Flipping the torus coordinates

Question

Here's the path on the torus parametrized by $(t,t/12)$ (i.e. twelve revolutions in one directions for each revolution in the other direction):

And here's the same path on a 3D embedding of the torus:

Now, if we flip the coordinates (by reflecting over the diagonal), the path looks like this when plotted on the unit square:

It is very obvious that this path is related in a simple way to the one shown before.

In contrast, the 3D representation of the second path looks very different from the one shown before:

It was not obvious to me, when I first saw them, that the two 3D paths shown above were isomorphic.

Q: Is it possible to deform one of the 3D representations to resemble the second one (in the way that the two flat representations resemble each other)?

Answer 1

I assume that by deforming you mean finding a homotopy between these two loops. Then the answer is no, because they represent different elements in the fundamental group of a torus $\pi(\mathbb{T}^2)=\mathbb{Z\oplus Z}$. One curve corresponds to $(1,12)$ and another to $(12,1)$. In fact, the fundamental group is defined exactly as the set of all loops on the space up to deformation. The reason, why $\pi(\mathbb{T}^2)=\mathbb{Z\oplus Z}$, is based on the fact that $\pi(\mathbb{S}^1)=\mathbb{Z}$ which uses some covering arguments and another fact $\pi(X\times Y)=\pi(X)\times\pi(Y)$, here take $X=Y=\mathbb{S}^1$. The details can be found in the Algebraic Topology book of Hatcher.

Answer 2

Cut a small hole in the torus and pull it through the hole to turn it inside out. Then glue the removed part back. Done.

See animation at Wikipedia
article: https://en.wikipedia.org/wiki/Torus#Topology
image: https://commons.wikimedia.org/wiki/File:Inside-out_torus_(animated,_small).gif