Holomorphic function on projective line. Why is essential that if we have a complex-valued holomorphic function on $P^1$ defined as $f[z:w] = \dfrac{p(z,w)}{q(z,w)}$ where $p$ and $q$ are homogeneous with $q \neq 0 $ they need to be of the same degree  ?
I remember that a complex-valued function $f$ is holomorphic on a point $p$ of a Riemann Surface $X$ when $f(\phi^{-1})$ is holomorphic in the complex way on $\phi(p)$ where $\phi$ is a chart on X.  
I tried using the inverse of a chart on $P^1$ such as the function sending $w$ to $[1,w]$ then composing with $f$ I obtain a ratio of polynomials in w which is always holomorphic. So, please help me to find the flaw in my reasoning.
Thank you.
 A: Notice that a holomorphic complex-valued function in $\mathbb{P}^1$ must be constant, due to the maximum principle. I will assume you meant a function which takes values in $\mathbb{P}^1$ instead. User90189's comment says that for a quotient of polynomials in homogeneous coordinates to yield such a map, it is necessary that p and q have the same degree, therefore answering your question. 
The homogeneity condition on the polynomials $p,q$ is essential not only for the function to be well-defined, as pointed out above. In fact, it is essential for the function to be holomorphic as well. One can prove that every holomorphic map from $\mathbb{P}^1$ to itself must be such a quotient. More precisely, any non-constant holomorphic map from $\mathbb{P}^1$ to itself can be described in homogeneous coordinates as $f([z,w])=[p(z,w),q(z,w)]$, where: 


*

*$p,q$ are homogeneous polynomials in $\mathbb{C}^2$, with the same degree;

*$p,q$ do not vanish simultaneously at any point in $\mathbb{C}^2 \setminus \{(0,0)\}$. 


This is shown by using the following fact: a surjective holomorphic map between connected, compact Riemann surfaces is a smooth (possibly branched) finite-sheeted covering. Indeed, such a map is a local diffeomorphism (by the local form of holomorphic functions) away from the set where the derivative of the map is zero (a finite subset of the domain where branching occurs). 
If a holomorphic map $f \colon \mathbb{P}^1 \longrightarrow \mathbb{P}^1$ is non-constant, then it must be surjective (otherwise it would contradict the maximum principle again), thus a (possibly branched) cover as above. Identifying the domain and target $\mathbb{P}^1$ with the extended complex plane via homogeneous coordinates, we can define as usual the poles of $f$ in the affine chart $U=\{[z,w] \in \mathbb{P}^1 | w \neq 0\}$ in the domain as the elements elements which are mapped to $\infty = [0,1]$ in the target. The orders of such poles are defined as usual, in terms of the order of vanishing of $1/f$ on the affine variable $x=z/w$ at these poles. 
Multiplying $f$ by a suitably chosen polynomial $\widehat{q}(x)$ on $U$ (vanishing at the poles with their respective orders) yields an entire function $g$ on $U$, which extends to a holomorphic function from $\mathbb{P}^1$ to itself by analytic continuation. Looking at this function in the other affine chart, i.e. replacing $x$ by $1/x$, yields a meromorphic function at $0$, thus $g$ must in fact have been a polynomial, say $\widehat{p}(x)$. 
One gets the above result by reconstructing $f$ from the auxiliary polynomials $\widehat{p}$ and $\widehat{q}$ and defining the (homogeneous) polynomials $p(z,w)=\widehat{p}(z/w)$, $q(z,w)=\widehat{q}(z/w)$ on $\mathbb{C}^2$.    
