**Proof.**
Via the identifications $\mathop{N\! L}\nolimits _{B/A} = \tau _{\geq -1}L_{B/A}$ (Lemma 89.10.3) and $H^0(L_{B/A}) = \Omega _{B/A}$ (Lemma 89.4.5) we have seen parts (2) and (3) in Deformation Theory, Lemmas 88.2.1 and 88.2.3.

Proof of (1). We will use the results of Deformation Theory, Lemma 88.2.4 without further mention. Let $\alpha \in \mathop{\mathrm{Ext}}\nolimits ^1_ A(\mathop{N\! L}\nolimits _{A/\mathbf{Z}}, I)$ be the element corresponding to the isomorphism class of $A'$. The existence of $B'$ corresponds to an element $\beta \in \mathop{\mathrm{Ext}}\nolimits _ B^1(\mathop{N\! L}\nolimits _{B/\mathbf{Z}}, N)$ which maps to the image of $\alpha $ in $\mathop{\mathrm{Ext}}\nolimits ^1_ A(\mathop{N\! L}\nolimits _{A/\mathbf{Z}}, N)$. Note that

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

and

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

by Lemma 89.10.3. Since the distinguished triangle (89.7.0.1) for $\mathbf{Z} \to A \to B$ gives rise to a long exact sequence

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

we obtain the result with $\xi $ the image of $\alpha $.
$\square$

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