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The exercise asks to prove the projection formula;

Let $f: X \to Y$ be a morphism of ringed spaces, let $\mathscr{F}$ be an $\mathscr{O}_X$-module, and let $\mathscr{E}$ be a locally free $\mathscr{O}_Y$-module of finite rank. Prove the projection formula $$ R^i f_* ( \mathscr{F} \otimes f^* \mathscr{E}) \cong R^i f_* (\mathscr{F}) \otimes \mathscr{E}. $$

In my solution, I needed to refer to another projection formula which is stated as exercise II.5.1.(d). Is there a way to prove the statement without using Ex. II.5.1.(d)?

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    $\begingroup$ To be frank, I don't know a proof without using that exercise. Then again, and I apologize if this was trivially obvious to you, but you do realize that the other exercise is the $i=0$ case of the exercise you state? So, you're asking whether you can prove it for all $i$ in one go? $\endgroup$ – Alex Youcis Mar 6 '16 at 6:41
  • $\begingroup$ @AlexYoucis In my solution, I am using the previous exercise to identify $f_* ( \mathscr{I}_j \otimes f^* \mathscr{E}) \cong f_* (\mathscr{I}_j) \otimes \mathscr{E}$, where $\mathscr{I}_j$ is the $j$th component in the injective resolution of $\mathscr{F}$. Once this is done, I believe the rest is taking homologies of the two complexes. I am curious if there is a solution using the case $i =0$ in a different way. $\endgroup$ – Youngsu Mar 6 '16 at 21:30
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I'm not sure what you are looking for, but the following argument from EGAIII$_1$, Prop. 12.2.3 uses the case $i = 0$ in a "different way".

First, let $\mathscr{F}$ and $\mathscr{G}$ be two $\mathcal{O}_X$-modules. There is then a cup product $$H^p(f^{-1}(V),\mathscr{F}) \otimes_{\Gamma(f^{-1}(V),\mathcal{O}_X)} H^q(f^{-1}(V),\mathscr{G}) \overset{\smile}{\longrightarrow} H^{p+q}(f^{-1}(V),\mathscr{F} \otimes_{\mathcal{O}_X} \mathscr{G})$$ on cohomology, which is compatible with restriction [Godement, II, 6.6; Stacks, Tag 01FP]. There is moreover a morphism $$H^p(f^{-1}(V),\mathscr{F}) \otimes_{\Gamma(V,\mathcal{O}_Y)} H^q(f^{-1}(V),\mathscr{G}) \longrightarrow H^p(f^{-1}(V),\mathscr{F}) \otimes_{\Gamma(f^{-1}(V),\mathcal{O}_X)} H^q(f^{-1}(V),\mathscr{G})$$ obtained via restricting scalars through the map $\Gamma(V,\mathcal{O}_Y) \to \Gamma(V,f_*\mathcal{O}_X) = \Gamma(f^{-1}(V),\mathcal{O}_X)$. By applying Prop. III.8.1 and the definition of sheafification, this gives a morphism of sheaves $$R^pf_*(\mathscr{F}) \otimes_{\mathcal{O}_Y} R^qf_*(\mathscr{G}) \longrightarrow R^{p+q}(\mathscr{F} \otimes_{\mathcal{O}_X} \mathscr{G})$$ for each $p,q$.

Now let $\mathscr{G} = f^*(\mathscr{E})$, and let $q = 0$. We then obtain a morphism $$R^pf_*(\mathscr{F}) \otimes_{\mathcal{O}_Y} \mathscr{E} \longrightarrow R^pf_*(\mathscr{F}) \otimes_{\mathcal{O}_Y} f_*(f^*(\mathscr{E})) \longrightarrow R^{p}(\mathscr{F} \otimes_{\mathcal{O}_X} f^*(\mathscr{E}))$$ where the first map is the identity on the first factor and the canonical map $\mathscr{E} \to f_*(f^*(\mathscr{E}))$ in the second factor.

Now to show this isomorphism, as in what is probably your proof for Exercise II.5.1(d), we can show it locally on an open cover trivializing $\mathscr{E}$, and moreover reduce to the case $\mathscr{E} = \mathcal{O}_Y$ since all of our functors are additive. But if $\mathscr{E} = \mathcal{O}_Y$, the composition above is $$R^pf_*(\mathscr{F}) \otimes_{\mathcal{O}_Y} \mathcal{O}_Y \longrightarrow R^{p}(\mathscr{F} \otimes_{\mathcal{O}_X} f^*(\mathcal{O}_Y))$$ which is an isomorphism since $f^*(\mathcal{O}_Y) = \mathcal{O}_X$.

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