Let $S$ be a quasi compact scheme and $X = \mathbb{P}(\mathcal{E})$ be its associated projective bundle corresponding to locally free sheaf $\mathcal{E}$ of finite rank $r$ on $S$ and $f : X \rightarrow S$ be the structure map. Let $\mathcal{F}$ be a quasi coherent sheaf on $X$. I want to prove that the natural map $f_{\ast}f^{\ast}f_{\ast}(\mathcal{F}) \rightarrow f_{\ast}(\mathcal{F})$ is an isomorphism.

I proved that this map is surjective, since if we look this map locally, it is clear that it is surjective. But I don't know how to prove that it is isomorphism. Any help would be great.

  • $\begingroup$ Do you know the projection formula? Under your hypothesis, it will say that for any sheaf $F$ on $S$, $f_*f^*F=F\otimes_S f_*\mathcal{O}_X$ and since $f_*\mathcal{O}_X=\mathcal{O}_S$, you are done. $\endgroup$
    – Mohan
    Aug 30, 2019 at 1:32
  • $\begingroup$ I didn't this version of projection formula. I know only ($\mathcal{R}^{q}f_{\ast}\mathcal{F}) \otimes \mathcal{E'} \cong \mathcal{R}^{q}f_{\ast}(\mathcal{F} \otimes f^{\ast}\mathcal{E'})$ for all $q \geq 0$ for any locally free sheaf $\mathcal{E'}$. Can you explain how did you get that? $\endgroup$
    – Sunny
    Aug 30, 2019 at 5:08
  • $\begingroup$ This is the same formula! For $q=0$, and $\mathcal{F}=\mathcal{O}_X$. $\endgroup$
    – Ahr
    Aug 30, 2019 at 8:44
  • $\begingroup$ According to your substitution, I get $f_{\ast}f^{\ast} \mathcal{E'} \cong \mathcal{E'}$, but this is not what we want? Can you explain a little bit? $\endgroup$
    – Sunny
    Aug 30, 2019 at 8:50

1 Answer 1


Let $\mathcal{G}=f_*(\mathcal{F})$, a sheaf on S.

Then $f_*f^*\mathcal{G}=f_*(f^*\mathcal{G}\otimes\mathcal{O}_X)=\mathcal{G}\otimes f_*\mathcal{O}_X=\mathcal{G}\otimes \mathcal{O}_S=\mathcal{G},$ using the projection formula.


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