The Stacks project

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20.45 Projection formula

In this section we collect variants of the projection formula. The most basic version is Lemma 20.45.2. After we state and prove it, we discuss a more general version involving perfect complexes.

Lemma 20.45.1. Let $X$ be a ringed space. Let $\mathcal{I}$ be an injective $\mathcal{O}_ X$-module. Let $\mathcal{E}$ be an $\mathcal{O}_ X$-module. Assume $\mathcal{E}$ is finite locally free on $X$, see Modules, Definition 17.14.1. Then $\mathcal{E} \otimes _{\mathcal{O}_ X} \mathcal{I}$ is an injective $\mathcal{O}_ X$-module.

Proof. This is true because under the assumptions of the lemma we have

\[ \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{E} \otimes _{\mathcal{O}_ X} \mathcal{I}) = \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}( \mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{E}^\vee , \mathcal{I}) \]

where $\mathcal{E}^\vee = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{E}, \mathcal{O}_ X)$ is the dual of $\mathcal{E}$ which is finite locally free also. Since tensoring with a finite locally free sheaf is an exact functor we win by Homology, Lemma 12.24.2. $\square$

Lemma 20.45.2. Let $f : X \to Y$ be a morphism of ringed spaces. Let $\mathcal{F}$ be an $\mathcal{O}_ X$-module. Let $\mathcal{E}$ be an $\mathcal{O}_ Y$-module. Assume $\mathcal{E}$ is finite locally free on $Y$, see Modules, Definition 17.14.1. Then there exist isomorphisms

\[ \mathcal{E} \otimes _{\mathcal{O}_ Y} R^ qf_*\mathcal{F} \longrightarrow R^ qf_*(f^*\mathcal{E} \otimes _{\mathcal{O}_ X} \mathcal{F}) \]

for all $q \geq 0$. In fact there exists an isomorphism

\[ \mathcal{E} \otimes _{\mathcal{O}_ Y} Rf_*\mathcal{F} \longrightarrow Rf_*(f^*\mathcal{E} \otimes _{\mathcal{O}_ X} \mathcal{F}) \]

in $D^{+}(Y)$ functorial in $\mathcal{F}$.

Proof. Choose an injective resolution $\mathcal{F} \to \mathcal{I}^\bullet $ on $X$. Note that $f^*\mathcal{E}$ is finite locally free also, hence we get a resolution

\[ f^*\mathcal{E} \otimes _{\mathcal{O}_ X} \mathcal{F} \longrightarrow f^*\mathcal{E} \otimes _{\mathcal{O}_ X} \mathcal{I}^\bullet \]

which is an injective resolution by Lemma 20.45.1. Apply $f_*$ to see that

\[ Rf_*(f^*\mathcal{E} \otimes _{\mathcal{O}_ X} \mathcal{F}) = f_*(f^*\mathcal{E} \otimes _{\mathcal{O}_ X} \mathcal{I}^\bullet ). \]

Hence the lemma follows if we can show that $f_*(f^*\mathcal{E} \otimes _{\mathcal{O}_ X} \mathcal{F}) = \mathcal{E} \otimes _{\mathcal{O}_ Y} f_*(\mathcal{F})$ functorially in the $\mathcal{O}_ X$-module $\mathcal{F}$. This is clear when $\mathcal{E} = \mathcal{O}_ Y^{\oplus n}$, and follows in general by working locally on $Y$. Details omitted. $\square$

Let $f : X \to Y$ be a morphism of ringed spaces. Let $E \in D(\mathcal{O}_ X)$ and $K \in D(\mathcal{O}_ Y)$. Without any further assumptions there is a map
\begin{equation} \label{cohomology-equation-projection-formula-map} Rf_*E \otimes ^\mathbf {L}_{\mathcal{O}_ Y} K \longrightarrow Rf_*(E \otimes ^\mathbf {L}_{\mathcal{O}_ X} Lf^*K) \end{equation}

Namely, it is the adjoint to the canonical map

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

coming from the map $Lf^*Rf_*E \to E$ and Lemmas 20.28.3 and 20.29.1. A reasonably general version of the projection formula is the following.

Lemma 20.45.3. Let $f : X \to Y$ be a morphism of ringed spaces. Let $E \in D(\mathcal{O}_ X)$ and $K \in D(\mathcal{O}_ Y)$. If $K$ is perfect, then

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

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

Proof. To check ( is an isomorphism we may work locally on $Y$, i.e., we have to find a covering $\{ V_ j \to Y\} $ 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 $\mathcal{O}_ Y$-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 $\mathcal{O}_ Y$-modules. In this case a simple argument involving stupid truncations reduces the statement to the case where $K$ is represented by a finite free $\mathcal{O}_ Y$-module. Since the statement is invariant under finite direct summands in the $K$ variable, we conclude it suffices to prove it for $K = \mathcal{O}_ Y[n]$ in which case it is trivial. $\square$

Here is a case where the projection formula is true in complete generality.

Lemma 20.45.4. Let $f : X \to Y$ be a morphism of ringed spaces such that $f$ is a homeomorphism onto a closed subset. Then ( is an isomorphism always.

Proof. Since $f$ is a homeomorphism onto a closed subset, the functor $f_*$ is exact (Modules, Lemma 17.6.1). Hence $Rf_*$ is computed by applying $f_*$ to any representative complex. Choose a K-flat complex $\mathcal{K}^\bullet $ of $\mathcal{O}_ Y$-modules representing $K$ and choose any complex $\mathcal{E}^\bullet $ of $\mathcal{O}_ X$-modules representing $E$. Then $Lf^*K$ is represented by $f^*\mathcal{K}^\bullet $ which is a K-flat complex of $\mathcal{O}_ X$-modules (Lemma 20.27.7). Thus the right hand side of ( is represented by

\[ f_*\text{Tot}(\mathcal{E}^\bullet \otimes _{\mathcal{O}_ X} f^*\mathcal{K}^\bullet ) \]

By the same reasoning we see that the left hand side is represented by

\[ \text{Tot}(f_*\mathcal{E}^\bullet \otimes _{\mathcal{O}_ Y} \mathcal{K}^\bullet ) \]

Since $f_*$ commutes with direct sums (Modules, Lemma 17.6.3) it suffices to show that

\[ f_*(\mathcal{E} \otimes _{\mathcal{O}_ X} f^*\mathcal{K}) = f_*\mathcal{E} \otimes _{\mathcal{O}_ Y} \mathcal{K} \]

for any $\mathcal{O}_ X$-module $\mathcal{E}$ and $\mathcal{O}_ Y$-module $\mathcal{K}$. We will check this by checking on stalks. Let $y \in Y$. If $y \not\in f(X)$, then the stalks of both sides are zero. If $y = f(x)$, then we see that we have to show

\[ \mathcal{E}_ x \otimes _{\mathcal{O}_{X, x}} (\mathcal{O}_{X, x} \otimes _{\mathcal{O}_{Y, y}} \mathcal{F}_ y) = \mathcal{E}_ x \otimes _{\mathcal{O}_{Y, y}} \mathcal{F}_ y \]

(using Sheaves, Lemma 6.32.1 and Lemma 6.26.4). This equality holds and therefore the lemma has been proved. $\square$

Remark 20.45.5. The map ( is compatible with the base change map of Remark 20.29.3 in the following sense. Namely, suppose that

\[ \xymatrix{ X' \ar[r]_{g'} \ar[d]_{f'} & X \ar[d]^ f \\ Y' \ar[r]^ g & Y } \]

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

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

is commutative. Here arrows labeled $t$ are gotten by an application of Lemma 20.28.3, arrows labeled $b$ by an application of Remark 20.29.3, arrows labeled $p$ by an application of (, and $c$ comes from $L(g')^* \circ Lf^* = L(f')^* \circ Lg^*$. We omit the verification.

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