I am trying to understand the proof of Koebe's Kreisnormierungsproblem given in Conway's Functions of One Complex Variable II. The theorem states:
If $G$ is a non-degenerate finitely connected domain, $A$ is a component of the extended boundary of $G$, and a $\in G$, then there is a unique circular domain $\Omega$ and a unique conformal mapping $f: G \to \Omega$ such that $f$ associates $A$ with $\partial \mathbb{D}$, $f(a) = 0$ and $f^\prime(a) > 0$.
In the first part uniqueness is shown by assuming that there are two different conformal mappings $f_j: G \to \Omega_j$, $j=1,2$. Then, we consider the composition $g = f_2 \circ f_1^{-1}: \Omega_1 \to \Omega_2$ satisfying $g(\partial \mathbb{D}) = \partial \mathbb{D}$, $g(0) = 0$ and $g^\prime(0) > 0$.
Then the auther says that this already implies that $g(\mathbb{D}) = \mathbb{D}$, which is used to show that $g$ is the identity. Why exactly?
My idea: We know that $g$ is a Möbius transformation by another theorem in this book. As $g$ maps $\partial \mathbb{D}$ onto $\partial \mathbb{D}$ and fixes the origin, it has to map everything within $\partial \mathbb{D}$ to everything within $\partial \mathbb{D}$?
Second question: Later, in the existence part. The circular domains (domains whose boundaries only consist of a finite number of disjoint circles) are topologized, i.e., if $C_1, \dots , C_n$ are the circles that form the boundaries of the bounded components of the complement of $G$. Then each circle $C_j$ is determined by its center $z_j = a_j + ib_j$ and its radius $r_j$. Thus every circular domain $G$ can be identified with a point in $\mathbb{R}^{3n}$ with coordinates $(a_1,b_1,r_1, \dots, a_n,b_n,r_n)$. The set of all points that are obtained so is a proper subset of $$ \{ (a_1,b_1,r_1, \dots ,a_n,b_n,r_n) \vert 0 < a_j^2 + b_j^2 < 1, 0 < r_j <1 \}. $$ Then he claims that the set of all points that are obtained so is an open and connected subset of $\mathbb{R}^{3n}$.
My idea for open: We can always change the position of the center and the radius by at least a small bit such that the resulting domain is still circular. We only have to take care that the circle do not touch. This shows that any point in $\mathbb{R}^{3n}$ obtained from a circular domain is an inner point. Hence, the set is open. In order to show that the set is connected: I thought about trying to show that it is even path-connected but did not come up with a solution yet.