hcju/stacksqa huggingface dataset:
Consider the map (\ref{equation-define-module-differentials}). There is a similar map of presheaves whose value on the open $U$ is $$ \mathcal{O}_2(U)[\mathcal{O}_2(U) \times \mathcal{O}_2(U)] \oplus \mathcal{O}_2(U)[\mathcal{O}_2(U) \times \mathcal{O}_2(U)] \oplus \mathcal{O}_2(U)[\mathcal{O}_1(U)] \longrightarrow \mathcal{O}_2(U)[\mathcal{O}_2(U)] $$ The cokernel of this map has value $\Omega_{\mathcal{O}_2(U)/\mathcal{O}_1(U)}$ over $U$ by the construction of the module of differentials in  Algebra, Definition \ref{algebra-definition-differentials}. On the other hand, the sheaves in (\ref{equation-define-module-differentials}) are the sheafifications of the presheaves above. Thus the result follows as sheafification is exact.
This is a special case of Lemma \ref{lemma-pullback-differentials} for the inclusion map $\{x\} \to X$. An alternative proof is to use Lemma \ref{lemma-differentials-sheafify}, Sheaves, Lemma \ref{sheaves-lemma-stalk-sheafification}, and Algebra, Lemma \ref{algebra-lemma-colimit-differentials}
For a local section $f$ of $\mathcal{I}$ denote $\overline{f}$ the image of $f$ in $\mathcal{I}/\mathcal{I}^2$. To show that the map $\overline{f} \mapsto \text{d}f \otimes 1$ is well defined we just have to check that $\text{d} f_1f_2 \otimes 1 = 0$ if $f_1, f_2$ are local sections of $\mathcal{I}$. And this is clear from the Leibniz rule $\text{d} f_1f_2 \otimes 1 = (f_1 \text{d}f_2 + f_2 \text{d} f_1 )\otimes 1 = \text{d}f_2 \otimes f_1 + \text{d}f_1 \otimes f_2 = 0$. A similar computation show this map is $\mathcal{O}'_2 = \mathcal{O}_2/\mathcal{I}$-linear. The map on the right is the one from Lemma \ref{lemma-functoriality-differentials}. To see that the sequence is exact, we can check on stalks (Lemma \ref{lemma-abelian}). By Lemma \ref{lemma-stalk-module-differentials} this follows from Algebra, Lemma \ref{algebra-lemma-differential-seq}.
The existence follows from general category theoretic arguments (insert future reference here), but we will also give a direct construction as this construction will be useful in the future proofs. We will freely use the notation introduced in the proof of Lemma \ref{lemma-universal-module}. Given any differential operator $D : \mathcal{F} \to \mathcal{G}$ we obtain an $\mathcal{O}_2$-linear map $L_D : \mathcal{O}_2[\mathcal{F}] \to \mathcal{G}$ sending $[m]$ to $D(m)$. If $D$ has order $0$ then $L_D$ annihilates the local sections $$ [m + m'] - [m] - [m'],\quad g_0[m] - [g_0m] $$ where $g_0$ is a local section of $\mathcal{O}_2$ and $m, m'$ are local sections of $\mathcal{F}$. If $D$ has order $1$, then $L_D$ annihilates the local sections $$ [m + m' - [m] - [m'],\quad f[m] - [fm], \quad g_0g_1[m] - g_0[g_1m] - g_1[g_0m] + [g_1g_0m] $$ where $f$ is a local section of $\mathcal{O}_1$, $g_0, g_1$ are local sections of $\mathcal{O}_2$, and $m, m'$ are local sections of $\mathcal{F}$. If $D$ has order $k$, then $L_D$ annihilates the local sections $[m + m'] - [m] - [m']$, $f[m] - [fm]$, and the local sections $$ g_0g_1\ldots g_k[m] - \sum g_0 \ldots \hat g_i \ldots g_k[g_im] + \ldots +(-1)^{k + 1}[g_0\ldots g_km] $$ Conversely, if $L : \mathcal{O}_2[\mathcal{F}] \to \mathcal{G}$ is an $\mathcal{O}_2$-linear map annihilating all the local sections listed in the previous sentence, then $m \mapsto L([m])$ is a differential operator of order $k$. Thus we see that $\mathcal{P}^k_{\mathcal{O}_2/\mathcal{O}_1}(\mathcal{F})$ is the quotient of $\mathcal{O}_2[\mathcal{F}]$ by the $\mathcal{O}_2$-submodule generated by these local sections.
Observe that $\NL_{\mathcal{B}/\mathcal{A}} = \NL(\text{id}_\mathcal{B})$. Thus it suffices to show that given two maps $\alpha_i : \mathcal{E}_i \to \mathcal{B}$ as above, there is a canonical quasi-isomorphism $\NL(\alpha_1) = \NL(\alpha_2)$ in $D(\mathcal{B})$. To see this set $\mathcal{E} = \mathcal{E}_1 \amalg \mathcal{E}_2$ and $\alpha = \alpha_1 \amalg \alpha_2 : \mathcal{E} \to \mathcal{B}$. Set $\mathcal{J}_i = \Ker(\mathcal{A}[\mathcal{E}_i] \to \mathcal{B})$ and $\mathcal{J} = \Ker(\mathcal{A}[\mathcal{E}] \to \mathcal{B})$. We obtain maps $\mathcal{A}[\mathcal{E}_i] \to \mathcal{A}[\mathcal{E}]$ which send $\mathcal{J}_i$ into $\mathcal{J}$. Thus we obtain canonical maps of complexes $$ \NL(\alpha_i) \longrightarrow \NL(\alpha) $$ and it suffices to show these maps are quasi-isomorphism. To see this it suffices to check on stalks (Lemma \ref{lemma-abelian}). If $x \in X$ then the stalk of $\NL(\alpha)$ is the complex $\NL(\alpha_x)$ of Algebra, Section \ref{algebra-section-netherlander} associated to the presentation $\mathcal{A}_x[\mathcal{E}_x] \to \mathcal{B}_x$ coming from the map $\alpha_x : \mathcal{E}_x \to \mathcal{B}_x$. (Some details omitted; use Lemma \ref{lemma-stalk-module-differentials} to see compatibility of forming differentials and taking stalks.) We conclude the result holds by Algebra, Lemma \ref{algebra-lemma-NL-homotopy}.
To give the map $c$ we have to give a map $c_1 : \NL_{\mathcal{B}/\mathcal{A}} \otimes_\mathcal{B} \mathcal{C} \to \NL_{\mathcal{C}/\mathcal{A}}$ and an explicity homotopy between the composition $$ \NL_{\mathcal{B}/\mathcal{A}} \otimes_\mathcal{B} \mathcal{C} \to \NL_{\mathcal{C}/\mathcal{A}} \to \NL_{\mathcal{C}/\mathcal{B}} $$ and the zero map, see Derived Categories, Lemma \ref{derived-lemma-map-from-cone}. For $c_1$ we use the functoriality described above for the obvious diagram. For the homotopy we use the map $$ \NL_{\mathcal{B}/\mathcal{A}}^0 \otimes_\mathcal{B} \mathcal{C} \longrightarrow \NL_{\mathcal{C}/\mathcal{B}}^{-1},\quad \text{d}[b] \otimes 1 \longmapsto [\varphi(b)] - b[1] $$ where $\varphi : \mathcal{B} \to \mathcal{C}$ is the given map. Please compare with Algebra, Remark \ref{algebra-remark-composition-homotopy-equivalent-to-zero}. To see the consequence for cohomology sheaves, it suffices to show that $H^0(c)$ is an isomorphism and $H^{-1}(c)$ surjective. To see this we can look at stalks, see Lemma \ref{lemma-stalk-NL}, and then we can use the corresponding result in commutative algebra, see Algebra, Lemma \ref{algebra-lemma-exact-sequence-NL}. Some details omitted.
Write $B = A[x_1, \ldots, x_n]/J$. Let $J' \subset A'[x_1, \ldots, x_n]$ be the inverse image of $J$. Denote $I[x_1, \ldots, x_n]$ the kernel of $A'[x_1, \ldots, x_n] \to A[x_1, \ldots, x_n]$. By More on Algebra, Lemma \ref{more-algebra-lemma-conormal-sequence-H1-regular-ideal} we have $I[x_1, \ldots, x_n] \cap (J')^2 = J'I[x_1, \ldots, x_n] = JI[x_1, \ldots, x_n]$. Hence we obtain a short exact sequence $$ 0 \to I \otimes_A B \to J'/(J')^2 \to J/J^2 \to 0 $$ Since $J/J^2$ is projective (More on Algebra, Lemma \ref{more-algebra-lemma-quasi-regular-ideal-finite-projective}) we can choose a splitting of this sequence $$ J'/(J')^2 = I \otimes_A B \oplus J/J^2  $$ Let $(J')^2 \subset J'' \subset J'$ be the elements which map to the second summand in the decomposition above. Then $$ 0 \to I \otimes_A B \to A'[x_1, \ldots, x_n]/J'' \to B \to 0 $$ is a solution to (\ref{equation-to-solve}) with $N = I \otimes_A B$. The general case is obtained by doing a pushout along the given map $I \otimes_A B \to N$.
We observe right away that given two solutions $B'_1$ and $B'_2$ to (\ref{equation-to-solve}) we obtain by Lemma \ref{lemma-huge-diagram} an obstruction element $o(B'_1, B'_2) \in \Ext^1_B(\NL_{B/A}, N)$ to the existence of a map $B'_1 \to B'_2$. Clearly, this element is the obstruction to the existence of an isomorphism, hence separates the isomorphism classes. To finish the proof it therefore suffices to show that given a solution $B'$ and an element $\xi \in \Ext^1_B(\NL_{B/A}, N)$ we can find a second solution $B'_\xi$ such that $o(B', B'_\xi) = \xi$. \medskip\noindent Let $E = B$ viewed as a set. Consider the surjection $A[E] \to B$ with kernel $J$ used to define the naive cotangent complex by the formula $$ \NL_{B/A} = (J/J^2 \to \Omega_{A[E]/A} \otimes_{A[E]} B) $$ in Algebra, Section \ref{algebra-section-netherlander}. Since $\Omega_{A[E]/A} \otimes B$ is a free $B$-module we have $$ \Ext^1_B(\NL_{B/A}, N) = \frac{\Hom_B(J/J^2, N)} {\Hom_B(\Omega_{A[E]/A} \otimes B, N)} $$ Thus we may represent $\xi$ as the class of a morphism $\delta : J/J^2 \to N$. \medskip\noindent For every $e \in E$ denote $x_e \in A[E]$ the corresponding variable. Choose a lift $y_e \in B'$ of the image of $x_e$ in $B$. These choices determine an $A'$-algebra map $\varphi : A'[E] \to B'$. Let $J' = \Ker(A'[E] \to B)$. Observe that $\varphi$ induces a map $\varphi|_{J'} : J' \to N$ and that $B'$ is the pushout, as in the following diagram $$ \xymatrix{ 0 \ar[r] & N \ar[r] & B' \ar[r] & B \ar[r] & 0 \\ 0 \ar[r] & J' \ar[u]^{\varphi|_{J'}} \ar[r] & A'[E] \ar[u] \ar[r] & B \ar[u]_{=} \ar[r] & 0 } $$ Let $\psi : J' \to N$ be the sum of the map $\varphi|_{J'}$ and the composition $$ J' \to J'/(J')^2 \to J/J^2 \xrightarrow{\delta} N. $$ Then the pushout along $\psi$ is an other ring extension $B'_\xi$ fitting into a diagram as above. A calculation shows that $o(B', B'_\xi) = \xi$ as desired.
To prove this we apply the previous results where we work over $0 \to 0 \to \mathbf{Z} \to \mathbf{Z} \to 0$, in order words, we work over the extension of $\mathbf{Z}$ by $0$. Part (1) follows from Lemma \ref{lemma-choices} and the fact that there exists a solution, namely $I \oplus A$. Part (2) follows from Lemma \ref{lemma-huge-diagram} and a compatibility between the constructions in the proofs of Lemmas \ref{lemma-choices} and \ref{lemma-huge-diagram} whose statement and proof we omit.
Part (1) is immediate from the description of the maps. For (2), if $\varphi'$ and $\varphi''$ are two maps $\mathcal{F}' \to \mathcal{G}'$ compatible with $\varphi$ and $\psi$, then $\varphi' - \varphi''$ factors as $$ \mathcal{F}' \to \mathcal{F} \to \mathcal{L} \to \mathcal{G}' $$ The map in the middle comes from a unique element of $\Hom_{\mathcal{O}_X}(\mathcal{F}, \mathcal{L})$ by Modules, Lemma \ref{modules-lemma-i-star-equivalence}. Conversely, given an element $\alpha$ of this group we can add the composition (as displayed above with $\alpha$ in the middle) to $\varphi'$. Some details omitted.
Assume given extensions $$ 0 \to \mathcal{K} \to \mathcal{F}'_1 \to \mathcal{F} \to 0 \quad\text{and}\quad 0 \to \mathcal{K} \to \mathcal{F}'_2 \to \mathcal{F} \to 0 $$ with $c_{\mathcal{F}'_1} = c_{\mathcal{F}'_2} = c$. Then the difference (in the extension group, see Homology, Section \ref{homology-section-extensions}) is an extension $$ 0 \to \mathcal{K} \to \mathcal{E} \to \mathcal{F} \to 0 $$ where $\mathcal{E}$ is annihilated by $\mathcal{I}$ (local computation omitted). Hence the sequence is an extension of $\mathcal{O}_X$-modules, see Modules, Lemma \ref{modules-lemma-i-star-equivalence}. Conversely, given such an extension $\mathcal{E}$ we can add the extension $\mathcal{E}$ to the $\mathcal{O}_{X'}$-extension $\mathcal{F}'$ without affecting the map $c_{\mathcal{F}'}$. Some details omitted.
This is a special case of Lemma \ref{lemma-inf-map} but we also give a direct proof. We have short exact sequences of modules $$ 0 \to \mathcal{I} \to \mathcal{O}_{X'} \to \mathcal{O}_X \to 0 \quad\text{and}\quad 0 \to \mathcal{I}\mathcal{G}' \to \mathcal{G}' \to \mathcal{G} \to 0 $$ and similarly for $\mathcal{F}'$. Since $\mathcal{I}$ has square zero the $\mathcal{O}_{X'}$-module structure on $\mathcal{I}$ and $\mathcal{I}\mathcal{G}'$ comes from a unique $\mathcal{O}_X$-module structure. It follows that $$ \Hom_{\mathcal{O}_{X'}}(\mathcal{F}', \mathcal{I}\mathcal{G}') = \Hom_{\mathcal{O}_X}(\mathcal{F}, \mathcal{I}\mathcal{G}') \quad\text{and}\quad \Hom_{\mathcal{O}_{X'}}(\mathcal{F}', \mathcal{G}) = \Hom_{\mathcal{O}_X}(\mathcal{F}, \mathcal{G}) $$ The lemma now follows from the exact sequence $$ 0 \to \Hom_{\mathcal{O}_{X'}}(\mathcal{F}', \mathcal{I}\mathcal{G}') \to \Hom_{\mathcal{O}_{X'}}(\mathcal{F}', \mathcal{G}') \to \Hom_{\mathcal{O}_{X'}}(\mathcal{F}', \mathcal{G}) $$ see Homology, Lemma \ref{homology-lemma-check-exactness}.
The map $f^*\mathcal{J} \to \mathcal{I}$ is surjective as $(f, f')$ is a strict morphism of thickenings. Hence the final statement is a consequence of (2). \medskip\noindent Proof of the equivalence of (1) and (2). We may check these conditions at stalks. Let $x \in X \subset X'$ be a point with image $s = f(x) \in S \subset S'$. Set $A' = \mathcal{O}_{S', s}$, $B' = \mathcal{O}_{X', x}$, $A = \mathcal{O}_{S, s}$, and $B = \mathcal{O}_{X, x}$. Then $A = A'/J$ and $B = B'/I$ for some square zero ideals. Since $(f, f')$ is a strict morphism of thickenings we have $I = JB'$. Let $M' = \mathcal{F}'_x$ and $M = \mathcal{F}_x$. Then $M'$ is a $B'$-module and $M$ is a $B$-module. Since $\mathcal{F} = i^*\mathcal{F}'$ we see that the kernel of the surjection $M' \to M$ is $IM' = JM'$. Thus we have a short exact sequence $$ 0 \to JM' \to M' \to M \to 0 $$ Using Sheaves, Lemma \ref{sheaves-lemma-stalk-pullback-modules} and Modules, Lemma \ref{modules-lemma-stalk-tensor-product} to identify stalks of pullbacks and tensor products we see that the stalk at $x$ of the canonical map of the lemma is the map $$ (J \otimes_A B) \otimes_B M = J \otimes_A M = J \otimes_{A'} M' \longrightarrow JM' $$ The assumption that $\mathcal{F}$ is flat over $S$ signifies that $M$ is a flat $A$-module. \medskip\noindent Assume (1). Flatness implies $\text{Tor}_1^{A'}(M', A) = 0$ by Algebra, Lemma \ref{algebra-lemma-characterize-flat}. This means $J \otimes_{A'} M' \to M'$ is injective by Algebra, Remark \ref{algebra-remark-Tor-ring-mod-ideal}. Hence $J \otimes_A M \to JM'$ is an isomorphism. \medskip\noindent Assume (2). Then $J \otimes_{A'} M' \to M'$ is injective. Hence $\text{Tor}_1^{A'}(M', A) = 0$ by Algebra, Remark \ref{algebra-remark-Tor-ring-mod-ideal}. Hence $M'$ is flat over $A'$ by Algebra, Lemma \ref{algebra-lemma-what-does-it-mean}.
It is clear from the proof of Lemma \ref{lemma-inf-map-special} that the vanishing of the boundary of $\varphi$ via the map $$ \Hom_{\mathcal{O}_X}(\mathcal{F}, \mathcal{G}) = \Hom_{\mathcal{O}_{X'}}(\mathcal{F}', \mathcal{G}) \longrightarrow \Ext^1_{\mathcal{O}_{X'}}(\mathcal{F}', \mathcal{I}\mathcal{G}') $$ is a necessary and sufficient condition for the existence of a lift. We conclude as $$ \Ext^1_{\mathcal{O}_{X'}}(\mathcal{F}', \mathcal{I}\mathcal{G}') = \Ext^1_{\mathcal{O}_X}(Li^*\mathcal{F}', \mathcal{I}\mathcal{G}') $$ the adjointness of $i_* = Ri_*$ and $Li^*$ on the derived category (Cohomology, Lemma \ref{cohomology-lemma-adjoint}).
[First proof] This follows from Lemma \ref{lemma-inf-obs-map-special} as we claim that under the assumptions of the lemma we have $$ \Ext^1_{\mathcal{O}_X}(Li^*\mathcal{F}', \mathcal{I}\mathcal{G}') = \Ext^1_{\mathcal{O}_X}(\mathcal{F}, \mathcal{G} \otimes_{\mathcal{O}_X} f^*\mathcal{J}) $$ Namely, we have $\mathcal{I}\mathcal{G}' = \mathcal{G} \otimes_{\mathcal{O}_X} f^*\mathcal{J}$ by Lemma \ref{lemma-deform-module}. On the other hand, observe that $$ H^{-1}(Li^*\mathcal{F}') = \text{Tor}_1^{\mathcal{O}_{X'}}(\mathcal{F}', \mathcal{O}_X) $$ (local computation omitted). Using the short exact sequence $$ 0 \to \mathcal{I} \to \mathcal{O}_{X'} \to \mathcal{O}_X \to 0 $$ we see that this $\text{Tor}_1$ is computed by the kernel of the map $\mathcal{I} \otimes_{\mathcal{O}_X} \mathcal{F} \to \mathcal{I}\mathcal{F}'$ which is zero by the final assertion of Lemma \ref{lemma-deform-module}. Thus $\tau_{\geq -1}Li^*\mathcal{F}' = \mathcal{F}$. On the other hand, we have $$ \Ext^1_{\mathcal{O}_X}(Li^*\mathcal{F}', \mathcal{I}\mathcal{G}') = \Ext^1_{\mathcal{O}_X}(\tau_{\geq -1}Li^*\mathcal{F}', \mathcal{I}\mathcal{G}') $$ by the dual of Derived Categories, Lemma \ref{derived-lemma-negative-vanishing}.
If we assume there exists one such module, then the canonical map $$ f^*\mathcal{J} \otimes_{\mathcal{O}_X} \mathcal{F} \to \mathcal{I} \otimes_{\mathcal{O}_X} \mathcal{F} $$ is an isomorphism by Lemma \ref{lemma-deform-module}. Apply Lemma \ref{lemma-inf-ext} with $\mathcal{K} =  \mathcal{I} \otimes_{\mathcal{O}_X} \mathcal{F}$ and $c = 1$. By Lemma \ref{lemma-deform-module} the corresponding extensions $\mathcal{F}'$ are all flat over $S'$.
Note that both $\pi^*\mathcal{F}$ and $h^*\mathcal{F}'_2$ restrict to $\mathcal{F}$ on $X$ and that the kernels of $\pi^*\mathcal{F} \to \mathcal{F}$ and $h^*\mathcal{F}'_2 \to \mathcal{F}$ are given by $f^*\mathcal{J}_1 \otimes_{\mathcal{O}_X} \mathcal{F}$. Hence flatness by Lemma \ref{lemma-deform-module}. Taking the boundary makes sense as the sequence of modules $$ 0 \to f^*\mathcal{J}_3 \otimes_{\mathcal{O}_X} \mathcal{F} \to f^*\mathcal{J}_2 \otimes_{\mathcal{O}_X} \mathcal{F} \to f^*\mathcal{J}_1 \otimes_{\mathcal{O}_X} \mathcal{F} \to 0 $$ is short exact due to the assumptions in Situation \ref{situation-ses-flat-thickenings} and the fact that $\mathcal{F}$ is flat over $S$. The statement on the obstruction class is a direct translation of the result of Remark \ref{remark-complex-thickenings-and-ses-modules} to this particular situation.
Recall that given $\alpha : \mathcal{E} \to \mathcal{B}$ such that $\mathcal{A}[\mathcal{E}] \to \mathcal{B}$ is surjective with kernel $\mathcal{I}$ the complex $\NL(\alpha) = (\mathcal{I}/\mathcal{I}^2 \to  \Omega_{\mathcal{A}[\mathcal{E}]/\mathcal{A}} \otimes_{\mathcal{A}[\mathcal{E}]} \mathcal{B})$ is canonically isomorphic to $\NL_{\mathcal{B}/\mathcal{A}}$, see Modules, Lemma \ref{modules-lemma-NL-up-to-qis}. Observe moreover, that $\Omega = \Omega_{\mathcal{A}[\mathcal{E}]/\mathcal{A}} \otimes_{\mathcal{A}[\mathcal{E}]} \mathcal{B}$ is the sheaf associated to the presheaf $U \mapsto \bigoplus_{e \in \mathcal{E}(U)} \mathcal{B}(U)$. In other words, $\Omega$ is the free $\mathcal{B}$-module on the sheaf of sets $\mathcal{E}$ and in particular there is a canonical map $\mathcal{E} \to \Omega$. \medskip\noindent Having said this, pick some $\mathcal{E}$ (for example $\mathcal{E} = \mathcal{B}$ as in the definition of the naive cotangent complex). The obstruction to writing $\xi$ as the class of a map $\mathcal{I}/\mathcal{I}^2 \to \mathcal{G}$ is an element in $\Ext^1_\mathcal{B}(\Omega, \mathcal{G})$. Say this is represented by the extension $0 \to \mathcal{G} \to \mathcal{H} \to \Omega \to 0$ of $\mathcal{B}$-modules. Consider the sheaf of sets $\mathcal{E}' = \mathcal{E} \times_\Omega \mathcal{H}$ which comes with an induced map $\alpha' : \mathcal{E}' \to \mathcal{B}$. Let $\mathcal{I}' = \Ker(\mathcal{A}[\mathcal{E}'] \to \mathcal{B})$ and $\Omega' = \Omega_{\mathcal{A}[\mathcal{E}']/\mathcal{A}} \otimes_{\mathcal{A}[\mathcal{E}']} \mathcal{B}$. The pullback of $\xi$ under the quasi-isomorphism $\NL(\alpha') \to \NL(\alpha)$ maps to zero in $\Ext^1_\mathcal{B}(\Omega', \mathcal{G})$ because the pullback of the extension $\mathcal{H}$ by the map $\Omega' \to \Omega$ is split as $\Omega'$ is the free $\mathcal{B}$-module on the sheaf of sets $\mathcal{E}'$ and since by construction there is a commutative diagram $$ \xymatrix{ \mathcal{E}' \ar[r] \ar[d] & \mathcal{E} \ar[d] \\ \mathcal{H} \ar[r] & \Omega } $$ This finishes the proof.
We observe right away that given two solutions $X'_1$ and $X'_2$ to (\ref{equation-to-solve-ringed-spaces}) we obtain by Lemma \ref{lemma-huge-diagram-ringed-spaces} an obstruction element $o(X'_1, X'_2) \in \Ext^1_{\mathcal{O}_X}(\NL_{X/S}, \mathcal{G})$ to the existence of a map $X'_1 \to X'_2$. Clearly, this element is the obstruction to the existence of an isomorphism, hence separates the isomorphism classes. To finish the proof it therefore suffices to show that given a solution $X'$ and an element $\xi \in \Ext^1_{\mathcal{O}_X}(\NL_{X/S}, \mathcal{G})$ we can find a second solution $X'_\xi$ such that $o(X', X'_\xi) = \xi$. \medskip\noindent Pick $\alpha : \mathcal{E} \to \mathcal{O}_X$ as in Lemma \ref{lemma-NL-represent-ext-class} for the class $\xi$. Consider the surjection $f^{-1}\mathcal{O}_S[\mathcal{E}] \to \mathcal{O}_X$ with kernel $\mathcal{I}$ and corresponding naive cotangent complex $\NL(\alpha) = (\mathcal{I}/\mathcal{I}^2 \to \Omega_{f^{-1}\mathcal{O}_S[\mathcal{E}]/f^{-1}\mathcal{O}_S} \otimes_{f^{-1}\mathcal{O}_S[\mathcal{E}]} \mathcal{O}_X)$. By the lemma $\xi$ is the class of a morphism $\delta : \mathcal{I}/\mathcal{I}^2 \to \mathcal{G}$. After replacing $\mathcal{E}$ by $\mathcal{E} \times_{\mathcal{O}_X} \mathcal{O}_{X'}$ we may also assume that $\alpha$ factors through a map $\alpha' : \mathcal{E} \to \mathcal{O}_{X'}$. \medskip\noindent These choices determine an $f^{-1}\mathcal{O}_{S'}$-algebra map $\varphi : \mathcal{O}_{S'}[\mathcal{E}] \to \mathcal{O}_{X'}$. Let $\mathcal{I}' = \Ker(\varphi)$. Observe that $\varphi$ induces a map $\varphi|_{\mathcal{I}'} : \mathcal{I}' \to \mathcal{G}$ and that $\mathcal{O}_{X'}$ is the pushout, as in the following diagram $$ \xymatrix{ 0 \ar[r] & \mathcal{G} \ar[r] & \mathcal{O}_{X'} \ar[r] & \mathcal{O}_X \ar[r] & 0 \\ 0 \ar[r] & \mathcal{I}' \ar[u]^{\varphi|_{\mathcal{I}'}} \ar[r] & f^{-1}\mathcal{O}_{S'}[\mathcal{E}] \ar[u] \ar[r] & \mathcal{O}_X \ar[u]_{=} \ar[r] & 0 } $$ Let $\psi : \mathcal{I}' \to \mathcal{G}$ be the sum of the map $\varphi|_{\mathcal{I}'}$ and the composition $$ \mathcal{I}' \to \mathcal{I}'/(\mathcal{I}')^2 \to \mathcal{I}/\mathcal{I}^2 \xrightarrow{\delta} \mathcal{G}. $$ Then the pushout along $\psi$ is an other ring extension $\mathcal{O}_{X'_\xi}$ fitting into a diagram as above. A calculation (omitted) shows that $o(X', X'_\xi) = \xi$ as desired.
To prove this we apply the previous results where we work over the base ringed space $(*, \mathbf{Z})$ with trivial thickening. Part (1) follows from Lemma \ref{lemma-choices-ringed-spaces} and the fact that there exists a solution, namely $\mathcal{J} \oplus \mathcal{O}_S$. Part (2) follows from Lemma \ref{lemma-huge-diagram-ringed-spaces} and a compatibility between the constructions in the proofs of Lemmas \ref{lemma-choices-ringed-spaces} and \ref{lemma-huge-diagram-ringed-spaces} whose statement and proof we omit.
First we observe that thickenings of schemes as defined in More on Morphisms, Section \ref{more-morphisms-section-thickenings} are the same things as morphisms of schemes which are thickenings in the sense of Section \ref{section-thickenings-spaces}. We may think of $X$ as a closed subscheme of $X'$ so that $(f, f') : (X \subset X') \to (S \subset S')$ is a morphism of first order thickenings. Then we see from More on Morphisms, Lemma \ref{more-morphisms-lemma-deform} (or from the more general Lemma \ref{lemma-deform-module}) that the ideal sheaf of $X$ in $X'$ is equal to $f^*\mathcal{C}_{S/S'}$. Hence we have a commutative diagram $$ \xymatrix{ 0 \ar[r] & f^*\mathcal{C}_{S/S'} \ar[r] & \mathcal{O}_{X'} \ar[r] & \mathcal{O}_X \ar[r] & 0 \\ 0 \ar[r] & \mathcal{C}_{S/S'} \ar[u] \ar[r] & \mathcal{O}_{S'} \ar[u] \ar[r] & \mathcal{O}_S \ar[u] \ar[r] & 0 } $$ where the vertical arrows are $f$-maps; please compare with (\ref{equation-to-solve-ringed-spaces}). Thus part (1) follows from Lemma \ref{lemma-choices-ringed-spaces} and part (2) from part (2) of Lemma \ref{lemma-huge-diagram-ringed-spaces}. (Note that $\NL_{X/S}$ as defined for a morphism of schemes in More on Morphisms, Section \ref{more-morphisms-section-netherlander} agrees with $\NL_{X/S}$ as used in Section \ref{section-deformations-ringed-spaces}.)
Part (1) is immediate from the description of the maps. For (2), if $\varphi'$ and $\varphi''$ are two maps $\mathcal{F}' \to \mathcal{G}'$ compatible with $\varphi$ and $\psi$, then $\varphi' - \varphi''$ factors as $$ \mathcal{F}' \to \mathcal{F} \to \mathcal{L} \to \mathcal{G}' $$ The map in the middle comes from a unique element of $\Hom_\mathcal{O}(\mathcal{F}, \mathcal{L})$ by Modules on Sites, Lemma \ref{sites-modules-lemma-i-star-equivalence}. Conversely, given an element $\alpha$ of this group we can add the composition (as displayed above with $\alpha$ in the middle) to $\varphi'$. Some details omitted.