algebraic-topologygeneral-topology
The problem concerns a two-holed torus and an infinite length of string that
passes through one of the holes. The object is to use continuous transformations
on either the 2-hold torus or the infinite piece of string so that the string
threads the two holes of the torus, i.e., it enters one hole and leaves
through the other. This is a problem posed in one of the video lectures of
N J Wildberger on youtube algebraic topology lectures.
After a lot of head scratching, I am at a loss.
This would be easiest to describe with a series of pictures but I don't have the time to draw them at the moment. Imagine your 2-holed torus is actually a 1-holed torus with another handle attached to one side along two disks. Now move one of these disks so that it circumambulates the hole of the central torus, moving around the perimeter until it comes back to near where it started. Now both of the torus's holes are penetrated by the line, kind of like two concentric circles which are tangent at a point. If you want you can push one of these circles up a bit so that the line goes through each hole at a different height. I hope this helps!
Update: here is a hand-drawn sequence of pictures which I took a photo of with my cellphone.
Actually that's correct. The fundamental group of the torus is $\mathbb{Z} \times \mathbb{Z}$, and, for example, the subgroup generated by $(1,0) $ and $(0,5)$ is an index 5 subgroup. Indeed it is easy to come up with a covering map $p:T^2 \to T^2$ of arbitrary (finite) degree. As you mention, the group of deck transformations in this picture can be taken to be a cyclic group of rotations.
Best Answer
You can do a continuous transformation of a genus-$2$ surface so that it has obvious three-fold rotational symmetry. Once it is in this form, it's hard to say whether the string passes through one or two holes, or even how many holes a genus-$2$ surface is meant to have!
David Richeson made a claymation video illustrating the transformation: https://www.youtube.com/watch?v=S5fPwE7GQOA