I am looking for a certain way of proving the following :
Let $f. X \rightarrow S$ be a morphism of schemes. Suppose that f is quasiseparated and quasicompact, or that X is noetherian. Let $\mathcal{G}$ be a locally free sheaf on S and $\mathcal{F}$ a quasicoherent sheaf on X. Show that we have a canonical isomorphism:
$$f_\ast \mathcal{F} \otimes_{\mathcal{O}_S} \mathcal{G} \cong f_\ast (\mathcal{F} \otimes_{\mathcal{O}_X} f^\ast \mathcal{G}).$$
I an prove this by using a gluing argument on affines , but I would want a proof that is more global, but finishes it off with checking the isomorphism on stalks Let me show you what I mean. We have a morphism
$$\alpha: f_\ast \mathcal{F} \otimes_{\mathcal{O}_S} \mathcal{G} \rightarrow f_\ast \mathcal{F} \otimes_{\mathcal{O}_S} f_\ast f^\ast \mathcal{G}$$ given by tensoring the canonical morphism $\eta: \mathcal{G} \rightarrow f_\ast f^\ast \mathcal{G}$ with the identity on $f_\ast \mathcal{F}$. It can be shown that we also have a canonical morphism:
$$\beta: f_\ast \mathcal{F} \otimes_{\mathcal{O}_S} f_\ast f^\ast \mathcal{G} \rightarrow f_\ast (\mathcal{F} \otimes_{\mathcal{O}_X} f^\ast \mathcal{G})$$
and composing $\alpha$ and $\beta$ we get our desired morphism. Is it, by only this, possible to check this isomorphism on the stalks? If so – how could we do it? What properties do we use of the stalks of $f_\ast (\mathcal{F} \otimes_{\mathcal{O}_X} f^\ast \mathcal{G})$ in that case?
Thanks for help!
Best Answer
The projection formula has nothing to do with schemes and quasi-coherent modules. It holds for arbitrary ringed spaces and sheaves of modules on them. This can be found in EGA I or the Stacks project. In fact we may replace spaces by arbitrary sites, so that it really becomes a global statement about ringed topoi. In particular we don't need to look at the stalks, which would be clumsy anyway.
Let $f : X \to Y$ be a morphism of ringed spaces, $F$ a sheaf of modules on $X$ and $G$ a sheaf of modules on $Y$. There is a canonical homomorphism $\alpha : f_* F \otimes G \to f_* (F \otimes f^* G)$, which under the adjunction $f^* \dashv f_*$ corresponds to $f^*(f_* F \otimes G) \cong f^* f_* F \otimes f^* G \to F \otimes f^* G$.
Now we claim that $\alpha$ is an isomorphism when $G$ is locally free of finite rank. Using the usual restriction properties of $f_*$ and $f^*$, we may work locally on $Y$. Besides, if the claim holds for $G_1$ and $G_2$, then also for $G_1 \oplus G_2$. Thus we may assume $G=\mathcal{O}_Y$. But then $\alpha$ equals with the composite of the canonical isomorphisms $f_* F \otimes \mathcal{O}_Y \cong f_* F \cong f_* (F \otimes \mathcal{O}_X) \cong f_* (F \otimes f^* \mathcal{O}_Y)$.