Global generation of $\mathcal{O}_{\mathbb{P}(\mathcal{E})}(1)$ and $\mathcal{E}$ I'm trying to prove that $\mathcal{O}_{\mathbb{P}(\mathcal{E})}(1)$ is generated by global sections on $\mathbb{P}(\mathcal{E})$ if and only if $\mathcal{E}$ is generated by global sections ($\pi: \mathbb{P}(\mathcal{E}) \to X$ is projective bundle associated to locally free sheaf $\mathcal{E}$ on $X$). 
If $\mathcal{E}$ is gbgs, we can pullback the surjection $\mathcal{O}_X^n \to \mathcal{E}$ to obtain a surjection $\mathcal{O}_{\mathbb{P}(\mathcal{E})}^n \to \pi^* \mathcal{E}$ (as pullback is right exact), which we can then compose with natural surjection $\pi^* \mathcal{E} \to \mathcal{O}_{\mathbb{P}(\mathcal{E})}(1)$.
I have trouble with opposite direction, though. Suppose we have a surjection $\mathcal{O}_{\mathbb{P}(\mathcal{E})}^n \to \mathcal{O}_{\mathbb{P}(\mathcal{E})}(1)$. We can pushforward it via $\pi$ to obtain a map $\pi_* \mathcal{O}_{\mathbb{P}(\mathcal{E})}^n \to \pi_*\mathcal{O}_{\mathbb{P}(\mathcal{E})}(1)$, but since $\pi_* \mathcal{O}_{\mathbb{P}(\mathcal{E})}^n \simeq \mathcal{O}_X^n, \pi_*\mathcal{O}_{\mathbb{P}(\mathcal{E})}(1) \simeq \mathcal{E}$, we have a map $\mathcal{O}_X^n \to \mathcal{E}$. As pushforward is only left exact, we cannot really conclude that it will be surjective as well. Also, $\pi$ is not affine, so higher direct images don't have to vanish.
Any hints?
 A: The relative cohomology $R^1\pi_* K$ is the sheafification of $U \mapsto H^1(\pi^{-1}(U), K|_{\pi^{-1}(U)})$ so to show that $R^1$ vanishes it suffices to show that the above $H^1$ vanishes on small enough affine open subsets.  
Pick $U= Spec A$ affine open and small enough that $\mathcal E|_U$ is trivial, so that $\pi^{-1}(U) = \mathbb P(\Gamma(\mathcal E|_U)) \times U = \mathbb P^r_A$. On this space we have the short exact sequence $0 \to K|_{\pi ^{-1}(U)} \to \mathcal O_{\mathbb P^r_A}^n \to \mathcal O_{\mathbb P^r_A}(1)\to 0$,
By direct inspection, you can see that any proper $A$ submodule  $M \subsetneq H^0(\mathcal O_{\mathbb P_A^r}(1))$ has a base point. (Roughly, there is a point   $P \in Spec \ A$ such that at $M_P \neq H^0(\mathcal O_{\mathbb P_A^r}(1))_P$.  Then at $\mathbb P^r \times \{P\} \subset \mathbb P^r_A$ we have $M \otimes k(P) \subsetneq H^0(\mathcal O_{\mathbb P^r_{k(P)}}(1))$, and in $\mathbb P^r_{k(P)}$ (over a field now) it is clear that a proper subspace cannot be base point free, since there is a point contained in the intersection of $r-1$ hyperplanes. )
Hence the map on global sections $H^0( \mathcal O^n)   \to H^0( \mathcal O(1))$ must be actually surjective.  But from the long exact sequence in cohomology and the fact that $H^1(\mathcal O_{\mathbb P^r_A}(1)) = 0$, we have that  $H^1(K|_{\pi^{-1}(U)})$ is the cokernel of that map.  Hence the desired relative cohomology vanishes, and so $
\mathcal E$ must be globally generated.
