# Isomorphism of sheaf

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.

• 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. Aug 30, 2019 at 1:32
• 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? Aug 30, 2019 at 5:08
• This is the same formula! For $q=0$, and $\mathcal{F}=\mathcal{O}_X$.
– Ahr
Aug 30, 2019 at 8:44
• 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? Aug 30, 2019 at 8:50

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.