Missing Link in the Proof of Riemann Mapping Theorem (Conway) I was going through the proof of the Riemann Mapping theorem. I got the idea and saw where everything is headed. I could not see two things. I need help in them. The first one is :
$$F=\{f \in H(G): f \text{ is one-one}, f(a)=0, f'(a) \gt 0, f(G) \subset D\}$$
If we assume that $F \ne \varnothing$ and $\bar{F}=F \cup \{0\}$  and consider the function $f \to f'(a)$ of $H(G) \to \mathbb{C}$. I don't see why this is a continuous function. Moreover it says that $\bar{F}$ is compact, there is an $f \in \bar{F}$ with $f'(a) \ge g'(a)$ for all $g \in F$. I also don't see why this is true. I guess these are the consequences of Montel's theorem but I am not able to see them. 
The second thing: Let $$U=B(-g(a),r)$$It says that there is a Möbius Transformation $T$ such that $T(\mathbb{C_{\infty}}-\bar{U})=D$. I could not find an explicit one. 
For all the above cases here $D$ is the unit disk
Thanks for the help!!
 A: Choose $\rho > 0$ small enough that the closed disk with radius $\rho$ and centre $a$ is contained in $G$. Then for every $f\in H(G)$ we have
$$f'(a) = \frac{1}{2\pi i} \int_{\lvert z-a\rvert = \rho} \frac{f(z)}{(z-a)^2}\,dz\tag{1}$$
by Cauchy's integral formula. The circle $C = \{z : \lvert z-a\rvert = \rho\}$ is a compact subset of $G$, hence if $f_n \to f$ in $H(G)$, then $f_n \to f$ uniformly on $C$, and hence $f_n'(a) \to f'(a)$ by $(1)$, so $f \mapsto f'(a)$ is continuous.
Since the unit disk is bounded, the set $F$ (more, even the set of all holomorphic $h\colon G \to D$) is relatively compact in $H(G)$ by Montel's theorem. By the continuity of $f \mapsto f'(a)$, we have $g'(a) \geqslant 0$ for all $g \in \overline{F}$. By Hurwitz's theorem, any locally uniform limit of injective holomorphic functions is either injective or constant, which shows that for $g \in \overline{F}\setminus F$ we must have $g = 0$. For if $f_n \to g$ with $f_n \in F$ and $g'(a) \neq 0$, then $g\in F$ by Hurwitz's theorem, and if $g'(a) = 0$ then $g$ is constant, and since $g(a) = \lim f_n(a) = 0$ it follows that $g\equiv 0$. Thus $\overline{F} = F\cup \{0\}$.
Then by continuity of $f\mapsto f'(a)$, the set
$$\{ f'(a) : f \in \overline{F}\} \subset [0,+\infty)$$
is compact, hence contains a maximum, i.e. there is a $g\in \overline{F}$ with
$$g'(a) = \max \{ f'(a) : f \in \overline{F}\}.$$
For the second thing, note that
$$z \mapsto \frac{z + g(a)}{r}$$
is a Möbius transformation mapping $U$ conformally to $D$. Then compose with the inversion $z \mapsto 1/z$ which swaps the unit disk and its exterior to obtain the Möbius transformation
$$T\colon z \mapsto \frac{r}{z + g(a)}$$
that maps $\mathbb{C}_\infty \setminus \overline{U}$ conformally to the unit disk.
