Linear system of divisors Let $F:X \rightarrow \mathbb{C}_{\infty}$ be a nonconstant holomorphic map with X a Riemann surface.. How could I show that the divisors $F*(q)$ for $q \in \mathbb{C}_{\infty}$ form a pencil? $F*(q)$ here is the inverse image divisor. I am not quite sure how to do this to be honest, any help would be appreciated.
 A: Let $P$ (a projective space) be the complete linear system on $\mathbb{C}_{\infty}$ consisting of those effective divisors linearly equivalent to $q$.  Note that $P$ is a projective space of dimension one (hence a pencil), and its divisors are precisely those of the form $p$, for each $p \in \mathbb{C}_{\infty}$.  
Let $Q$ be the complete linear system on $X$ consisting of all effective divisors linearly equivalent to $F^*q$.  Then $F^*P$ is a linear subspace of the projective space $Q$; we want to show that $F^*P$ has dimension one.  Since 
$$F^* \colon P \to Q$$
has domain of dimension one, its image $F^* P$ has dimension at most one.  Thus, $F^* P$ is either a pencil (dimension one) or a point (dimension zero).  
If $F^* P$ is a point, then $F^*p = F^* q$ for every $p, q \in \mathbb{C}_{\infty}$. In particular, $F^{-1}(\{p\}) = F^{-1}(\{q\})$.  Since every point of $X$ lies in $F^{-1}(\{p\})$ for some $p$, every point of $X$ lies in $F^{-1}(\{q\})$.  That is, every point of $X$ is mapped to $q$.  But this is impossible since $F$ was assumed nonconstant.
Thus, $F^*P$ must be a pencil, as desired. 
