Let $\Omega\subset\mathbb{R}^2$ be a disk with $n\geq0$ holes, and $\partial\Omega$ its boundary, which consists of $n+1$ disjoint loops. Assume we are given $2k$ distinct points $p_1,q_1,\ldots,p_k,q_k\in\partial\Omega$.
I'm interested in constructing disjoint continuous paths connecting each pair of points $(p_i,q_i)$. Concretely, I'm looking for $\gamma_i : (0,1)\to\Omega$ with $\gamma_i(0)=p_i$ and $\gamma_i(1)=q_i$ such that the images of any two paths $\gamma_i,\gamma_j$ with $i\neq j$ are disjoint.
Solution for disk without holes (suggested by Karl in the comments): If $\Omega$ is a disk with no holes, we can tell by the ordering of the points on the boundary loop whether the problem is feasible. This is the case iff in the ordered set of $2k$ points, a subsequence of the form $(\ldots,p_i,\ldots,p_j,\ldots,q_i,\ldots,q_j,\ldots)$ never occurs. Equivalently, the ordered set can be partitioned in such a way that each subset looks like $p_i,a_1,\ldots,a_{2m},q_i$, such that the $(2m)$-tuple $a_1,\ldots,a_{2m}$ contains $p_j$ iff it contains $q_j$. This condition has to hold recursively for the subsequence $a_1,\ldots,a_{2m}$. Examples of feasible orderings with $k=3$ are $(p_1,p_2,p_3,q_3,q_2,q_1)$ and $(p_1,p_2,q_2,p_1,p_3,q_3)$ and $(p_1,q_1,p_2,q_2,p_3,q_3)$ and $(p_1,p_2,q_2,p_3,q_3,q_1)$.
Is there a similar ordering condition that tells us whether the problem is feasible when points are placed on multiple boundary loops?
I am guessing that this problem can be approached with tools from homology theory. I'd be glad for any concrete pointers that could help me tackle it.
Some thoughts: I am hoping that it is possible to reduce the problem inductively by adding a path between points on two different boundary loops, and thus effectively reducing the number of boundary components by one. However, does the choice of path matter when considering whether there exists a solution to connecting the remaining $k-1$ pairs?
My main difficulty is that in the presence of holes, there are countably infinitely many non-homotopic paths between two points on different boundary loops. (E.g. for a disk with one hole, we can circle the inner loop any number of times when connecting a point on the inner loop to a point on the outer loop.) It is not obvious to me whether adding many extra turns may be necessary to construct a solution to certain inputs.
A superficially similar problem is the disjoint paths problem in graph theory: Given $k$ vertex pairs on a graph $G$, can we connect all pairs with disjoint paths? This problem is NP-hard (if $k$ is part of the input), but I am hoping that the problem I posted will not be.
I think the general problem can be reduced to the no-holes version as follows (along the lines of your idea):
Whenever $p_i$ and $q_i$ are on different boundary components, if there is a solution, we can cut along the path from $p_i$ to $q_i$ to join the two holes (thinking of the outside of the disk as just another hole in the sphere). The order of vertices around the resulting larger hole doesn't depend on the path, so we can perform this operation on the input data.
Repeat this until only same-hole-edges remain. Now notice that there is a global solution if and only if there is a solution for each individual hole. (For the forward direction, just ignore parts of the global solution; for the backward direction, place small isolated solutions on the sphere.)