Lemma 20.52.2. 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 20.52.1. 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. $\square$

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