Lemma 91.7.3. If there exists a solution to (91.7.0.1), then the set of isomorphism classes of solutions is principal homogeneous under $\mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(\mathop{N\! L}\nolimits _{X/S}, \mathcal{G})$.
Proof. We observe right away that given two solutions $X'_1$ and $X'_2$ to (91.7.0.1) we obtain by Lemma 91.7.1 an obstruction element $o(X'_1, X'_2) \in \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(\mathop{N\! L}\nolimits _{X/S}, \mathcal{G})$ 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 $\xi \in \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(\mathop{N\! L}\nolimits _{X/S}, \mathcal{G})$ we can find a second solution $X'_\xi $ such that $o(X', X'_\xi ) = \xi $.
Pick $\alpha : \mathcal{E} \to \mathcal{O}_ X$ as in Lemma 91.7.2 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 $\mathop{N\! L}\nolimits (\alpha ) = (\mathcal{I}/\mathcal{I}^2 \to \Omega _{f^{-1}\mathcal{O}_ S[\mathcal{E}]/f^{-1}\mathcal{O}_ S} \otimes _{f^{-1}\mathcal{O}_ S[\mathcal{E}]} \mathcal{O}_ X)$. 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 $\alpha ' : \mathcal{E} \to \mathcal{O}_{X'}$.
These choices determine an $f^{-1}\mathcal{O}_{S'}$-algebra map $\varphi : \mathcal{O}_{S'}[\mathcal{E}] \to \mathcal{O}_{X'}$. Let $\mathcal{I}' = \mathop{\mathrm{Ker}}(\varphi )$. 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
Let $\psi : \mathcal{I}' \to \mathcal{G}$ be the sum of the map $\varphi |_{\mathcal{I}'}$ and the composition
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. $\square$
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