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.
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.