A homeomorphism of the unit disk that cannot be extended to the boundary of its domain I have a problem with the following exercise: 

Let $D^2$ be the unit disc and S^1 be the unit circle. Show that the function
$
h: {D^2\setminus{S^1}} \to {D^2\setminus{S^1}} \\
h(re^{it})=
\begin{cases}
 0 &\text{if}\, r=0 \\
    r \cdot e^{i(t+\frac{2\pi r}{1-r})} &\text{else}
\end{cases}
$ 
cannot be extended to a homeomorphism $D^2 \to D^2$. 

I started as follows: 
Suppose that h can be extended to a homeomorphism $g: D^2 \to D^2$. 
Define a sequence via 
$
r_n=
\begin{cases}
 \frac{n+1}{n} &\text{if n even} \\
    \frac{n}{n+2} &\text{if n odd}
\end{cases}
$ 
This implies 
$\frac{r_n}{1-r_n}= -(n+1)$ for n even and $\frac{r_n}{1-r_n}=\frac{n}{2}$ for n odd. 
Hence 
$e^{i\frac{2\pi r_n}{1-r_n}}$=1 if n even and $e^{i\frac{2\pi r_n}{1-r_n}}$=-1 if n odd. If I had $\lim_{n \to \infty} r_n = 1$ then for the sequence of points $(r_n,t)$ where $t=0$ I would get 
$lim_{n \to \infty}g(r_n)\neq g(1)$ since $g(r_n)$ does not converge but alternates between 1 and -1. The problem is that currently the sequence $(r_n)$ does not converge, since the subsequences for even and odd n have different limits. Is it possible to define a sequence $(r_n)$ such that the conditions 
$\lim_{n \to \infty} r_n = 1$
$
e^{i\frac{2\pi r_n}{1-r_n}}=
\begin{cases}
 \  1 \ \text{if n even} \\
 \ -1 \ \text{if n odd} \\
\end{cases}
$
are satisfied?

Observe
$\quad \frac{2 \pi r}{1-r} = 1 \pi \text{ implies } r = \frac{1}{3}$
$\quad \frac{2 \pi r}{1-r} = 3 \pi \text{ implies } r = \frac{3}{5}$
$\quad \frac{2 \pi r}{1-r} = 5 \pi \text{ implies } r = \frac{5}{7}$
$\quad \frac{2 \pi r}{1-r} = 7 \pi \text{ implies } r = \frac{7}{9}$
etc.
Observe
$\quad \frac{2 \pi r}{1-r} = 2 \pi \text{ implies } r = \frac{1}{2}$
$\quad \frac{2 \pi r}{1-r} = 4 \pi \text{ implies } r = \frac{2}{3}$
$\quad \frac{2 \pi r}{1-r} = 6 \pi \text{ implies } r = \frac{3}{4}$
$\quad \frac{2 \pi r}{1-r} = 8 \pi \text{ implies } r = \frac{4}{5}$
etc.
