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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