A ramified branched covering of degree $d$ is equivalent to $z^d$

Question

Let $f:N_{z_0}\to N_{f(z_0)}$ be a holomorphic branched covering of degree $d>1$ ramified only at $z_0$ between open neighbourhoods of $z_0$ and $f(z_0)$ respectively (i.e. such that $f:N_{z_0}\setminus \{z_0\}\to N_{f(z_0)}\setminus \{f(z_0)\}$ is biholomorphic and the multiplicity of $z_0$ as a preimage by $f$ of $f(z_0)$ is exactly $d$).

By the Riemann mapping theorem, there exists a biholomorphism $\phi:N_{f(z_0)}\to \mathbb{D}$ such that $\phi(f(z_0)) = 0$.

I want to know what results (theorem, proposition or whatever) are applied in these arguments:

Then, $\phi$ can be lifted to a conformal map $\widetilde{\phi}:N_{z_0}\to\mathbb{D}$ and we have the following commutative diagram: $\require{AMScd}$ \begin{CD} N_{z_0} @>f>> N_{f(z_0)}\\ @V \widetilde{\phi} V V= @VV \phi V\\ \mathbb{D} @>>z\mapsto z^d> \mathbb{D} \end{CD}

Thank you!

Answer 1

I got it! Here is the general result that I found:

Proposition.

Given a holomorphic map $f$ between Riemann surfaces $X$ and $Y$, there is a unique integer $m$ such that there are local coordinates near $p$ and $f(p)$ with $f$ having the form $z\mapsto z^m$.

Proof

We can choose two charts $\psi:U_1\subset X\to U_2\subset \mathbb{C}$ and $\phi:V_1\subset Y\to V_2\subset\mathbb{C}$ such that $\psi(p) = 0$ and $\phi(f(p)) = 0$, so that $P := \psi\circ f\circ \phi^{-1}$ is analytic. Since $P(0) = 0$, we can write $P(\omega) = \omega^m g(\omega)$ with $g(\omega) = a_0 + a_1\omega + a_2\omega^2 + \dots$ and choose a holomorphic branch $r$ of the $m$-th root so that $(r(\omega))^m = g(\omega)$. Let us consider $\nu(\omega) := \omega r(\omega)$, that satisfy $\nu'(0)\neq 0$ because $r(0)\neq 0$ and hence is locally invertible. Finally, setting $\widetilde{\phi} := \nu\circ \phi$, we obtain \begin{equation*} (\psi\circ f \circ \widetilde{\phi}^{-1})(z) = (\psi\circ f\circ \phi^{-1}\circ\nu^{-1})(z) = (P\circ \nu)(z) = z^m. \end{equation*}