Lemma 91.2.2. If there exists a solution to (91.2.0.1), then the set of isomorphism classes of solutions is principal homogeneous under $\mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}, N)$.

**Proof.**
We observe right away that given two solutions $B'_1$ and $B'_2$ to (91.2.0.1) we obtain by Lemma 91.2.1 an obstruction element $o(B'_1, B'_2) \in \mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}, N)$ 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 $\xi \in \mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}, N)$ we can find a second solution $B'_\xi $ such that $o(B', B'_\xi ) = \xi $.

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

in Algebra, Section 10.134. Since $\Omega _{A[E]/A} \otimes B$ is a free $B$-module we have

Thus we may represent $\xi $ as the class of a morphism $\delta : J/J^2 \to N$.

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 $J' = \mathop{\mathrm{Ker}}(A'[E] \to B)$. Observe that $\varphi $ induces a map $\varphi |_{J'} : J' \to N$ and that $B'$ is the pushout, as in the following diagram

Let $\psi : J' \to N$ be the sum of the map $\varphi |_{J'}$ and the composition

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. $\square$

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