# Vector Bundle contains the Tautological Line Bundle

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:

1. 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?

1. Does $\mathcal{O}_{\mathbb{P}_n}(-1) \subset E$ hold & why?

2. 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 $p$)
• Hi. Thank you for the enlightening answer. One question: Do you have a nice reference for the construction of the projective bundle of $E$? The wiki explanation en.wikipedia.org/wiki/Projective_bundle using universal property don't helps to understand the local behavior of $\mathbb{P}(E)$ – KarlPeter Apr 7 '18 at 13:36
• If $(l,v)\in \mathbb{P}(E)\times E$ with $p(l)=\pi(v)$, how do you mean $v \in l$? Do you mean with $v \in l$ more precisely $[v] = l$ as equivalence class since $\mathbb{P}(E) = E \backslash s(X) / \sim$ where the equilvalence relation is defined via $v_1 \sim v_2 \Leftrightarrow \pi(v_1) = \pi(v_2) \text{ and there exist } c \in k \setminus \{0\} \text{ with } v_2 = c v_1$? – KarlPeter Apr 7 '18 at 17:19
• Yes, the equivalence class brackets are sometimes omitted, i.e. $v\in l$ means that the $v$ lies in the line that is "spanned" by $l$ (since each $l$ represents a one dimensional subspace) – Notone Apr 7 '18 at 17:38