Lemma 21.50.1. Let $f : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O}_\mathcal {C}) \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{D}), \mathcal{O}_\mathcal {D})$ be a morphism of ringed topoi. Let $E \in D(\mathcal{O}_\mathcal {C})$ and $K \in D(\mathcal{O}_\mathcal {D})$. If $K$ is perfect, then

\[ Rf_*E \otimes ^\mathbf {L}_{\mathcal{O}_\mathcal {D}} K = Rf_*(E \otimes ^\mathbf {L}_{\mathcal{O}_\mathcal {C}} Lf^*K) \]

in $D(\mathcal{O}_\mathcal {D})$.

**Proof.**
To check (21.50.0.1) is an isomorphism we may work locally on $\mathcal{D}$, i.e., for any object $V$ of $\mathcal{D}$ we have to find a covering $\{ V_ j \to V\} $ such that the map restricts to an isomorphism on $V_ j$. By definition of perfect objects, this means we may assume $K$ is represented by a strictly perfect complex of $\mathcal{O}_\mathcal {D}$-modules. Note that, completely generally, the statement is true for $K = K_1 \oplus K_2$, if and only if the statement is true for $K_1$ and $K_2$. Hence we may assume $K$ is a finite complex of finite free $\mathcal{O}_\mathcal {D}$-modules. In this case a simple argument involving stupid truncations reduces the statement to the case where $K$ is represented by a finite free $\mathcal{O}_\mathcal {D}$-module. Since the statement is invariant under finite direct summands in the $K$ variable, we conclude it suffices to prove it for $K = \mathcal{O}_\mathcal {D}[n]$ in which case it is trivial.
$\square$

## Comments (2)

Comment #7139 by Hao Peng on

Comment #7292 by Johan on