Lemma 91.6.1. Assume (91.6.0.1) induces a quasi-isomorphism $B \otimes _ A^\mathbf {L} A' = B'$. Then, with notation as in (91.6.0.2) and $\mathcal{F}' \in \textit{Ab}(\mathcal{C}')$, we have $L\pi _!(g^{-1}\mathcal{F}') = L\pi '_!(\mathcal{F}')$.

Proof. We use the results of Remark 91.5.5 without further mention. We will apply Cohomology on Sites, Lemma 21.39.8. Let $P_\bullet \to B$ be a resolution. If we can show that $u(P_\bullet ) = P_\bullet \otimes _ A A' \to B'$ is a quasi-isomorphism, then we are done. The complex of $A$-modules $s(P_\bullet )$ associated to $P_\bullet$ (viewed as a simplicial $A$-module) is a free $A$-module resolution of $B$. Namely, $P_ n$ is a free $A$-module and $s(P_\bullet ) \to B$ is a quasi-isomorphism. Thus $B \otimes _ A^\mathbf {L} A'$ is computed by $s(P_\bullet ) \otimes _ A A' = s(P_\bullet \otimes _ A A')$. Therefore the assumption of the lemma signifies that $\epsilon ' : P_\bullet \otimes _ A A' \to B'$ is a quasi-isomorphism. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).