Lemma 17.31.2. In the situation above there is a canonical isomorphism $\mathop{N\! L}\nolimits (\alpha ) = \mathop{N\! L}\nolimits _{\mathcal{B}/\mathcal{A}}$ in $D(\mathcal{B})$.

**Proof.**
Observe that $\mathop{N\! L}\nolimits _{\mathcal{B}/\mathcal{A}} = \mathop{N\! L}\nolimits (\text{id}_\mathcal {B})$. Thus it suffices to show that given two maps $\alpha _ i : \mathcal{E}_ i \to \mathcal{B}$ as above, there is a canonical quasi-isomorphism $\mathop{N\! L}\nolimits (\alpha _1) = \mathop{N\! L}\nolimits (\alpha _2)$ in $D(\mathcal{B})$. To see this set $\mathcal{E} = \mathcal{E}_1 \amalg \mathcal{E}_2$ and $\alpha = \alpha _1 \amalg \alpha _2 : \mathcal{E} \to \mathcal{B}$. Set $\mathcal{J}_ i = \mathop{\mathrm{Ker}}(\mathcal{A}[\mathcal{E}_ i] \to \mathcal{B})$ and $\mathcal{J} = \mathop{\mathrm{Ker}}(\mathcal{A}[\mathcal{E}] \to \mathcal{B})$. We obtain maps $\mathcal{A}[\mathcal{E}_ i] \to \mathcal{A}[\mathcal{E}]$ which send $\mathcal{J}_ i$ into $\mathcal{J}$. Thus we obtain canonical maps of complexes

and it suffices to show these maps are quasi-isomorphism. To see this it suffices to check on stalks (Lemma 17.3.1). If $x \in X$ then the stalk of $\mathop{N\! L}\nolimits (\alpha )$ is the complex $\mathop{N\! L}\nolimits (\alpha _ x)$ of Algebra, Section 10.134 associated to the presentation $\mathcal{A}_ x[\mathcal{E}_ x] \to \mathcal{B}_ x$ coming from the map $\alpha _ x : \mathcal{E}_ x \to \mathcal{B}_ x$. (Some details omitted; use Lemma 17.28.7 to see compatibility of forming differentials and taking stalks.) We conclude the result holds by Algebra, Lemma 10.134.2. $\square$

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