I'm trying to solve Exercise 5.1 of Chapter II of Hartshorne – Algebraic Geometry.
I'm fine with the first $3$ parts, but I'm having troubles with the very last part, which asks to prove the projection formula:
Let $f:X\to Y$ be a morphism of ringed spaces, $\mathscr{F}$ an $\mathcal{O}_X$-module and $\mathcal{E}$ a locally free $\mathcal{O}_Y$-module of finite rank. Then there is a natural isomorphism
$$ f_*(\mathscr{F}\otimes f^*\mathcal{E}) \;\cong\; f_*(\mathscr{F})\otimes \mathcal{E} $$
After thinking quite a long time about it, I checked on the internet and I found the following solution:
$$
\begin{eqnarray}
f_*(\mathscr{F}\otimes f^*\mathcal{E})
&\;\cong\;& f_*(\mathscr{F}\otimes \mathcal{O}_X^{\,n}) \\\\
&\;\cong\;& f_*(\mathscr{F}\otimes \mathcal{O}_X)^{n} \\\\
&\;\cong\;& f_*(\mathscr{F})^{n} \\\\
&\;\cong\;& f_*(\mathscr{F})\otimes \mathcal{O}_Y^{\,n} \\\\
&\;\cong\;& f_*(\mathscr{F})\otimes \mathcal{E} \\\\
\end{eqnarray}
$$
Is this correct? If it is, could you explain me why do we have the isomorphism
$$ f_*(\mathscr{F}\otimes \mathcal{O}_X^{\,n})
\;\cong\; f_*(\mathscr{F}\otimes \mathcal{O}_X)^{n} \quad ? $$
Best Answer
As said by Martin, first you have to find a morphism between the two sheaves, then you can use that solution locally. So, here is the morphism.
Consider on $Y$ the presheaf $P$ with sections $V\mapsto \mathscr{F}(f^{-1}(V))\otimes\mathcal{E}(V)$ for all $V\subseteq Y$ open. The sheafification of $P$ is $f_*(\mathscr{F})\otimes\mathcal{E}$.
Similarly, consider $P'$ the presheaf on $X$ with sections $U\mapsto\mathscr{F}(U)\otimes f^*\mathcal{E}(U)$. The sheafification of $P'$ is $\mathscr{F}\otimes f^*\mathcal{E}$.
Now, for all open $V\subseteq Y$, we have a moprhism $\mathcal{E}(V)\to f^*\mathcal{E}(f^{-1}(V))$, this gives a morphism
$$P(V)=\mathscr{F}(f^{-1}(V))\otimes\mathcal{E}(V)\to\mathscr{F}(f^{-1}(V))\otimes f^*\mathcal{E}(f^{-1}(V))=f_*P'(V)$$
Hence I have a natural morphism $\phi:P\to f_*P'$. Now, I have the sheafication morphism $P'\to P'^{sh}=\mathscr{F}\otimes f^*\mathcal{E}$, hence a morphism $f_*P'\to f_*(\mathscr{F}\otimes f^*\mathcal{E})$ that, composed with $\phi$, gives a natural morphism $P\to f_*(\mathscr{F}\otimes f^*\mathcal{E})$. Finally, passing to the sheafification of $P$, I get a morphism
$$\psi:f_*(\mathscr{F})\otimes\mathcal{E}\to f_*(\mathscr{F}\otimes f^*\mathcal{E})$$
What a mess! Fortunately, when we restrict to an open set where $\mathcal{E}$ is free, everything looks nicer.
So, restrict to an open set $W\subseteq Y$ where $\mathcal{E}$ is free. Now, $\mathcal{E}$ and $f^*\mathcal{E}$ are free, hence you can easily check that $P$ and $P'$ are already sheaves, so $\psi=\phi=\operatorname{id}\otimes\gamma$ where we have $\gamma:\mathcal{E}\to f_*f^*\mathcal{E}$. But, for $\mathcal{E}$ free, you can easily check that this is an isomorphism.
p.s: the passage in the proof you have found not clear maybe it's simpler if viewed in this way: $f_*(\mathscr{F}\otimes \mathcal{O}_X^{\,n})\;\cong\; f_*(\mathscr{F}^{n})\cong f_*(\mathscr{F})^n$, and you can "take the $n$ out" just applying the definition of $f_*$.