Lemma 90.6.2. If (90.6.0.1) induces a quasi-isomorphism $B \otimes _ A^\mathbf {L} A' = B'$, then the functoriality map induces an isomorphism

$L_{B/A} \otimes _ B^\mathbf {L} B' \longrightarrow L_{B'/A'}$

Proof. We will use the notation introduced in Equation (90.6.0.2). We have

$L_{B/A} \otimes _ B^\mathbf {L} B' = L\pi _!(\Omega _{\mathcal{O}/A} \otimes _\mathcal {O} \underline{B}) \otimes _ B^\mathbf {L} B' = L\pi _!(Lh^*(\Omega _{\mathcal{O}/A} \otimes _\mathcal {O} \underline{B}))$

the first equality by Lemma 90.4.3 and the second by Cohomology on Sites, Lemma 21.38.6. Since $\Omega _{\mathcal{O}/A}$ is a flat $\mathcal{O}$-module, we see that $\Omega _{\mathcal{O}/A} \otimes _\mathcal {O} \underline{B}$ is a flat $\underline{B}$-module. Thus $Lh^*(\Omega _{\mathcal{O}/A} \otimes _\mathcal {O} \underline{B}) = \Omega _{\mathcal{O}/A} \otimes _\mathcal {O} \underline{B'}$ which is equal to $g^{-1}(\Omega _{\mathcal{O}'/A'} \otimes _{\mathcal{O}'} \underline{B'})$ by inspection. we conclude by Lemma 90.6.1 and the fact that $L_{B'/A'}$ is computed by $L\pi '_!(\Omega _{\mathcal{O}'/A'} \otimes _{\mathcal{O}'} \underline{B'})$. $\square$

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