Let $E$ be complex vector bundle of rank $r$ over a top. space $X$ and $p:\mathbb{P}(E) \to X$ the associated projective bundle of lines in $E$.
Why $p^*E$ containes the linebundle $L = \{(l,u)\in \mathbb{P}(E) \times E \vert v \in l\}$?
Ideas: because of similarities to $\mathcal{O}_{\mathbb{P}_n}(-1) = \{(l,u)\in \mathbb{P}_n \times E \vert v \in l\}$ (by the way: how to see that this equation holds?), I suppose that $L = p^*\mathcal{O}_{\mathbb{P}_n}(-1)$.
But here there occure four problems to me:
- If we have a morphism between schemes $f:X \to Y$ I know how the induced functor $f^*$ maps a sheaf $\mathcal{G}$ on $Y$ to a sheaf on $X$ via $\mathcal{G} \mapsto f^{-1}\mathcal{G} \otimes{f^{-1}\mathcal{O}_Y} \mathcal{O}_X$.
But how to interpret $f^*$ if $f$ is just a morphism between vector bundle and a topological space?
Does $\mathcal{O}_{\mathbb{P}_n}(-1) \subset E$ hold & why?
And if 2. holds why does it imply $p^* \mathcal{O}_{\mathbb{P}_n}(-1) \subset p^*E$. The functor $p^*$ is just right exact, so why does it preserve injectivity?
Background of my question:
"Vector Bundles on Complex Projective Spaces" von Okonek, Schneider and Spindler, page 7:
In general you can define the pullback $f^*E$ of a vector bundle $\pi:E\to B$ along a map $f:C\to B$ in the following way: $$ f^*E=\{(c,e)\in C\times E| f(c)=\pi(E)\}\subset C\times E$$ (See also here https://en.m.wikipedia.org/wiki/Pullback_bundle)
Now consider $p:\mathbb{P}(E)\to X$ , then we get: $$ p^*E=\{(l,v)\in \mathbb{P}(E)\times E| p(l)=\pi(v)\}\subset \mathbb{P}(E)\times E$$ Now check that $v\in l$ will imply $p(l)=\pi(v)$, which is a simple consequence of the construction of the projective bundle (and the map $p$)