A homeomorphism of the unit disk that cannot be extended to the boundary of its domain

Question

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?

Answer 1

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.