On Griffiths and Harris, Principles of Algebraic Geometry, p.138, right after the proof of the Bertini's theorem. It says:
The essential point here is that a pencil $\{D_\lambda\}_{\lambda\in\mathbb{P}^1}$ with base locus $B$ gives a holomorphic mapping $M\setminus B\to\mathbb{P}^1$ since by linearity every $p\in M]\setminus B$ is on a unique $D_\lambda$.
My questions are:
How to see the divisors in $\{D_\lambda\}_{\lambda\in\mathbb{P}^1}$ cover the whole $M$? In other words, to any $p\in M\setminus B$, why is there always a divisor from the pencil s.t. $p$ is on it? (I can see the uniqueness though)
Why is the map $M\setminus B\to\mathbb{P}^1$ sending each point to the unique divisor containing it a holomorphic map? I don't see how is that related to Bertini's theorem as well as its proof.