# Blowing up vector bundles in the zero section

Assume we are given a scheme $$X$$ (feel free to add all the needed hypotheses, at this point I’m working with smooth schemes, but the fewer is needed, the better) and a vector bundle $$E$$ over $$X$$. I will denote $$|E|$$ the total space of this vector bundle. I know that if we blow up $$|E|$$ in the zero section, we get that $$Bl_B|E| = |\mathcal{O}_{\mathbb{P}(E)}(-1)|$$ (if possible, I would like a proof or reference for this fact). The question is whether there is any relation between $$\mathcal{O}_g(-1)$$ and $$p^{*}(\mathcal{O}_{\mathbb{P}(E)}(-1))$$, where $$g$$ is the map of the blow up, and $$p$$ is the map from the total space of $$\mathcal{O}_{\mathbb{P}(E)}(-1)$$ to $$\mathbb{P}(E)$$. I would hope them to be isomorphic, but I don’t have any clue on how to prove it.

i) $$\mathrm{Pic}(|\mathcal{O}_{\mathbb{P}(E)}(-1)|) = \mathrm{Pic}(\mathbb{P}(E))$$
ii) $$\mathrm{Pic}(\mathbb{P}(E)) = \mathrm{Pic}(X) \oplus \mathbb{Z}. \mathcal{O}_{\mathbb{P}(E)}(-1)$$
iii) $$\mathrm{Pic}(\mathrm{Bl}_{0}(|E|)) = \mathrm{Pic}(|E|) \oplus \mathbb{Z}.\mathcal{O}_g(1)$$
iv) $$\mathrm{Pic}(|E|) = \mathrm{Pic}(X)$$.
The identification $$\mathrm{Bl}_0(|E|) = |\mathcal{O}_{\mathbb{P}(E)}(-1)|$$ identifies the zero section of $$\mathcal{O}_{\mathbb{P}(E)}(-1)$$ with the exceptional divisor or $$\mathrm{Bl}_0(|E|)$$. Since the restriction of both $$\mathcal{O}_g(-1)$$ and $$p^*\mathcal{O}_{\mathbb{P}(E)}(-1)$$ to $$\mathbb{P}(E)$$ (seen as the zero section) is the relative tautological bundle, these two line bundles coincide up to the pull back of a line bundle on $$X$$. The formula in ii), ii) and iv) show that this line bundle is trivial