Remark 90.2.8. Observe that in the situation of Lemma 90.2.7 we have

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

The first equality by More on Algebra, Lemma 15.60.3 and the second by More on Algebra, Lemma 15.85.1. We have maps of complexes

$\mathop{N\! L}\nolimits _{A'/A} \otimes _ A B \to \mathop{N\! L}\nolimits _{B/A'} \to \mathop{N\! L}\nolimits _{B/A}$

which is close to being a distinguished triangle, see Algebra, Lemma 10.134.4. If it were a distinguished triangle we would conclude that the image of $\xi$ in $\mathop{\mathrm{Ext}}\nolimits ^2_ B(\mathop{N\! L}\nolimits _{B/A}, N)$ would be the obstruction to the existence of a solution to (90.2.0.1).

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).