**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.
$\square$

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