# Line bundle determined by a linear surjection $V \to V/W$

I am new to things like line bundles and have just been reading page 10 of this document here. Now I have some elementary questions:

1. Given a finite dimensional complex vector space $V$ and a one - dimensional vector space $W$, we can form $V/W$ together with $\pi : V \to V/W$ the canonical projection. Now I understand that upon taking projectivizations we have a map $\Bbb{P}(\pi) : \Bbb{P}(V) - (\Bbb{P}(W) = \{\ast\}) \longrightarrow \Bbb{P}(V/W)$. But why is this latter thing a line bundle? I can see that given a line $L \subseteq V/W$, the inverse image $\pi^{-1}(L)$ consists of all two dimensional subspaces $K$ in $V$ that contain $W$. But now how can I see that the fibre $\Bbb{P}(\pi)^{-1}([L])$ "is" a one-dimensional vector space? Also, how does one verify the local triviality condition?

2. Given a section $s$ of the map $\pi : V \to V/W$, we have a section $\Bbb{P}(s)$ of the map $\Bbb{P}(\pi)$. However it is claimed in the document I linked to that the image of $\Bbb{P}(V/W)$ in $\Bbb{P}(V)\setminus \{\ast\}$ is a hypersurface. Why is this so?

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Your description of the fiber is not correct. –  Martin Brandenburg Aug 17 '13 at 15:18
Instead, the fiber of a line $U/W$ (i.e. $\dim(U)=2$ and $W \subseteq U$) consists of all complements of $W$ in $U$. –  Martin Brandenburg Aug 17 '13 at 20:59
@user38268: What happened to your username? –  Michael Joyce Aug 18 '13 at 10:33

Conceptually, I think it is easier to think of the setup as a short exact sequence: $$0 \rightarrow W \rightarrow V \xrightarrow{\pi} V' \rightarrow 0,$$ where $V' = V/W$.
(1) First, identify lines in $V'$ with planes in $V$ that contain $W$. Then the preimage of a plane $K$ with $W \subset K \subset V$ consists of all lines in $K$ except for $W$. That is, $$\pi^{-1}([K/W]) = \mathbb{P}(K) \setminus [W] \cong \mathbb{P}^1 \setminus \{\text{pt}\} \cong \mathbb{A}^1.$$ As I see it, this gives you an $\mathbb{A}^1$-bundle structure only, not a line bundle structure. I hate doing local triviality checks (sorry!), but I expect that if you explicitly write the above out on the standard open cover, you'll find that it is trivial there.
(2) Since $\dim \mathbb{P}(V') = \dim \mathbb{P}(V) - 1$, clearly the codimension of the image is at least 1. On the other hand, since $\pi \circ s = \text{id}$, the image of $\mathbb{P}(s)$ must map onto a variety of dimension $\dim \mathbb{P}(V')$, which is therefore a lower bound for the dimension of the image of $\mathbb{P}(s)$. These two facts together say that $\dim \text{image}(\mathbb{P}(s)) = \dim \mathbb{P}(V) - 1$, so the image is a hypersurface.