Lemma 15.79.1. Let $R \to S$ and $S \to S'$ be ring maps. The canonical map $\mathop{N\! L}\nolimits _{S/R} \otimes _ S^\mathbf {L} S' \to \mathop{N\! L}\nolimits _{S/R} \otimes _ S S'$ induces an isomorphism $\tau _{\geq -1}(\mathop{N\! L}\nolimits _{S/R} \otimes _ S^\mathbf {L} S') \to \mathop{N\! L}\nolimits _{S/R} \otimes _ S S'$ in $D(S')$. Similarly, given a presentation $\alpha $ of $S$ over $R$ the canonical map $\mathop{N\! L}\nolimits (\alpha ) \otimes _ S^\mathbf {L} S' \to \mathop{N\! L}\nolimits (\alpha ) \otimes _ S S'$ induces an isomorphism $\tau _{\geq -1}(\mathop{N\! L}\nolimits (\alpha ) \otimes _ S^\mathbf {L} S') \to \mathop{N\! L}\nolimits (\alpha ) \otimes _ S S'$ in $D(S')$.

**Proof.**
Let $K^\bullet $ be a complex of $S$-modules with $K^ n = 0$ for $n \not\in [-1, 0]$ and $K^0$ flat, for example $K^\bullet = \mathop{N\! L}\nolimits _{S/R}$ or $K^\bullet = \mathop{N\! L}\nolimits (\alpha )$. Then we have a distinguished triangle

in $D(S)$. This determines a map of distinguished triangles

The left and right vertical arrows are isomorphisms as $K^0$ is flat. Since $K^{-1} \otimes _ S^\mathbf {L} S' \to K^{-1} \otimes _ S S'$ is an isomorphism on cohomology in degree $0$ we conclude. $\square$

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