Lemma 18.35.2. 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 we argue as follows. First, observe that $H^0(\mathop{N\! L}\nolimits (\alpha _ i)) = \Omega _{\mathcal{B}/\mathcal{A}}$ and $H^0(\mathop{N\! L}\nolimits (\alpha )) = \Omega _{\mathcal{B}/\mathcal{A}}$ by Lemma 18.33.8 hence the map is an isomorphism on cohomology sheaves in degree $0$. Similarly, we claim that $H^{-1}(\mathop{N\! L}\nolimits (\alpha _ i))$ and $H^{-1}(\mathop{N\! L}\nolimits (\alpha ))$ are the sheaves associated to the presheaf $U \mapsto H_1(L_{\mathcal{B}(U)/\mathcal{A}(U)})$ where $H_1(L_{-/-})$ is as in Algebra, Definition 10.134.1. If the claim holds, then the proof is finished.

Proof of the claim. Let $\alpha : \mathcal{E} \to \mathcal{B}$ be as above. Let $\mathcal{B}' \subset \mathcal{B}$ be the subpresheaf of $\mathcal{A}$-algebras whose value on $U$ is the image of $\mathcal{A}(U)[\mathcal{E}(U)] \to \mathcal{B}(U)$. Let $\mathcal{I}'$ be the presheaf whose value on $U$ is the kernel of $\mathcal{A}(U)[\mathcal{E}(U)] \to \mathcal{B}(U)$. Then $\mathcal{I}$ is the sheafification of $\mathcal{I}'$ and $\mathcal{B}$ is the sheafification of $\mathcal{B}'$. Similarly, $H^{-1}(\mathop{N\! L}\nolimits (\alpha ))$ is the sheafification of the presheaf

by Lemma 18.33.4. By Algebra, Lemma 10.134.2 we conclude $H^{-1}(\mathop{N\! L}\nolimits (\alpha ))$ is the sheaf associated to the presheaf $U \mapsto H_1(L_{\mathcal{B}'(U)/\mathcal{A}(U)})$. Thus we have to show that the maps $H_1(L_{\mathcal{B}'(U)/\mathcal{A}(U)}) \to H_1(L_{\mathcal{B}(U)/\mathcal{A}(U)})$ induce an isomorphism $\mathcal{H}'_1 \to \mathcal{H}_1$ of sheafifications.

Injectivity of $\mathcal{H}'_1 \to \mathcal{H}_1$. Let $f \in H_1(L_{\mathcal{B}'(U)/\mathcal{A}(U)})$ map to zero in $\mathcal{H}_1(U)$. To show: $f$ maps to zero in $\mathcal{H}'_1(U)$. The assumption means there is a covering $\{ U_ i \to U\} $ such that $f$ maps to zero in $H_1(L_{\mathcal{B}(U_ i)/\mathcal{A}(U_ i)})$ for all $i$. Replace $U$ by $U_ i$ to get to the point where $f$ maps to zero in $H_1(L_{\mathcal{B}(U)/\mathcal{A}(U)})$. By Algebra, Lemma 10.134.9 we can find a finitely generated subalgebra $\mathcal{B}'(U) \subset B \subset \mathcal{B}(U)$ such that $f$ maps to zero in $H_1(L_{B/\mathcal{A}(U)})$. Since $\mathcal{B} = (\mathcal{B}')^\# $ we can find a covering $\{ U_ i \to U\} $ such that $B \to \mathcal{B}(U_ i)$ factors through $\mathcal{B}'(U_ i)$. Hence $f$ maps to zero in $H_1(L_{\mathcal{B}'(U_ i)/\mathcal{A}(U_ i)})$ as desired.

The surjectivity of $\mathcal{H}'_1 \to \mathcal{H}_1$ is proved in exactly the same way. $\square$

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