# The projection formula for quasicoherent sheaves.

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!

• Come thinking of it: I realize that it might be futile to check for isomorphism on the stalks, since it will be hard to say something meaningful about the stals of the pushforward. But maybe I am wrong in this special case, and I would be happy to be! Mar 27, 2013 at 21:37
• (Although I wouldn't mind being right either). Mar 27, 2013 at 21:37
• In fact the stalks of pushforwards are not easy to compute. Mar 27, 2013 at 22:00

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 $f_* F \otimes \mathcal{O}_Y \cong f_* F \cong f_* (F \otimes \mathcal{O}_X) \cong f_* (F \otimes f^* \mathcal{O}_Y)$.
• A quick question: How do you check that $\alpha$ equals the composite you are talking about? Mar 28, 2013 at 13:33
• can we say anything about the kernel and cokernel of $\alpha$ when $G$ is any sheaf of modules, not necessarily locally free? Nov 16, 2020 at 15:46
• In the last line, how do we know that $f_*(F\otimes \mathcal{O}_X)\simeq f_*(F\otimes f^*\mathcal{O}_Y)$? Jul 14, 2021 at 14:08