otimes_K K \to L(K)$ given by multiplication by $K$. For general $V$, it is an isomorphism by the case $V = K$ and the fact that $L$ commutes with finite products (Remark \ref{remark-linear-functor}).
Note that $R \cong R[0]$ and $R[M] \times_R R[N] \cong R[M \times N]$ hence $R$ and $R[M] \times_R R[N]$ are objects of $\mathcal{C}$ by our assumptions on $\mathcal{C}$. Thus the conditions on $F$ make sense. The functor $\text{Mod}_R \to S\text{-Alg}/R$ of Lemma \ref{lemma-preserves-products} restricts to a functor $\text{Mod}^{fg}_R \to \mathcal{C}$ by the assumption on $\mathcal{C}$. Let $L$ be the composition $\text{Mod}^{fg}_R \to \mathcal{C} \to \textit{Sets}$, i.e., $L(M) = F(R[M])$. Then $L$ preserves finite products by Lemma \ref{lemma-preserves-products} and the assumption on $F$. Hence Lemma \ref{lemma-linear-functor} shows that $L(M) = F(R[M])$ has a natural $R$-module structure for any $M \in \Ob(\text{Mod}^{fg}_R)$.
Let us write $F = \overline{\mathcal{F}} : \mathcal{C}_\Lambda \to \textit{Sets}$. This is a predeformation functor and $F$ satisfies (S2). By Lemma \ref{lemma-S2-extensions} (and the translation of Remark \ref{remark-compare-S1-S2-schlessinger}) we see that $$ F(A \times_k k[V]) \longrightarrow F(A) \times F(k[V]) $$ is a bijection for every finite dimensional vector space $V$ and every $A \in \Ob(\mathcal{C}_\Lambda)$. In particular, if $A = k[W]$ then we see that $F(k[W] \times_k k[V]) = F(k[W]) \times F(k[V])$. In other words, the hypotheses of Lemma \ref{lemma-tangent-space-functor} hold and we see that $TF = T \mathcal{F}$ has a natural $k$-vector space structure. The final assertion follows from Lemma \ref{lemma-tangent-space-tensor}.
In the proof of Lemma \ref{lemma-tangent-space-vector-space} we have seen that the functor $V \mapsto \overline{\mathcal{F}}(k[V])$ transforms $0$ to a singleton and products to products. The same is true of the functor $V \mapsto \text{Der}_\Lambda(k, V)$. Hence $\gamma_V$ is linear by Lemma \ref{lemma-morphism-linear-functors}. Let $D : k \to V$ be a $\Lambda$-derivation. Set $D_1 : k \to V^{\oplus 2}$ equal to $a \mapsto (D(a), 0)$. Then $$ \xymatrix{ k[V \times V] \ar[r]_{+} \ar[d]^{f_{1, D_1}} & k[V] \ar[d]^{f_{1, D}} \\ k[V \times V] \ar[r]^{+} & k[V] } $$ commutes. Unwinding the definitions and using that $\overline{F}(V \times V) = \overline{F}(V) \times \overline{F}(V)$ this means that $a_D(x_1) + x_2 = a_D(x_1 + x_2)$ for all $x_1, x_2 \in \overline{F}(V)$. Thus it suffices to show that $a_V(D, 0) = 0 + \gamma_V(D)$ where $0 \in \overline{F}(V)$ is the zero vector. By definition this is the element $f_{0, *}(x_0)$. Since $f_D = f_{1, D} \circ f_0$ the desired result follows.
Let $\xi$ be a versal formal object of $\mathcal{F}$. Let $$ \xymatrix{            & x_2 \ar[d] \\ x_1 \ar[r] & x } $$ be a diagram in $\mathcal{F}$ such that $x_2 \to x$ lies over a surjective ring map. Since the natural morphism $\widehat{\mathcal{F}}|_{\mathcal{C}_\Lambda} \xrightarrow{\sim} \mathcal{F}$ is an equivalence (see Remark \ref{remark-restrict-completion}), we can consider this diagram also as a diagram in $\widehat{\mathcal{F}}$. By Lemma \ref{lemma-versal-object-quasi-initial} there exists a morphism $\xi \to x_1$, so by Remark \ref{remark-versal-object} we also get a morphism $\xi \to x_2$ making the diagram $$ \xymatrix{ \xi \ar[r] \ar[d]          & x_2 \ar[d] \\ x_1 \ar[r] & x } $$ commute. If $x_1 \to x$ and $x_2 \to x$ lie above ring maps $A_1 \to A$ and $A_2 \to A$ then taking the pushforward of $\xi$ to $A_1 \times_A A_2$ gives an object $y$ as required by (S1).
If $\xi$ is versal then (1) holds by Lemma \ref{lemma-smooth-morphism-essentially-surjective} and (2) holds by Remark \ref{remark-versal-object}. Assume (1) and (2) hold. By Remark \ref{remark-versal-object} we must show that given a diagram in $\widehat{\mathcal{F}}$ as in (2), there exists $\xi \to y$ such that $$ \xymatrix{             &  y \ar[d] \\ \xi \ar[ur] \ar[r]  &  x } $$ commutes. Let $b : R \to B$ be the map guaranteed by (2). Denote $y' = b_*\xi$ and choose a factorization $\xi \to y' \to x$ lying over $R \to B \to A$ of the given morphism $\xi \to x$. By (S1) we obtain a commutative diagram $$ \vcenter{ \xymatrix{ z  \ar[r] \ar[d]          &  y \ar[d] \\ y' \ar[r]  &  x } } \quad\text{lying over}\quad \vcenter{ \xymatrix{ B \times_A B \ar[d] \ar[r] &   B  \ar[d]^{f} \\ B \ar[r]^{f} &   A . } } $$ Set $I = \Ker(f)$. Let $\overline{g} : B \times_A B \to k[I]$ be the ring map $(u, v) \mapsto \overline{u} \oplus (v - u)$, cf.\ Lemma \ref{lemma-lifting-along-small-extension}. By (1) there exists a morphism $\xi \to \overline{g}_*z$ which lies over a ring map $i : R \to k[\epsilon]$. Choose an Artinian quotient $b_1 : R \to B_1$ such that both $b : R \to B$ and $i : R \to k[\epsilon]$ factor through $R \to B_1$, i.e., giving $h : B_1 \to B$ and $i' : B_1 \to k[\epsilon]$. Choose a pushforward $y_1 = b_{1, *}\xi$, a factorization $\xi \to y_1 \to y'$ lying over $R \to B_1 \to B$ of $\xi \to y'$, and a factorization $\xi \to y_1 \to \overline{g}_*z$ lying over $R \to B_1 \to k[\epsilon]$ of $\xi \to \overline{g}_*z$. Applying (S1) once more we obtain $$ \vcenter{ \xymatrix{ z_1  \ar[r] \ar[d] & z \ar[r] \ar[d] & y \ar[d] \\ y_1 \ar[r] & y' \ar[r] &  x } } \quad\text{lying over}\quad \vcenter{ \xymatrix{ B_1 \times_A B \ar[d] \ar[r] & B \times_A B \ar[r] \ar[d] & B \ar[d]^{f} \\ B_1 \ar[r]  & B \ar[r] &  A . } } $$ Note that the map $g : B_1 \times_A B \to k[I]$ of Lemma \ref{lemma-lifting-along-small-extension} (defined using $h$) is the composition of $B_1 \times_A B \to B \times_A B$ and the map $\overline{g}$ above. By construction there exists a morphism $y_1 \to g_*z_1 \cong \overline{g}_*z$! Hence Lemma \ref{lemma-lifting-along-small-extension} applies (to the outer rectangles in the diagrams above) to give a morphism $y_1 \to y$ and precomposing with $\xi \to y_1$ gives the desired morphism $\xi \to y$.
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 \ref{lemma-smallest-where-descends} (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 $R' = \lim_n R'_n$. Then $R' \to R$ is injective. \medskip\noindent 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 \ref{lemma-versal-object-quasi-initial}. 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 \ref{lemma-limit-artinian} implies $R'$ is Noetherian, i.e., an object of $\widehat{\mathcal{C}}_\Lambda$. By Lemma \ref{lemma-formal-objects-different-filtration} (plus the result on filtrations of Lemma \ref{lemma-limit-artinian}) the sequence of elements $\xi'_n$ defines a formal object $\xi'$ over $R'$ and we have a map $\xi' \to \xi$. \medskip\noindent 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 \ref{lemma-smallest-where-descends} (1). \medskip\noindent 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).
By Lemma \ref{lemma-versal-object-quasi-initial} there exists a morphism $\xi \to \xi'$. By Lemma \ref{lemma-smallest-where-descends-versal} the corresponding map $f : R \to R'$ induces a surjection $f|_{R'} : R' \to R'$. This is an isomorphism by Algebra, Lemma \ref{algebra-lemma-surjective-endo-noetherian-ring-is-iso}. Hence $I = \Ker(f)$ 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 $y \in \Ob(\mathcal{F}(B)$, resp.\ $x \in \Ob(\mathcal{F}(A))$ 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 $$ \vcenter{ \xymatrix{ & y \ar[d] \\ \xi \ar[r] \ar@{..>}[ru] & x } } \quad\text{lying over}\quad \vcenter{ \xymatrix{ & S/\mathfrak m_S^n \ar[d] \\ R \ar[r] \ar@{..>}[ru] & R/\mathfrak m_R^n } } $$ Because $\xi$ is versal, using Remark \ref{remark-versal-object} 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 \ref{lemma-power-series} that $S \to R$ is an isomorphism. \medskip\noindent As $\xi$ is a pushforward of $\xi'$ along $R' \to R$ we obtain from Remark \ref{remark-formal-objects-yoneda-map} a commutative diagram $$ \xymatrix{ \underline{R}|_{\mathcal{C}_\Lambda} \ar[rr] \ar[rd]_{\underline{\xi}} & & \underline{R'}|_{\mathcal{C}_\Lambda} \ar[ld]^{\underline{\xi'}} \\ & \mathcal{F} } $$ 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 \ref{lemma-smooth-properties} we see that $\underline{\xi'}$ is smooth, i.e., $\xi'$ is a versal formal object.
Suppose $\mathcal{F}$ has a versal formal object $\xi$ over $R$. Then it satisfies (S1), see Lemma \ref{lemma-versal-object-S1}. Let $\xi' \to \xi$ over $R' \subset R$ be any of the morphisms constructed in Lemma \ref{lemma-smallest-where-descends-versal}. By Lemma \ref{lemma-descends-versal} we see that $\xi'$ is versal, hence it is a minimal versal formal object (by construction). This proves (1). Also, $R \cong R'[[x_1, \ldots, x_n]]$ which proves (3). \medskip\noindent Suppose that $\xi_i/R_i$ are two minimal versal formal objects. By Lemma \ref{lemma-versal-object-quasi-initial} there exist morphisms $\xi_1 \to \xi_2$ and $\xi_2 \to \xi_1$. The corresponding ring maps $f : R_1 \to R_2$ and $g : R_2 \to R_1$ are surjective by minimality. Hence the compositions $g \circ f : R_1 \to R_1$ and $f \circ g : R_2 \to R_2$ are isomorphisms by Algebra, Lemma \ref{algebra-lemma-surjective-endo-noetherian-ring-is-iso}. Thus $f$ and $g$ are isomorphisms whence the maps $\xi_1 \to \xi_2$ and $\xi_2 \to \xi_1$ are isomorphisms (because $\widehat{\mathcal{F}}$ is cofibred in groupoids by Lemma \ref{lemma-completion-cofibred}). This proves (2) and finishes the proof of the lemma.
If $\xi$ is not minimal, then there exists a morphism $\xi' \to \xi$ lying over $R' \to R$ such that $R = R'[[x_1, \ldots, x_n]]$ with $n > 0$, see Lemma \ref{lemma-minimal-versal}. Thus $d\underline{\xi}$ factors as $$ \text{Der}_\Lambda(R, k) \to \text{Der}_\Lambda(R', k) \to T\mathcal{F} $$ and we see that (\ref{equation-bijective-orbits}) cannot hold because $D : f \mapsto \partial/\partial x_1(f) \bmod \mathfrak m_R$ is an element of the kernel of the first arrow which is not in the image of $\text{Der}_\Lambda(k, k) \to \text{Der}_\Lambda(R, k)$.
Condition (S1) holds by Lemma \ref{lemma-versal-object-S1}. The first part of (S2) holds since (S1) holds. Let $$ \vcenter{ \xymatrix{ y \ar[r]_c \ar[d]_a & x_\epsilon \ar[d]^e \\ x \ar[r]^d          & x_0 } } \quad\text{and}\quad \vcenter{ \xymatrix{ y' \ar[r]_{c'} \ar[d]_{a'} & x_\epsilon \ar[d]^e \\ x \ar[r]^d                 & x_0 } } \quad\text{lying over}\quad \vcenter{ \xymatrix{ A \times_k k[\epsilon] \ar[r] \ar[d] & k[\epsilon] \ar[d] \\ A  \ar[r] & k } } $$ be diagrams as in the second part of (S2). As above we can find morphisms $b : \xi \to y$ and $b' : \xi \to y'$ such that $$ \xymatrix{ \xi \ar[r]^{b'} \ar[d]_b          & y' \ar[d]^{a'} \\ y \ar[r]^{a} & x } $$ commutes. Let $p : \mathcal{F} \to \mathcal{C}_\Lambda$ denote the structure morphism. Say $\widehat{p}(\xi) = R$, i.e., $\xi$ lies over $R \in \Ob(\widehat{\mathcal{C}}_\Lambda)$. We see that the pushforward of $\xi$ via $p(c) \circ p(b)$ is $x_\epsilon$ and that the pushforward of $\xi$ via $p(c') \circ p(b')$ is $x_\epsilon$. Since $\xi$ satisfies (\ref{equation-bijective}), we see that $p(c) \circ p(b) = p(c') \circ p(b')$ as maps $R \to k[\epsilon]$. Hence $p(b) = p(b')$ as maps from $R \to A \times_k k[\epsilon]$. Thus we see that $y$ and $y'$ are isomorphic to the pushforward of $\xi$ along this map and we get a unique morphism $y \to y'$ over $A \times_k k[\epsilon]$ compatible with $b$ and $b'$ as desired. \medskip\noindent Finally, by Example \ref{example-tangent-space-prorepresentable-functor} we see $\dim_k T\mathcal{F} = \dim_k T\underline{R}|_{\mathcal{C}_\Lambda}$ is finite.
Let $\xi$ be a minimal versal formal object for $\mathcal{F}$, see Lemma \ref{lemma-minimal-versal}. Say $\xi$ lies over $R \in \Ob(\widehat{\mathcal{C}}_\Lambda)$. In order to parse (\ref{equation-bijective-orbits}) we point out that $T\mathcal{F}$ has a natural $k$-vector space structure (see Lemma \ref{lemma-tangent-space-vector-space}), that $d\underline{\xi} : \text{Der}_\Lambda(R, k) \to T\mathcal{F}$ is linear (see Lemma \ref{lemma-k-linear-differential}), and that the action of $\text{Der}_\Lambda(k, k)$ is given by addition (see Lemma \ref{lemma-action-linear}). Consider the diagram $$ \xymatrix{ & \Hom_k(\mathfrak m_R/\mathfrak m_R^2, k) \\ K \ar[r] & \text{Der}_\Lambda(R, k) \ar[r]^{d\underline{\xi}} \ar[u] & T\mathcal{F} \\ & \text{Der}_\Lambda(k, k) \ar[u] \ar[ru] } $$ The vector space $K$ is the kernel of $d\underline{\xi}$. Note that the middle column is exact in the middle as it is dual to the sequence (\ref{equation-sequence}). If (\ref{equation-bijective-orbits}) fails, then we can find a nonzero element $D \in K$ which does not map to zero in $\Hom_k(\mathfrak m_R/\mathfrak m_R^2, k)$. This means there exists an $t \in \mathfrak m_R$ such that $D(t) = 1$. Set $R' = \{a \in R \mid D(a) = 0\}$. As $D$ is a derivation this is a subring of $R$. Since $D(t) = 1$ we see that $R' \to k$ is surjective (compare with the proof of Lemma \ref{lemma-essential-surjection}). Note that $\mathfrak m_{R'} = \Ker(D : \mathfrak m_R \to k)$ is an ideal of $R$ and $\mathfrak m_R^2 \subset \mathfrak m_{R'}$. Hence $$ \mathfrak m_R/\mathfrak m_R^2 = \mathfrak m_{R'}/\mathfrak m_R^2 + k\overline{t} $$ which implies that the map $$ R'/\mathfrak m_R^2 \times_k k[\epsilon] \to R/\mathfrak m_R^2 $$ sending $\epsilon$ to $\overline{t}$ is an isomorphism. In particular there is a map $R/\mathfrak m_R^2 \to R'/\mathfrak m_R^2$. \medskip\noindent Let $\xi \to y$ be a morphism lying over $R \to R/\mathfrak m_R^2$. Let $y \to x$ be a morphism lying over $R/\mathfrak m_R^2 \to R'/\mathfrak m_R^2$. Let $y \to x_\epsilon$ be a morphism lying over $R/\mathfrak m_R^2 \to k[\epsilon]$. Let $x_0$ be the unique (up to unique isomorphism) object of $\mathcal{F}$ over $k$. By the axioms of a category cofibred in groupoids we obtain a commutative diagram $$ \vcenter{ \xymatrix{ y \ar[r] \ar[d] & x_\epsilon \ar[d] \\ x \ar[r]   & x_0 } } \quad\text{lying over}\quad \vcenter{ \xymatrix{ R'/\mathfrak m_R^2 \times_k k[\epsilon] \ar[r] \ar[d] & k[\epsilon] \ar[d] \\ R'/\mathfrak m_R^2 \ar[r] & k. } } $$ Because $D \in K$ we see that $x_\epsilon$ is isomorphic to $0 \in \mathcal{F}(k[\epsilon])$, i.e., $x_\epsilon$ is the pushforward of $x_0$ via $k \to k[\epsilon], a \mapsto a$. Hence by Lemma \ref{lemma-lifting-section} we see that there exists a morphism $x \to y$. Since $\text{length}_\Lambda(R'/\mathfrak m_R^2) < \text{length}_\Lambda(R/\mathfrak m_R^2)$ the corresponding ring map $R'/\mathfrak m_R^2 \to R/\mathfrak m_R^2$ is not surjective. This contradicts the minimality of $\xi/R$, see part (1) of Lemma \ref{lemma-smallest-where-descends-versal}. This contradiction shows that such a $D$ cannot exist, hence we win.