I am thinking about the following problem
Consider four disjoint disc of the form $D_k:=\{z\in \Bbb C:|z-a_k|\leq r_k\}$ for $ k=1,2,3,4$ in $\Bbb C$. Write $\text{int}(D_k):=\{z\in \Bbb C:|z-a_k|< r_k\}$ and $\partial D_k:=\{z\in \Bbb C:|z-a_k|= r_k\}$ for $ k=1,2,3,4$.
Does there exist a homeomorphism $f$ from $\displaystyle\Bbb C\backslash\big(\text{int}(D_1)\cup\text{int}(D_2)\cup\text{int}(D_3)\cup\text{int}(D_4)\big)$ to itself such that $f(\partial D_1)=\partial D_3$ and $f(\partial D_2)=\partial D_4$ with $\lim_{|z|\to \infty}\big|f(z)|=\infty$?
An alternative formulation of this question is that
Does there exist a homeomorphism $F$ from $\Bbb S^2\backslash\text{interior of four non-intersecting circles}$ to itself so that $F$ sends boundary one circle to the boundary of another circle?
Also what happens if I remove interior $2n$-many circles for any $n\geq 2$? Is it special for only $\Bbb S^2$, or it can be done for any manifold of dimension at least $2$, though in this case, we have to consider removing $n$-dimensional open ball from manifold.
Any help will be appreciated. Thanks in advance.
If you arrange the $4$ disks evenly around the equator of the unit $2$-sphere in $\mathbb{R}^3$, in the order $D_1,D_2,D_3,D_4$, then the map $x\mapsto -x$ will interchange $\partial D_1$ and$\partial D_3$ as well as interchange $\partial D_2$ and $\partial D_4$.
In general, if you have $n$ disks removed from a $2$-sphere, you can arrange all but $2$ of them along the equator, and the other two at the poles, and then reflect about the plane through the equator, swapping just two of them. Composing swaps will give you any permutation.
Note that applying an odd permutation in this way reverses the orientation of each boundary. To fix this, you can line all the disks along the equator and reflect about the plane through the equator.