Homeomorphism between two disks with a hole I need to show that $\mathbb{D}^2\setminus B(x_1,r_1)$ and $\mathbb{D}^2\setminus B(x_2,r_2)$ are homeomorphic. I have tried with a Möbius transformation of the disk but I can't show that it sends the first open ball onto the second one.
Could someone help me?
Thanks in advance.
 A: You can't expect to do this by Möbius transformations alone because a Riemann surface that is homeomorphic to an annulus has a positive real valued modulus which is preserved by conformal homeomorphisms, and all possible values of this modulus can occur. In particular, if $x_1=x_2=0$ and $r_1 \ne r_2$ then there is no Möbius transformation at all which works.
But that's okay, it just means that you'll need a nonconformal transformation to make this work.
First choose Möbius transformations $T_1,T_2$ of $\mathbb D^2$ so that $T_1(B(x_1,r_1)) = B(0,r'_1)$ for some radius $r'_1 > 0$, and so that $T_2(B(x_2,r_2)) = B(0,r'_2)$ for some radius $r'_2 > 0$. One can justify that these transformations $T_1,T_2$ and radii $r'_1,r'_2$ exist by concrete calculation as done in the answer of @ThomasAndrews, or more abstractly by using properties of the Poincaré metric on $\mathbb D^2$, also known as the hyperbolic metric (Euclidean circles in $\mathbb D$ are also hyperbolic circles; and hyperbolic circles of equal hyperbolic radius are congruent by some Möbius transformation, just as Euclidean circles of equal Euclidean radius are congruent by some Euclidean rigid motion).
Now construct a homeomorphism $g : [0,1] \to [0,1]$ such that $g(0)=0$, $g(r'_1)=r'_2$, and $g(1)=1$ (I'll leave this part as an exercise). Using this $g$, define a homeomorphism
$$G : \mathbb D \to \mathbb D
$$
by the formula
$$G(p) = \frac{g(|p|)}{|p|} \cdot p
$$
It follows that $G(B(0,r'_1)) = B(0,r'_2)$.
Finally, your desired homeomorphism
$$F : \mathbb D^2 \setminus B(x_1,r_1) \to \mathbb D^2 \setminus B(x_2,r_2)
$$
is given by
$$F(x) = T_2^{-1} \circ G \circ T_1
$$
A: There are three separate cases to show. Write the space $Y_{x,r}=\mathbb D^2\setminus B(x,r),$ when $r>0$ and $|x|+r<1.$

*

*Show $Y_{0,r_1}\cong Y_{0,r_2},$ for any $0<r_1,r_2<1.$ In this case, it is easy to come up with a homeomorphism, but harder to come up with one which is analytic.

*Given $x_1\in\mathbb R^+$ and $r_1,$ show $Y_{x_1,r_1}\equiv Y_{0,r_2}$ for some $r_2.$

*Every $Y_{x,r}$ is homeomorphic to a $Y_{x’,r}$ where $x’$ is a non-negative real.

For case 1: $$re^{i\theta}\mapsto e^{i\theta}\left(\frac{1-r_2}{1-r_1}\left(r-r_1\right)+r_2\right)$$
This is continuous on $Y_{0,r_1}$ with inverse gotten by switching $r_1,r_2.$ I’ll leave it to you to show this sends $Y_{0,r_1}\to Y_{0,r_2}.$
This is really just the hidden use of the homeomorphism $[r,1]\times S^1\to Y_{0,r},$ and the obvious homeomorphism between $[r_1,1]$ and $[r_2,1].$
Case 3. is easy - just rotate.
For case 2, we will use a Möbius transformation, as you guessed. This is tedious, but once you see the conditions needed, it just works.
We need $\phi(z)=\frac{az+b}{cz+d}$ such that $$\phi(x_1-r_1)=-\phi(x_1+r_1).\tag1$$ We can choose real $a,b,c,d,$ and we need $\phi(S^1)=S^1.$ This means either $a=d$ and $b=c$ or $a=-d, b=-c.$ We can choose the $a=d,b=c$ case because when the other case works, we can take $\rho(z)=-\phi(z).$
Choosing $\phi$ this way ensures that the image of the ball $B_(x_1,r_1)$ is a ball in $\mathbb D^2$ that is symmetric across the real line, which means its center must be on the real line. (1) then ensures that a diameter is centered at $0.$
We also need $\phi(0)\in \mathbb D^1,$ to ensure $\phi$ isn’t an inversion.
So if $u=x_1-r_1$ and $v=x_1+r_1,$ then $u,v$ are real and we need:
$$\frac{au+b}{bu+a}=-\frac{av+b}{bv+a}$$ It isn’t possible for $a=0,$ (or $\phi(z)=\frac1z$ would be an inversion,) so we can assume $a=1.$ Then $$(u+b)(bv+1)+(v+b)(bu+1)=0\\ (u+v)b^2+b(2uv+2)+(u+v)=0.$$
Since $u+v=2x_1\neq 0,$ and the discriminant is $$\begin{align} (2uv+2)^2-4(u+v)^2&=4(u^2v^2+2uv+1-(u^2+2uv+v^2))\\
&=4(u^2-1)(v^2-1)>0,
\end{align}$$
there are two real $b,$ and the product $b_1b_2=-1.$ We can’t have $b_1,b_2=\pm 1,$ or that would mean $uv=-1.$
So choose $b$ with $|b|<1.$ Then $\phi(0)=b\in \mathbb D^2$ and we are done.

More generally, let $U$ be a non-empty convex open subset of $\mathbb D^2$ with and $r<1$ that if $u\in U,$ $|u|<r.$ (So $U$ is bounded away from the boundary.) Then $X=\mathbb D^1\setminus U$ is homeomorphic to $Y_{0,1/2}.$
This amounts to picking $u_0\in U$ and seeing that for each $\theta,$ there are
$0<r_1(\theta)<r_2(\theta)$ such that for $r>0,$ $$u_0+re^{i\theta}\in X\iff r_1(\theta)\leq r_2(\theta).$$
This is because both $U$ and $\mathbb D^2$ are convex, and $U$ is open and $\mathbb D^2$ is compact in $\mathbb R^2.$
Since $u_0\in U,$ there is an$\epsilon>0$ such that $B(u_0,\epsilon)\subseteq U,$ and we see that $r_1(\theta)\geq\epsilon$ for all $\theta.$
Then our general theorem follows if we can show $r_1$ and $r_2$ are continuous.
This can be proved if we can prove:

If $V$ is open, convex and bounded on $\mathbb R^2$ and $v\in V.$ Then the function:
$$\rho:\theta\mapsto \inf\{ r\in\mathbb R^+\mid v+re^{i\theta}\notin V\}$$
is continuous.

Given any $\theta$ and let $x=v+\rho(\theta e^{i\theta}.$ Then for any $0<r<\rho(\theta),$ $v+re^{i\theta}\in V$ and any $r>\rho(\theta),$ $v+re^{i\theta}\in W=\mathbb R\setminus\overline V,$ where $\overline V$ is the closure of $V.$
Given $\epsilon>0$ choose $r_1=\rho(\theta)-\epsilon/2, r_2=\rho(\theta)+\epsilon/2.$ Then let $x_i=v+r_ie^{i\theta}.$ So we know that $x_1\in V$ and $x_2\in W.$ Find balls around $x_1,x_2$ that are contained in $V,W$ respectively.
Those balls must each contain an open arc $(\theta-\delta_i,\theta+\delta_)$ from the circles of radius $r_i,$ containing the angle $\theta.$ Picking the minimum $\delta$ between them, we see that if $|\theta_1-\theta|<\delta$ then $\rho(\theta)-\epsilon<\rho(\theta_1)<\rho(\theta)+\epsilon$ or $|\rho(\theta)-\rho(\theta_1)|<\epsilon.$
So $\rho$ is continuous.
[I kind of hand-waved to assert $x_2\in W.$ We know $x_2\notin V,$ but it seems like it takes some work to prove it is not in $\overline{V}.$ To get this, you need to know that if $v\in V$ and $x\in \overline V,$ then for all $t\in(0,1),$ $tx+(1-t)v\in V.$]
