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 $K \in D(\mathcal{O}_\mathcal {D})$. Without any further assumptions there is a map

21.50.0.1
\begin{equation} \label{sites-cohomology-equation-projection-formula-map} Rf_*E \otimes ^\mathbf {L}_{\mathcal{O}_\mathcal {D}} K \longrightarrow Rf_*(E \otimes ^\mathbf {L}_{\mathcal{O}_\mathcal {C}} Lf^*K) \end{equation}

Namely, it is the adjoint to the canonical map

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

coming from the map $Lf^*Rf_*E \to E$ and Lemmas 21.18.4 and 21.19.1. A reasonably general version of the projection formula is the following.

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 (0)