I'm trying to solve problem 21 from chapter 1 of M. A. Armstrong's Basic Topology:
Let $C$ denote the unit circle in the complex plane and $D$ the disc which it bounds. Given two points $x,y \in D - C$, find a homeomorphism from D to D which interchanges x and y and leaves all the points of C fixed.
I tried to find a function which interchanges two given points in the complex plane, because I know that interior of any disk is homeomorphic to the plane. Here is the details:
If $x,y$ are the points which we want to be interchanged, then define the two functions as below: $$f:D-C \rightarrow \mathbb{C}\, \,\,\,\,\,\, st.\,\,\,\,\,\, f(z) = \frac {1}{1 - \mid z\mid} z$$ $$g:\mathbb{C} \rightarrow \mathbb{C}\, \,\,\,\,\,\, st.\,\,\,\,\,\, g(z) = f(x)+f(y)-z $$ we also have: $$f^{-1}:\mathbb{C} \rightarrow D-C \, \,\,\,\,\,\, st.\,\,\,\,\,\, f(z) = \frac {1}{1 + \mid z\mid} z$$ So $h=f^{-1}\circ g\circ f$ has the property in the interior of disk, but I don't know how to extend $h$ to a homeomorphism like $\hat{h}$ in order to leaving all points of $S$ fixed.
I also saw this and it seems to be a good argument but don't know how to extend it to the unit circle again.
Thanks for any help.
It is not that easy. Let us first prove that your $h$ extends to $D$. To do so, we show that if $(z_n)$ is a sequence in $U = D \setminus C$ which converges to $\zeta \in C$, then $h(z_n) \to -\zeta$. Let $d = f(x)+f(y)$. An easy calculation shows that $$h(z) = \dfrac{d-\frac{z}{1-\lvert z \rvert}}{1+\lvert d -\frac{z}{1-\lvert z \rvert} \rvert} = \dfrac{(1- \lvert z \rvert)d - z}{1-\lvert z \rvert + \lvert (1-\lvert z \rvert)d - z\rvert} .$$
As $z_n \to \zeta$, we get $\lvert z_n \rvert \to 1$, thus $1-\lvert z_n \rvert \to 0$ and $(1-\lvert z_n \rvert)d \to 0$ and $(1-\lvert z_n \rvert)d - z_n \to -\zeta$ and $\lvert z_n + (1-\lvert z_n \rvert)d \rvert \to \lvert -\zeta \rvert = 1$. Therefore $h(z_n) \to -\zeta$.
Thus $h$ extends to a homeomorphism $H : D \to D$, but on the boundary $C$ it is the antipodal map which is the same as a rotation by the angle $\pi$. We have to correct this.
Let $r = \max(\lvert x \rvert, \rvert y \rvert)$. Clearly $r < 1$. Define $$G : D \to D, G(z) = \begin{cases} z & \lvert z \rvert \le r \\ e^{\frac{\lvert z \rvert -r}{1-r}\pi i}z & \lvert z \rvert \ge r \end{cases}$$ This map twists the annulus between $r$ and $1$. Now let $$F = G \circ H$$ which is again a homeomorphism. Then $F(x) = H(x) = y, F(y) = H(y) = x$ and for $z \in C$ we have $F(z) = G(-z) = e^{\pi i}(-z) = z$.