Lemma 91.13.3. If there exists a solution to (91.13.0.1), then the set of isomorphism classes of solutions is principal homogeneous under $\mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {O}( \mathop{N\! L}\nolimits _{\mathcal{O}/\mathcal{O}_\mathcal {B}}, \mathcal{G})$.

Proof. We observe right away that given two solutions $\mathcal{O}'_1$ and $\mathcal{O}'_2$ to (91.13.0.1) we obtain by Lemma 91.13.1 an obstruction element $o(\mathcal{O}'_1, \mathcal{O}'_2) \in \mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {O}( \mathop{N\! L}\nolimits _{\mathcal{O}/\mathcal{O}_\mathcal {B}}, \mathcal{G})$ to the existence of a map $\mathcal{O}'_1 \to \mathcal{O}'_2$. 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 $\mathcal{O}'$ and an element $\xi \in \mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {O}( \mathop{N\! L}\nolimits _{\mathcal{O}/\mathcal{O}_\mathcal {B}}, \mathcal{G})$ we can find a second solution $\mathcal{O}'_\xi$ such that $o(\mathcal{O}', \mathcal{O}'_\xi ) = \xi$.

Pick $\alpha : \mathcal{E} \to \mathcal{O}$ as in Lemma 91.13.2 for the class $\xi$. Consider the surjection $f^{-1}\mathcal{O}_\mathcal {B}[\mathcal{E}] \to \mathcal{O}$ with kernel $\mathcal{I}$ and corresponding naive cotangent complex $\mathop{N\! L}\nolimits (\alpha ) = (\mathcal{I}/\mathcal{I}^2 \to \Omega _{f^{-1}\mathcal{O}_\mathcal {B}[\mathcal{E}]/ f^{-1}\mathcal{O}_\mathcal {B}} \otimes _{f^{-1}\mathcal{O}_\mathcal {B}[\mathcal{E}]} \mathcal{O})$. By the lemma $\xi$ is the class of a morphism $\delta : \mathcal{I}/\mathcal{I}^2 \to \mathcal{G}$. After replacing $\mathcal{E}$ by $\mathcal{E} \times _\mathcal {O} \mathcal{O}'$ we may also assume that $\alpha$ factors through a map $\alpha ' : \mathcal{E} \to \mathcal{O}'$.

These choices determine an $f^{-1}\mathcal{O}_{\mathcal{B}'}$-algebra map $\varphi : \mathcal{O}_{\mathcal{B}'}[\mathcal{E}] \to \mathcal{O}'$. Let $\mathcal{I}' = \mathop{\mathrm{Ker}}(\varphi )$. Observe that $\varphi$ induces a map $\varphi |_{\mathcal{I}'} : \mathcal{I}' \to \mathcal{G}$ and that $\mathcal{O}'$ is the pushout, as in the following diagram

$\xymatrix{ 0 \ar[r] & \mathcal{G} \ar[r] & \mathcal{O}' \ar[r] & \mathcal{O} \ar[r] & 0 \\ 0 \ar[r] & \mathcal{I}' \ar[u]^{\varphi |_{\mathcal{I}'}} \ar[r] & f^{-1}\mathcal{O}_{\mathcal{B}'}[\mathcal{E}] \ar[u] \ar[r] & \mathcal{O} \ar[u]_{=} \ar[r] & 0 }$

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 $\mathcal{O}'_\xi$ fitting into a diagram as above. A calculation (omitted) shows that $o(\mathcal{O}', \mathcal{O}'_\xi ) = \xi$ as desired. $\square$

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