
## 21.46 Projection formula

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

21.46.0.1
$$\label{sites-cohomology-equation-projection-formula-map} Rf_*E \otimes ^\mathbf {L}_{\mathcal{O}_\mathcal {D}} K \longrightarrow Rf_*(E \otimes ^\mathbf {L}_{\mathcal{O}_\mathcal {C}} Lf^*K)$$

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 21.19.4 and 21.20.1. A reasonably general version of the projection formula is the following.

Lemma 21.46.1. 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 (21.46.0.1) 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. $\square$

Remark 21.46.2. The map (21.46.0.1) is compatible with the base change map of Remark 21.20.3 in the following sense. Namely, suppose that

$\xymatrix{ (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}'), \mathcal{O}_{\mathcal{C}'}) \ar[r]_{g'} \ar[d]_{f'} & (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O}_\mathcal {C}) \ar[d]^ f \\ (\mathop{\mathit{Sh}}\nolimits (\mathcal{D}'), \mathcal{O}_{\mathcal{D}'}) \ar[r]^ g & (\mathop{\mathit{Sh}}\nolimits (\mathcal{D}), \mathcal{O}_\mathcal {D}) }$

is a commutative diagram of ringed topoi. Let $E \in D(\mathcal{O}_\mathcal {C})$ and $K \in D(\mathcal{O}_\mathcal {D})$. Then the diagram

$\xymatrix{ Lg^*(Rf_*E \otimes ^\mathbf {L}_{\mathcal{O}_\mathcal {D}} K) \ar[r]_ p \ar[d]_ t & Lg^*Rf_*(E \otimes ^\mathbf {L}_{\mathcal{O}_\mathcal {C}} Lf^*K) \ar[d]_ b \\ Lg^*Rf_*E \otimes ^\mathbf {L}_{\mathcal{O}_{\mathcal{D}'}} Lg^*K \ar[d]_ b & Rf'_*L(g')^*(E \otimes ^\mathbf {L}_{\mathcal{O}_\mathcal {C}} Lf^*K) \ar[d]_ t \\ Rf'_*L(g')^*E \otimes ^\mathbf {L}_{\mathcal{O}_{\mathcal{D}'}} Lg^*K \ar[rd]_ p & Rf'_*(L(g')^*E \otimes ^\mathbf {L}_{\mathcal{O}_{\mathcal{D}'}} L(g')^*Lf^*K) \ar[d]_ c \\ & Rf'_*(L(g')^*E \otimes ^\mathbf {L}_{\mathcal{O}_{\mathcal{D}'}} L(f')^*Lg^*K) }$

is commutative. Here arrows labeled $t$ are gotten by an application of Lemma 21.19.4, arrows labeled $b$ by an application of Remark 21.20.3, arrows labeled $p$ by an application of (21.46.0.1), and $c$ comes from $L(g')^* \circ Lf^* = L(f')^* \circ Lg^*$. We omit the verification.

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