$\newcommand{\H}{\operatorname{Hom}{}}$ $\newcommand{\HH}{\mathscr{H}}$

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

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


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

but I don't find it.

Have already tried adjunction formula without success ...


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.