hcju/stacksqa huggingface dataset:
The subcategory $\text{Mod}^{fg}_{A, torsion}$ is a Serre subcategory of $\text{Mod}^{fg}_A$, see Homology, Definition \ref{homology-definition-serre-subcategory}. This is clear from the description of objects given above but it also follows from More on Algebra, Lemma \ref{more-algebra-lemma-I-power-torsion}. Hence the quotient category on the left of the arrow is defined in Homology, Lemma \ref{homology-lemma-serre-subcategory-is-kernel}. To define the functor of the proposition, it suffices to show that the functor $M \mapsto \widetilde M$ sends torsion modules to $0$. This is clear because for any $f \in A_+$ homogeneous the module $M_f$ is zero and hence the value $M_{(f)}$ of $\widetilde M$ on $D_+(f)$ is zero too. \medskip\noindent By Lemma \ref{lemma-coherent-on-proj} the proposed quasi-inverse makes sense. Namely, the lemma shows that $\mathcal{F} \longmapsto \bigoplus_{n \geq 0} \Gamma(X, \mathcal{F}(n))$ is a functor $\textit{Coh}(\mathcal{O}_X) \to \text{Mod}^{fg}_A$ which we can compose with the quotient functor $\text{Mod}^{fg}_A \to \text{Mod}^{fg}_A/\text{Mod}^{fg}_{A, torsion}$. \medskip\noindent By Lemma \ref{lemma-recover-tail-graded-module} the composite left to right to left is isomorphic to the identity functor. \medskip\noindent Finally, let $\mathcal{F}$ be a coherent $\mathcal{O}_X$-module. Set $M = \bigoplus_{n \in \mathbf{Z}} \Gamma(X, \mathcal{F}(n))$ viewed as a graded $A$-module, so that our functor sends $\mathcal{F}$ to $M_{\geq 0} = \bigoplus_{n \geq 0} M_n$. By Properties, Lemma \ref{properties-lemma-proj-quasi-coherent} the canonical map $\widetilde M \to \mathcal{F}$ is an isomorphism. Since the inclusion map $M_{\geq 0} \to M$ defines an isomorphism $\widetilde{M_{\geq 0}} \to \widetilde M$ we conclude that the composite right to left to right is isomorphic to the identity functor as well.
We have already seen that (\ref{equation-completion-functor}) is fully faithful in Lemma \ref{lemma-fully-faithful}. Thus it suffices to show that the functor is essentially surjective. \medskip\noindent Consider the collection $\Xi$ of quasi-coherent sheaves of ideals $\mathcal{K} \subset \mathcal{O}_X$ such that every object $(\mathcal{F}_n)$ annihilated by $\mathcal{K}$ is in the essential image. We want to show $(0)$ is in $\Xi$. If not, then since $X$ is Noetherian there exists a maximal quasi-coherent sheaf of ideals $\mathcal{K}$ not in $\Xi$, see Lemma \ref{lemma-acc-coherent}. After replacing $X$ by the closed subscheme of $X$ corresponding to $\mathcal{K}$ we may assume that every nonzero $\mathcal{K}$ is in $\Xi$. (This uses the correspondence by coherent modules annihilated by $\mathcal{K}$ and coherent modules on the closed subscheme corresponding to $\mathcal{K}$, see Lemma \ref{lemma-i-star-equivalence}.) Let $(\mathcal{F}_n)$ be an object of $\textit{Coh}(X, \mathcal{I})$. We will show that this object is in the essential image of the functor (\ref{equation-completion-functor}), thereby completion the proof of the proposition. \medskip\noindent Apply Chow's lemma (Lemma \ref{lemma-chow-Noetherian}) to find a proper surjective morphism $f : X' \to X$ which is an isomorphism over a dense open $U \subset X$ such that $X'$ is projective over $A$. Let $\mathcal{K}$ be the quasi-coherent sheaf of ideals cutting out the reduced complement $X \setminus U$. By the projective case of Grothendieck's existence theorem (Lemma \ref{lemma-existence-projective}) there exists a coherent module $\mathcal{F}'$ on $X'$ such that $(\mathcal{F}')^\wedge \cong (f^*\mathcal{F}_n)$. By Proposition \ref{proposition-proper-pushforward-coherent} the $\mathcal{O}_X$-module $\mathcal{H} = f_*\mathcal{F}'$ is coherent and by Lemma \ref{lemma-inverse-systems-push-pull} there exists a morphism $(\mathcal{F}_n) \to \mathcal{H}^\wedge$ of $\textit{Coh}(X, \mathcal{I})$ whose kernel and cokernel are annihilated by a power of $\mathcal{K}$. The powers $\mathcal{K}^e$ are all in $\Xi$ so that (\ref{equation-completion-functor}) is an equivalence for the closed subschemes $X_e = V(\mathcal{K}^e)$. We conclude by Lemma \ref{lemma-existence-tricky}.
Lemma \ref{lemma-miniversal-object-existence-1} shows that (1) $\Rightarrow$ (3). Lemmas \ref{lemma-versal-object-existence} and \ref{lemma-construct-bijective-orbits} show that (3) $\Rightarrow$ (2). If $k' \subset k$ is separable then $\text{Der}_\Lambda(k, k) = 0$ and we see that (\ref{equation-bijective}) $=$ (\ref{equation-bijective-orbits}), i.e., (1) is the same as (2). \medskip\noindent An alternative proof of (3) $\Rightarrow$ (1) in the classical case is to add a few words to the proof of Lemma \ref{lemma-versal-object-existence} to see that one can right away construct a versal object which satisfies (\ref{equation-bijective}) in this case. This avoids the use of Lemma \ref{lemma-versal-object-existence} in the classical case. Details omitted.
Assume $F$ is prorepresentable by $R \in \widehat{\mathcal{C}}_\Lambda$. We see $F$ is a deformation functor by Example \ref{example-prorepresentable-deformation-functor}. We see $\dim_k TF$ is finite by Example \ref{example-tangent-space-prorepresentable-functor}. Finally, $\text{Der}_\Lambda(k, k) \to TF$ is identified with $\text{Der}_\Lambda(k, k) \to \text{Der}_\Lambda(R, k)$ by Example \ref{example-tangent-space-map-prorepresentable-functor} which is injective because $R \to k$ is surjective. \medskip\noindent Conversely, assume (a), (b), and (c) hold. By Lemma \ref{lemma-RS-implies-S1-S2} we see that (S1) and (S2) hold. Hence by Theorem \ref{theorem-miniversal-object-existence} there exists a minimal versal formal object $\xi$ of $F$ such that (\ref{equation-bijective-orbits}) holds. Say $\xi$ lies over $R$. The map $$ d\underline{\xi} : \text{Der}_\Lambda(R, k) \to T\mathcal{F} $$ is bijective on $\text{Der}_\Lambda(k, k)$-orbits. Since the action of $\text{Der}_\Lambda(k, k)$ on the left hand side is free by (c) and Lemma \ref{lemma-action-linear} we see that the map is bijective. Thus we see that $\underline{\xi}$ is an isomorphism by Lemma \ref{lemma-minimal-smooth-morphism-functors}.
Recall that a prorepresentable functor is a deformation functor, see Example \ref{example-prorepresentable-deformation-functor}. Thus if $\mathcal{F}$ is equivalent to a smooth prorepresentable groupoid in functors, then conditions (1), (2), and (3) follow from Lemma \ref{lemma-deformation-groupoid-quotient} (1), (2), and (3). \medskip\noindent Conversely, assume conditions (1), (2), and (3) hold. Condition (1) implies that (S1) and (S2) are satisfied, see Lemma \ref{lemma-RS-implies-S1-S2}. By Lemma \ref{lemma-versal-object-existence} there exists a versal formal object $\xi$. Setting $U = \underline{R}|_{\mathcal{C}_\Lambda}$ the associated map $\underline{\xi} : U \to \mathcal{F}$ is smooth (this is the definition of a versal formal object). Let $(U, R, s, t, c)$ be the groupoid in functors constructed in Lemma \ref{lemma-presentation-construction} from the map $\underline{\xi}$. By Lemma \ref{lemma-smooth-groupoid-in-functors-construction} we see that $(U, R, s, t, c)$ is a smooth groupoid in functors and that $[U/R] \to \mathcal{F}$ is an equivalence. By Lemma \ref{lemma-prorepresentable-groupoid-in-functors-construction} we see that $(U, R, s, t, c)$ is prorepresentable. Hence $[U/R] \to \mathcal{F}$ is the desired presentation of $\mathcal{F}$.
We see that (1) implies (3) and is equivalent to (3) if $k' \subset k$ is separable from Lemma \ref{lemma-characterize-minimal-groupoid-in-functors}. It is clear that (3) implies (2). We see that (2) implies (4) by Theorem \ref{theorem-presentation-deformation-groupoid}. We see that (4) implies (3) by Lemma \ref{lemma-minimal-groupoid-in-functors-construction}. This proves all the implications. The final uniqueness statement follows from Lemma \ref{lemma-minimal-presentations-equivalent}.
Let $I$ be the kernel of $f$.  The maximal ideal $\mathfrak{m}_B$ is nilpotent since $B$ is Artinian, say $\mathfrak{m}_B^n = 0$. Hence we get a factorization $$ B = B/I\mathfrak{m}_B^{n-1} \to B/I\mathfrak{m}_B^{n-2} \to \ldots \to B/I \cong A $$ of $f$ into a composition of surjective maps whose kernels are annihilated by the maximal ideal.  Thus it suffices to prove the lemma when $f$ itself is such a map, i.e.\ when $I$ is annihilated by $\mathfrak{m}_B$. In this case $I$ is a $k$-vector space, which has finite dimension, see Algebra, Lemma \ref{algebra-lemma-artinian-finite-length}. Take a basis $x_1, \ldots, x_n$ of $I$ as a $k$-vector space to get a factorization $$ B \to B/(x_1) \to \ldots \to  B/(x_1, \ldots, x_n) \cong  A $$ of $f$ into a composition of small extensions.
If $M$ is a simple $A$-module then $M \cong k$ as an $A$-module, see Algebra, Lemma \ref{algebra-lemma-characterize-length-1}. In this case $\text{length}_A(M) = 1$ and $\text{length}_\Lambda(M) = [k' : k]$, see Algebra, Lemma \ref{algebra-lemma-dimension-is-length}. If $\text{length}_A(M)$ is finite, then the result follows on choosing a filtration of $M$ by $A$-submodules with simple quotients using additivity, see Algebra, Lemma \ref{algebra-lemma-length-additive}. If $\text{length}_A(M)$ is infinite, the result follows from the obvious inequality $\text{length}_A(M) \leq \text{length}_\Lambda(M)$.
For any ring map $f : A \to B$ in $\mathcal{C}_\Lambda$ we have $f(\mathfrak m_A) \subset \mathfrak m_B$ for example because $\mathfrak m_A$, $\mathfrak m_B$ is the set of nilpotent elements of $A$, $B$. Suppose $f$ is surjective. Let $y \in \mathfrak m_B$. Choose $x \in A$ with $f(x) = y$. Since $f$ induces an isomorphism $A/\mathfrak m_A \to B/\mathfrak m_B$ we see that $x \in \mathfrak m_A$. Hence the induced map $\mathfrak m_A/\mathfrak m_A^2 \to \mathfrak m_B/\mathfrak m_B^2$ is surjective. In this way we see that (1) implies (2). \medskip\noindent It is clear that (2) implies (3). The map $A \to B$ gives rise to a canonical commutative diagram $$ \xymatrix{ \mathfrak m_\Lambda/\mathfrak m_\Lambda^2 \otimes_{k'} k \ar[r] \ar[d] & \mathfrak m_A/\mathfrak m_A^2 \ar[r] \ar[d] & \mathfrak m_A/(\mathfrak m_\Lambda A + \mathfrak m_A^2) \ar[r] \ar[d] & 0 \\ \mathfrak m_\Lambda/\mathfrak m_\Lambda^2 \otimes_{k'} k \ar[r] & \mathfrak m_B/\mathfrak m_B^2 \ar[r] & \mathfrak m_B/(\mathfrak m_\Lambda B + \mathfrak m_B^2) \ar[r] & 0 } $$ with exact rows. Hence if (3) holds, then so does (2). \medskip\noindent Assume (2). To show that $A \to B$ is surjective it suffices by Nakayama's lemma (Algebra, Lemma \ref{algebra-lemma-NAK}) to show that $A/\mathfrak m_A \to B/\mathfrak m_AB$ is surjective. (Note that $\mathfrak m_A$ is a nilpotent ideal.) As $k = A/\mathfrak m_A = B/\mathfrak m_B$ it suffices to show that $\mathfrak m_AB \to \mathfrak m_B$ is surjective. Applying Nakayama's lemma once more we see that it suffices to see that $\mathfrak m_AB/\mathfrak m_A\mathfrak m_B \to \mathfrak m_B/\mathfrak m_B^2$ is surjective which is what we assumed.
The ring $A_1 \times_A A_2$ is a $\Lambda$-algebra via  the map $\Lambda \to A_1 \times_A A_2$ induced by the maps $\Lambda \to A_1$ and $\Lambda \to A_2$. It is a local ring with unique maximal ideal $$ \mathfrak m_{A_1} \times_{\mathfrak m_A} \mathfrak m_{A_2} = \Ker(A_1 \times_A A_2 \longrightarrow k) $$ A ring is Artinian if and only if it has finite length as a module over itself, see Algebra, Lemma \ref{algebra-lemma-artinian-finite-length}. Since $A_1$ and $A_2$ are Artinian, Lemma \ref{lemma-length} implies $\text{length}_\Lambda(A_1)$ and $\text{length}_\Lambda(A_2)$, and hence $\text{length}_\Lambda(A_1 \times A_2)$, are all finite.  As $A_1 \times_A A_2 \subset A_1 \times A_2$ is a $\Lambda$-submodule, this implies $\text{length}_{A_1 \times_A A_2}(A_1 \times_A A_2) \leq \text{length}_\Lambda(A_1 \times_A A_2)$ is finite. So $A_1 \times_A A_2$ is Artinian. Thus the only thing that is keeping $A_1 \times_A A_2$ from being an object of $\mathcal{C}_\Lambda$ is the possibility that its residue field maps to a proper subfield of $k$ via the map $A_1 \times_A A_2 \to A \to A/\mathfrak m_A = k$ above. \medskip\noindent Proof of (1). If $f_2$ is surjective, then the projection $A_1 \times_A A_2 \to A_1$ is surjective. Hence the composition $A_1 \times_A A_2 \to A_1 \to A_1/\mathfrak m_{A_1} = k$ is surjective and we conclude that $A_1 \times_A A_2$ is an object of $\mathcal{C}_\Lambda$. \medskip\noindent Proof of (2). If $f_2$ is a small extension then $A_2 \to A$ and $A_1 \times_A A_2  \to A_1$ are both surjective with the same kernel. Hence the kernel of $A_1 \times_A A_2  \to A_1$ is a $1$-dimensional $k$-vector space and we see that $A_1 \times_A A_2  \to A_1$ is a small extension. \medskip\noindent Proof of (3). Choose $\overline{x} \in k$ such that $k = k'(\overline{x})$ (see Fields, Lemma \ref{fields-lemma-primitive-element}). Let $P'(T) \in k'[T]$ be the minimal polynomial of $\overline{x}$ over $k'$. Since $k/k'$ is separable we see that $\text{d}P/\text{d}T(\overline{x}) \not = 0$. Choose a monic $P \in \Lambda[T]$ which maps to $P'$ under the surjective map $\Lambda[T] \to k'[T]$. Because $A, A_1, A_2$ are henselian, see Algebra, Lemma \ref{algebra-lemma-local-dimension-zero-henselian}, we can find $x, x_1, x_2 \in A, A_1, A_2$ with $P(x) = 0, P(x_1) = 0, P(x_2) = 0$ and such that the image of $x, x_1, x_2$ in $k$ is $\overline{x}$. Then $(x_1, x_2) \in A_1 \times_A A_2$ because $x_1, x_2$ map to $x \in A$ by uniqueness, see Algebra, Lemma \ref{algebra-lemma-uniqueness}. Hence the residue field of $A_1 \times_A A_2$ contains a generator of $k$ over $k'$ and we win.
Assume (3). Let $C \to B$ be a ring map in $\mathcal{C}_\Lambda$ such that $C \to A$ is surjective. Then $C \to A/(\mathfrak m_\Lambda A + \mathfrak m_A^2)$ is surjective too. We conclude that $C \to B/(\mathfrak m_\Lambda B + \mathfrak m_B^2)$ is surjective by our assumption. Hence $C \to B$ is surjective by applying Lemma \ref{lemma-surjective} (2 times). \medskip\noindent Assume (1). Let $C \to B/(\mathfrak m_\Lambda B + \mathfrak m_B^2)$ be a morphism of $\mathcal{C}_\Lambda$ such that $C \to A/(\mathfrak m_\Lambda A + \mathfrak m_A^2)$ is surjective. Set $C' = C \times_{B/(\mathfrak m_\Lambda B + \mathfrak m_B^2)} B$ which is an object of $\mathcal{C}_\Lambda$ by Lemma \ref{lemma-fiber-product-CLambda}. Note that $C' \to A/(\mathfrak m_\Lambda A + \mathfrak m_A^2)$ is still surjective, hence $C' \to A$ is surjective by Lemma \ref{lemma-surjective}. Thus $C' \to B$ is surjective by our assumption. This implies that $C' \to B/(\mathfrak m_\Lambda B + \mathfrak m_B^2)$ is surjective, which implies by the construction of $C'$ that $C \to B/(\mathfrak m_\Lambda B + \mathfrak m_B^2)$ is surjective. \medskip\noindent In the first paragraph we proved (3) $\Rightarrow$ (1) and in the second paragraph we proved (1) $\Rightarrow$ (3). The equivalence of (2) and (3) is a special case of the equivalence of (1) and (3), hence we are done.
Note that $H_1(L_{k'/\Lambda}) = \mathfrak m_\Lambda/\mathfrak m_\Lambda^2$ as $\Lambda \to k'$ is surjective with kernel $\mathfrak m_\Lambda$. The map arises from functoriality of the naive cotangent complex. If $k' \subset k$ is separable, then $k' \to k$ is an \'etale ring map, see Algebra, Lemma \ref{algebra-lemma-etale-over-field}. Thus its naive cotangent complex has trivial homology groups, see Algebra, Definition \ref{algebra-definition-etale}. Then Algebra, Lemma \ref{algebra-lemma-exact-sequence-NL} applied to the ring maps $\Lambda \to k' \to k$ implies that $\mathfrak m_\Lambda/\mathfrak m_\Lambda^2 \otimes_{k'} k = H_1(L_{k/\Lambda})$. We omit the proof of the final statement.
Note that for $n \geq 2$ we have the equality of relative cotangent spaces $$ \mathfrak m_R/(\mathfrak m_\Lambda R + \mathfrak m_R^2) = \mathfrak m_{R_n}/(\mathfrak m_\Lambda R_n + \mathfrak m_{R_n}^2) $$ and similarly for $S$. Hence by Lemma \ref{lemma-surjective} we see that $R_n \to S_n$ is surjective for all $n$. Now let $K_n$ be the kernel of $R_n \to S_n$.  Then the sequences $$ 0 \to K_n \to R_n \to S_n \to 0 $$ form an exact sequence of directed inverse systems. The system $(K_n)$ is Mittag-Leffler since each $K_n$ is Artinian. Hence by Algebra, Lemma \ref{algebra-lemma-ML-exact-sequence} taking limits preserves exactness. So $\lim R_n \to \lim S_n$ is surjective, i.e., $f$ is surjective.
Let $R \to S_1$ and $R \to S_2$ be morphisms of $\widehat{\mathcal{C}}_\Lambda$. Consider the ring $C = S_1 \otimes_R S_2$. This ring has a finitely generated maximal ideal $\mathfrak m = \mathfrak m_{S_1} \otimes S_2 + S_1 \otimes \mathfrak m_{S_2}$ with residue field $k$. Set $C^\wedge$ equal to the completion of $C$ with respect to $\mathfrak m$. Then $C^\wedge$ is a Noetherian ring complete with respect to the maximal ideal $\mathfrak m^\wedge = \mathfrak mC^\wedge$ whose residue field is identified with $k$, see Algebra, Lemma \ref{algebra-lemma-completion-Noetherian}. Hence $C^\wedge$ is an object of $\widehat{\mathcal{C}}_\Lambda$. Then $S_1 \to C^\wedge$ and $S_2 \to C^\wedge$ turn $C^\wedge$ into a pushout over $R$ in $\widehat{\mathcal{C}}_\Lambda$ (details omitted).
Let $R$ and $S$ be objects of $\widehat{\mathcal{C}}_\Lambda$. Consider the ring $C = R \otimes_\Lambda S$. There is a canonical surjective map $C \to R \otimes_\Lambda S \to k \otimes_\Lambda k \to k$ where the last map is the multiplication map. The kernel of $C \to k$ is a maximal ideal $\mathfrak m$. Note that $\mathfrak m$ is generated by $\mathfrak m_R C$, $\mathfrak m_S C$ and finitely many elements of $C$ which map to generators of the kernel of $k \otimes_\Lambda k \to k$. Hence $\mathfrak m$ is a finitely generated ideal. Set $C^\wedge$ equal to the completion of $C$ with respect to $\mathfrak m$. Then $C^\wedge$ is a Noetherian ring complete with respect to the maximal ideal $\mathfrak m^\wedge = \mathfrak mC^\wedge$ with residue field $k$, see Algebra, Lemma \ref{algebra-lemma-completion-Noetherian}. Hence $C^\wedge$ is an object of $\widehat{\mathcal{C}}_\Lambda$. Then $R \to C^\wedge$ and $S \to C^\wedge$ turn $C^\wedge$ into a coproduct in $\widehat{\mathcal{C}}_\Lambda$ (details omitted).
If $f$ is not surjective, then $\mathfrak m_S/(\mathfrak m_R S + \mathfrak m_S^2)$ is nonzero by Lemma \ref{lemma-surjective-cotangent-space}. Then also $Q = S/(f(R) + \mathfrak m_R S + \mathfrak m_S^2)$ is nonzero. Note that $Q$ is a $k = R/\mathfrak m_R$-vector space via $f$. We turn $Q$ into an $S$-module via $S \to k$. The quotient map $D : S \to Q$ is an $R$-derivation: if $a_1, a_2 \in S$, we can write $a_1 = f(b_1) + a_1'$ and $a_2 = f(b_2) + a_2'$ for some $b_1, b_2 \in R$ and $a_1', a_2' \in \mathfrak m_S$. Then $b_i$ and $a_i$ have the same image in $k$ for $i = 1, 2$ and \begin{align*} a_1a_2 & = (f(b_1) + a_1')(f(b_2) + a_2') \\ & = f(b_1)a_2' + f(b_2)a_1' \\ & = f(b_1)(f(b_2) + a_2') + f(b_2)(f(b_1) + a_1') \\ & = f(b_1)a_2 + f(b_2)a_1 \end{align*} in $Q$ which proves the Leibniz rule. Hence $D : S \to Q$ is a $\Lambda$-derivation which is zero on composing with $R \to S$. Since $Q \not = 0$ there also exist derivations $D : S \to k$ which are zero on composing with $R \to S$, i.e., $\text{Der}_\Lambda(S, k) \to \text{Der}_\Lambda(R, k)$ is not injective.
It is clear that $\mathfrak m_{R/J}^n \subset J_n/J$. Thus it suffices to show that for every $n$ there exists an $N$ such that $J_N/J \subset \mathfrak m_{R/J}^n$. This is equivalent to $J_N \subset \mathfrak m_R^n + J$. For each $n$ the ring $R/\mathfrak m_R^n$ is Artinian, hence there exists a $N_n$ such that $$ J_{N_n} + \mathfrak m_R^n = J_{N_n + 1} + \mathfrak m_R^n = \ldots $$ Set $E_n = (J_{N_n} + \mathfrak m_R^n)/\mathfrak m_R^n$. Set $E = \lim E_n \subset \lim R/\mathfrak m_R^n = R$. Note that $E \subset J$ as for any $f \in E$ and any $m$ we have $f \in J_m + \mathfrak m_R^n$ for all $n \gg 0$, so $f \in J_m$ by Artin-Rees, see Algebra, Lemma \ref{algebra-lemma-intersect-powers-ideal-module-zero}. Since the transition maps $E_n \to E_{n - 1}$ are all surjective, we see that $J$ surjects onto $E_n$. Hence for $N = N_n$ works.
We will use freely that the maps $S \to A_n$ are surjective for all $n$. Note that the maps $\mathfrak m_{A_{n + 1}}/\mathfrak m_{A_{n + 1}}^2 \to \mathfrak m_{A_n}/\mathfrak m_{A_n}^2$ are surjective, see Lemma \ref{lemma-surjective-cotangent-space}. Hence for $n$ sufficiently large the dimension $\dim_k (\mathfrak m_{A_n}/\mathfrak m_{A_n}^2)$ stabilizes to an integer, say $r$. Thus we can find $x_1, \ldots, x_r \in \mathfrak m_S$ whose images in $A_n$ generate $\mathfrak m_{A_n}$. Moreover, pick $y_1, \ldots, y_t \in S$ whose images in $k$ generate $k$ over $\Lambda$. Then we get a ring map $P = \Lambda[z_1, \ldots, z_{r + t}] \to S$, $z_i \mapsto x_i$ and $z_{r + j} \mapsto y_j$ such that the composition $P \to S \to A_n$ is surjective for all $n$. Let $\mathfrak m \subset P$ be the kernel of $P \to k$. Let $R = P^\wedge$ be the $\mathfrak m$-adic completion of $P$; this is an object of $\widehat{\mathcal{C}}_\Lambda$. Since we still have the compatible system of (surjective) maps $R \to A_n$ we get a map $R \to S$. Set $J_n = \Ker(R \to A_n)$. Set $J = \bigcap J_n$. By Lemma \ref{lemma-m-adic-topology} we see that $R/J = \lim R/J_n = \lim A_n = S$ and that the ideals $J_n/J = I_n$ define the $\mathfrak m$-adic topology. (Note that for each $n$ we have $\mathfrak m_R^{N_n} \subset J_n$ for some $N_n$ and not necessarily $N_n = n$, so a renumbering of the ideals $J_n$ may be necessary before applying the lemma.)
As $R = R' \oplus I$ we have $$ \mathfrak m_R/\mathfrak m_R^2 = \mathfrak m_{R'}/\mathfrak m_{R'}^2 \oplus I/\mathfrak m_RI $$ and similarly $$ \mathfrak m_S/\mathfrak m_S^2 = \mathfrak m_{R'}/\mathfrak m_{R'}^2 \oplus \bigoplus kX_i $$ Hence for $n > 1$ the map $S/\mathfrak m_S^n \to R/\mathfrak m_R^n$ induces an isomorphism on cotangent spaces. Thus a left inverse $h_n : R/\mathfrak m_R^n \to S/\mathfrak m_S^n$ is surjective by Lemma \ref{lemma-surjective-cotangent-space}. Since $h_n$ is injective as a left inverse it is an isomorphism. Thus the canonical surjections $S/\mathfrak m_S^n \to R/\mathfrak m_R^n$ are all isomorphisms and we win.
Let $B \to A$ be a surjective ring map in $\mathcal{C}_\Lambda$. Let $y \in \Ob(\mathcal{G}(B))$, $x \in \Ob(\mathcal{F}(A))$, and $y \to \varphi(x)$ be a morphism lying over $B \to A$. By Lemma \ref{lemma-factor-small-extension} 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 $$ \xymatrix{ \varphi(x_{n - 1}) \ar[r]_-a \ar[dr] & f_*y \ar[d] \\ & \varphi(x) } $$ 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 $$ \xymatrix{ \varphi(x_n) \ar[r] \ar[d] & y \ar[d] \\ \varphi(x_{n - 1}) \ar[r]^-a \ar[dr] & f_*y \ar[d] \\ & \varphi(x) } $$ 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.