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Let $f : X \to Y$ a quasi-compact separated morphism of schemes , F a quasi-coherent sheaf on X, $\mathcal{E}$ a locally free sheaf on Y.

In order to prove the projection formula $$f_{*}F\otimes {\mathcal {E}}\to f_{*}(F\otimes f^{*}{\mathcal {E}})$$

I encounter following problem:

Indeed, if I have a concrete map then I can show locally that it's a isomorphism, since locally I could assump $\mathcal{E}= \mathcal{O}_Y ^n$ as free and therefore conclude

$\begin{eqnarray*} f_*(F\otimes f^*E) &\;\cong\;& f_*(F\otimes \mathcal{O}_X^{\,n}) &\;\cong\;& f_*(F\otimes \mathcal{O}_X)^{n} &\;\cong\;& f_*(F)^{n} &\;\cong\;& f_*(F)\otimes \mathcal{O}_Y^{\,n} &\;\cong\;& f_*(F)\otimes E \end{eqnarray*}$

But here occurs my problem:

I need for doing such an argument a concrete morphism $f_{*}F\otimes {\mathcal {E}}\to f_{*}(F\otimes f^{*}{\mathcal {E}})$ which I can't find.

Surely, it suffice to find such one between the presheaves

$U \to f_{*}F(U)\otimes {\mathcal {E}}(U)$

and

$V \to f_{*}(F\otimes f^{*}{\mathcal {E}})(V)$

but I don't find it.

Have already tried adjunction formula without success ...

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The map $f_*F\otimes\mathcal{E}\to f_*(F\otimes f^*\mathcal{E})$ comes by adjunction from a map $f^*(f_*F\otimes\mathcal{E})\to F\otimes f^*\mathcal{E}$. But $f^*$ is monoidal, so $f^*(f_*F\otimes\mathcal{E})\simeq f^*f_*F\otimes f^*\mathcal{E}$. Then the map is induced by the counit $f^*f_*F\to F$. In conclusion, the map is :

$$ f_*F\otimes\mathcal{E}\xrightarrow{\eta} f_*f^*(f_*F\otimes\mathcal{E})\simeq f_*(f^*f_*F\otimes f^*\mathcal{E})\xrightarrow{\varepsilon}f_*(F\otimes f^*\mathcal{E}) $$

where $\eta:1\to f_*f^*$ is the unit and $\varepsilon:f^*f_*\to 1$ is the counit of the adjunction.

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