Automorphism classes of branched covers of disks over disks I was reading Hubbard's book (Teichmuller Theory Vol 2) and in a proof (9.3.2), he mentions that there is one branched cover of the disk over a disk with 1 ramification point (degree $k$) up to automorphism.
Why is this true? (I feel like I'm missing something rather obvious)
Many thanks!
 A: Let $D$ denote the open unit disk in the complex plane (centered at the origin) and $f: X\to D$ be a $k$-fold branched covering map with unique branch-point $p\in D$, where $X$ is a connected Riemann surface.
First of all you can assume that the unique branch-point $p$ is the center of the unit disk (by post-composing $f$ with a Moebius transformation of $D$.
Next, by the definition, the restriction of $f$ to $X\setminus f^{-1}(A)$ is a $k$-fold covering map to $Y=D\setminus A$, $A=\{0\}$. Bt the classification of covering maps, the isomorphism classes of such covering maps are in bijective correspondence with index $k$ subgroups of $\pi_1(D\setminus A)\cong {\mathbb Z}$, i.e. there is exactly one such subgroup (for each $k$). Thus, the covering map
$$
f: X\setminus f^{-1}(A)\to Y
$$
is isomorphic to the covering map $f_k: Y\to Y$ given by the restriction of the map
$$
F_k: z\mapsto z^k. 
$$
The latter map extends holomorphically to the origin, of course. Accordingly, $f^{-1}(A)$ is a singleton $\{b\}$ and the biholomorphic map
$$
h: X\setminus \{b\}\to Y
$$
(defining the isomorphism of covering maps $f$ and $f_k$)
extends holomorphically to $b$ (Riemann's extension theorem). The same for the inverse map, of course, thus, we have
$$
h\circ F_k= f. 
$$
Thus, $h$ defines an isomorphism of the branched covering map $f$ and the canonical branched covering map $F_k$. qed
