Lemma 68.4.4. With $S$, $X$, $Y$, $\pi $, $\mathcal{A}$, $\mathcal{B}$, $\varphi $, and $f$ as in Lemma 68.4.3 we have

in $D(\mathcal{B})$ for any $K \in D(\mathcal{B})$ and $M \in D(\mathcal{A})$.

Lemma 68.4.4. With $S$, $X$, $Y$, $\pi $, $\mathcal{A}$, $\mathcal{B}$, $\varphi $, and $f$ as in Lemma 68.4.3 we have

\[ K \otimes _\mathcal {B}^\mathbf {L} Rf_*M = Rf_*(Lf^*K \otimes _\mathcal {A}^\mathbf {L} M) \]

in $D(\mathcal{B})$ for any $K \in D(\mathcal{B})$ and $M \in D(\mathcal{A})$.

**Proof.**
Since $f_*$ is exact (Lemma 68.4.1) the functor $Rf_*$ is computed by applying $f_*$ to any representative complex. Choose a complex $\mathcal{K}^\bullet $ of $\mathcal{B}$-modules representing $K$ which is K-flat with flat terms, see Cohomology on Sites, Lemma 21.17.11. Then $f^*\mathcal{K}^\bullet $ is K-flat with flat terms, see Cohomology on Sites, Lemma 21.18.1. Choose any complex $\mathcal{M}^\bullet $ of $\mathcal{A}$-modules representing $M$. Then we have to show

\[ \text{Tot}(\mathcal{K}^\bullet \otimes _\mathcal {B} f_*\mathcal{M}^\bullet ) = f_*\text{Tot}(f^*\mathcal{K}^\bullet \otimes _\mathcal {A} \mathcal{M}^\bullet ) \]

because by our choices these complexes represent the right and left hand side of the formula in the lemma. Since $f_*$ commutes with direct sums (for example by the description of the stalks in Lemma 68.4.2), this reduces to the equalities

\[ \mathcal{K}^ n \otimes _\mathcal {B} f_*\mathcal{M}^ m = f_*(f^*\mathcal{K}^ n \otimes _\mathcal {A} \mathcal{M}^ m) \]

which are true by Lemma 68.4.3. $\square$

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