For a morphism $f:X\rightarrow Y$, locally free sheaf $\mathcal{G}$ on $Y$, and a quasi-coherent sheaf $\mathcal{G}$ on $X$, we have the projection formula $$f_*(\mathcal{F}\otimes_{\mathcal{O}_X}f^*\mathcal{G})\simeq f_*\mathcal{F}\otimes_{\mathcal{O}_Y}\mathcal{G}.$$

I don't have a problem proving this, but I have a hard time seeing it as a projection! This looks more like compatibility of pushforward with base change if we take $\mathcal{G}=\mathcal{O}_Y$. What is the motivation for calling it the projection formula?

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    $\begingroup$ $G$ doesn't have to be quasi-coherent. $\endgroup$ – Martin Brandenburg Jan 14 '14 at 20:40

In my case, for example, I need to apply this formula to the natural projection $$ \pi: X\times \operatorname{Div}_X \to \operatorname{Div}_X $$ where $X$ is a curve and $\operatorname{Div}_X$ denotes the set of effective Cartier divisors on $X$.

I guess the name comes from the fact that one often faces these kind of situations.

Intuitively, if you have a projection $\pi : X\times Y \to Y$ and sheaves $\mathscr{F}$ on the product and $\mathscr{G}$ on $Y$, you can consider the pullback $\pi^*\mathscr{G}$, which is constant on the fibers of the projection. You can tensor it with $\mathscr{F}$ above, obtaining $$ \mathscr{F}\otimes \pi^* \mathscr{G} $$ and wonder what happens if you push this down again to $Y$. Well, under the above mentioned hypothesis, this statement ensures that the pushforward sheaf is what you expect it to be: $$ \pi_* \mathscr{F} \otimes \mathscr{G} $$ i.e. the pushforward of $\mathscr{F}$ tensored with the plain $\mathscr{G}$.


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    $\begingroup$ thanks for the lovely picture! $\endgroup$ – user309475 Apr 23 '17 at 20:32

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