and moreover how unique the solution is (if it exists). More precisely, we look for a surjection of $A'$-algebras $B' \to B$ whose kernel is an ideal of square zero and is identified with $N$ such that $A' \to B'$ induces the given map $c$. We will say $B'$ is a *solution* to (91.16.0.1).

**Proof.**
Via the identifications $\mathop{N\! L}\nolimits _{B/A} = \tau _{\geq -1}L_{B/A}$ (Lemma 91.11.3) and $H^0(L_{B/A}) = \Omega _{B/A}$ (Lemma 91.4.5) we have seen parts (2) and (3) in Deformation Theory, Lemmas 90.2.1 and 90.2.2.

Proof of (1). Roughly speaking, this follows from the discussion in Deformation Theory, Remark 90.2.8 by replacing the naive cotangent complex by the full cotangent complex. Here is a more detailed explanation. By Deformation Theory, Lemma 90.2.7 and Remark 90.2.8 there exists an element

\[ \xi ' \in \mathop{\mathrm{Ext}}\nolimits ^1_ A(\mathop{N\! L}\nolimits _{A/A'}, N) = \mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{A/A'} \otimes _ A^\mathbf {L} B, N) = \mathop{\mathrm{Ext}}\nolimits ^1_ B(L_{A/A'} \otimes _ A^\mathbf {L} B, N) \]

(for the equalities see Deformation Theory, Remark 90.2.8 and use that $\mathop{N\! L}\nolimits _{A'/A} = \tau _{\geq -1} L_{A'/A}$) such that a solution exists if and only if this element is in the image of the map

\[ \mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A'}, N) = \mathop{\mathrm{Ext}}\nolimits ^1_ B(L_{B/A'}, N) \longrightarrow \mathop{\mathrm{Ext}}\nolimits ^1_ B(L_{A/A'} \otimes _ A^\mathbf {L} B, N) \]

The distinguished triangle (91.7.0.1) for $A' \to A \to B$ gives rise to a long exact sequence

\[ \ldots \to \mathop{\mathrm{Ext}}\nolimits ^1_ B(L_{B/A'}, N) \to \mathop{\mathrm{Ext}}\nolimits ^1_ B(L_{A/A'} \otimes _ A^\mathbf {L} B, N) \to \mathop{\mathrm{Ext}}\nolimits ^2_ B(L_{B/A}, N) \to \ldots \]

Hence taking $\xi $ the image of $\xi '$ works.
$\square$

## Comments (2)

Comment #7112 by Amnon Yekutieli on

Comment #7279 by Johan on