Make a complex polynomial a covering map

Question

Let $p:\mathbb{C}\to \mathbb{C}$ be a complex polynomial. Let $C:=\{p(z):p'(z)=0\}$ and $V:=\mathbb{C}\setminus C$. I want to show that $p:p^{-1}(V)\to V$ is a covering map. By inverse function theorem, for every $y\in V$ there exist an open neighborhood $B$ of $y$ and an open set $A$ in $p^{-1}(V)$ such that $p:A\to B$ is an homeomorphism. But how I can show that $B$ can be chosen such that $p^{-1}(B)$ is union of disjoint open sets that are mapped homeomorphically onto $B$ by $p$?

Answer 1

I think I have a complete proof.

Let $X:=p^{-1}(V)$. I only use that $p:X\to V$ is a local homeomorphism and a proper map. Let $y\in V$, $p^{-1}(y)$ is compact (since $p$ is proper) and discrete (since $p$ is a local homeomorphism), then $p^{-1}(y)=\{x_1, \ldots, x_n\}$ is finite. Choose $U_i$ an open neighborhood of $x_i$ such that it is mapped homeomorphically onto some open neighborhood $V_i$ of $y$ and such that $U_i$ are pairwise disjoint. Let $A=X\setminus(U_1\cup\cdots\cup U_n)$ and $B=(V_1\cap\cdots\cap V_n)\setminus p(A)$. $B$ is an open set, since $p(A)$ is closed ($p$ is a closed map) and it's a neighborhood of $y$. It is easy to show that $p^{-1}(B)=\bigcup_i(U_i\cap p^{-1}(B))$ and $U_i\cap p^{-1}(B)$ is homeomorphically mapped onto $B$ by $p$.