In Complex Geometry written by Huybrechts, it defines the tautological line bundle $\mathcal{O}(-1)$ as followed
$$ \mathcal{O}(-1):=\{(\ell,z)\in \mathbb{C}P^n \times \mathbb{C}^{n+1}: z\in \ell\} $$
and the projection map $\pi:(\ell,z)\mapsto \ell$, then I think for a standard open covering $\{U_i\}_{i=0}^n$ of $\mathbb{C}P^n$ where
$$ U_i:=\{ (z_0:\cdots:z_n): z_i\neq 0 \} $$
then $\pi^{-1}(U_i)\cong U_i \times \mathbb{C}^* $ where $\mathbb{C}^*=\mathbb{C}\backslash\{0\}$, then there is no zero in each fiber $\pi^{-1}(\ell)$, that is $\mathcal{O}(-1)$ is not a line bundle. So where am i wrong?
Best Answer
The fiber $\pi^{-1}(\ell)\subset \mathcal{O}(-1)$ above $\ell \in \Bbb CP^n$ is by definition $\{\ell\}\times \ell \simeq \ell$, which obviously has an origin $0$: it is the origin of $\Bbb C^{n+1}$. The zero section is just the map $\ell \in \Bbb CP^n \mapsto (\ell,0)\in \mathcal{O}(-1)\subset \Bbb CP^n \times \Bbb C^{n+1}$.