Partition of $\mathbb{C}$ into Borel sets for which a polynomial is a bijection with $\mathbb{C}$

Question

Background

Let $P$ be a polynomial of degree $n>0$ with coefficients in $\mathbb{C}$. By the Fundamental Theorem of Algebra, for any complex $y$, there exist $n$ complex numbers $z$ such that $P(z)=y$ (possibly with multiplicity). So, we can consider $P$ to be a sort of "$n$-injective" function, in that, for each output, there exist $n$ inputs that give that output.

Question

We can assign each complex $c$ with a "degree" $\mu(z)\leq n$ in accordance with the multiplicity of the root $z=c$ in the polynomial $P(z)-P(c)$.

Does there exist a (almost) partition of $\mathbb{C}$ into $n$ Borel sets such that:the polynomial $P$ is bijective from each set to $\mathbb{C}$ and each complex number $z$ appears in exactly $\mu(z)$ of the sets? This would almost be a partition since, for almost all $z$, $\mu(z)=1$ (in fact, $\mu(z)\neq 1$ only for a finite number of $z$ as that would require $P'(z)=0$).

I've chosen Borel as the qualifier as, as far as I can tell, it's the best way to define a "nice set" - obviously if the quantifier is taken away the problem is true - but I'm not sure if I'm using it properly.

Progress

The statement is trivially true if $n=1$ (all non-constant linear polynomials are bijective, so one can simply take the set to be $\mathbb{C}$). If $n=2$, the polynomial can be completed into $P(z)=a\left((z-b)^2+c\right)$ for some complex $a,b,c$, so we can choose our sets to be, if $b=x+iy$ where $x,y\in\mathbb{R}$,

1. The half-plane above $\Im(z)=y$ and the ray $\Re(z)\geq x, \Im(z)=y$

2. The half-plane below $\Im(z)=-y$ and the ray $\Re(z)\leq x, \Im(z)=y$.

However, this is all based on the fact that, as functions, all quadratic polynomials act like $z^2$. So this sort of thing is, as far as I can see, not readily generalizable to $n>2$.

Is this a known problem (and if so, where can I read more about it)? Am I missing something simple?

Answer 1

$P$ defines an $n$-to-$1$ map $X\to Y$ where $X=\mathbb C\setminus \{z\mid P'(z)=0\}$ and $Y=\mathbb C\setminus \{P(z)\mid P'(z)=0\}.$ Consider a simply-connected set $Y'\subseteq Y,$ a point $y\in Y',$ and a choice $x_j\in P^{-1}(\{y\}).$ The function $P$ can be inverted locally at $y$ so that $y$ is sent to $x_j,$ and this extends to an inverse on $Y'.$ Call the image $X_j.$ The images $X_1,\dots,X_n$ cannot intersect so they will be disjoint open sets in $X.$ If the closure of $Y'$ is $Y,$ then the closures of $X_1,\dots,X_n$ will cover $X.$

Restricting to a simply connected set $Y'$ ensures the extension of $P^{-1}$ is well-defined - it avoids the monodromy around the branch points $\{P(z)\mid P'(z)=0\}$.

Assuming all the branch points ($P(z)$ with $P'(z)=0$) have different imaginary part, we could define $Y'$ by branch cuts going in the direction of $+\infty,$ i.e. $Y'=\mathbb C\setminus \{P(z)+r \mid P'(z)=0, r\geq 0\}.$ Then $X$ gets carved up along the preimages of these branch cuts. To get the boundary partitioned we could specify that $X_j$ includes the boundary points coming from above the branch cuts in $Y'$ - the limits coming from a continuous choice of $P^{-1}(P(z)+r+i\epsilon)$ as $\epsilon\to 0$. In the simple case $P(z)=z^n,$ each set $X_j$ would then correspond to $\left\{z\in\mathbb C\mid z\neq 0, \arg(z)\in \left[\tfrac{2\pi (j-1)}n,\tfrac{2\pi j}n\right)\right\}.$

Alternatively we can consider the uniformization of $Y$: a map of Riemann surfaces $\pi:D\to Y$ and a group $G$ acting on the universal cover $D$ such that $Y$ is the quotient $D/G.$ For example in the case $Y=\mathbb C\setminus \{0\}$ we can think of $D$ as being $\mathbb C$, with $\pi=\exp$ and with $G$ acting by translations $z\mapsto z+2\pi i.$ In this context what we are looking for is a nice fundamental domain and this problem has several nice solutions, though the boundary is usually dealt with differently. The surface $D$ comes with a Riemannian metric and after choosing a point $z_0\in D$ we can define the Dirichlet polygon

$$F=\{z\in D :d(z,z_{0})<d(z,gz_{0})\;\;\forall g\in G,g\neq 1\}$$

whose image $\pi(F)$ serves as the set $Y'$ above. The geometry of $F$ is quite tame. See Alan F. Beardon's "The Geometry of Discrete Groups" for more. I'm sure the boundary given by these sets could also be partitioned with a bit more care.