I would like to understand why the torus with one hole is not homeomorphic to the torus with two holes.
I have a very basic understanding of the concepts (I know what an homeomorphism is but not much more).
My idea is that a torus with one hole can be disconnected by two loops, whereas the torus with two holes may not be disconnected by two loops.
Is this argument correct? Can it me formalized (even the part on connectedness).
Or maybe there is better argument
Best Answer
For the sake of getting some contradiction, let $f:S_1\to S_2$ be a homeomorphism. Here, $S_1$ is the one-holed torus and $S_2$ is the two-holed torus.
Now, define a non-separating curve $\gamma$ on a surface $S$ to be a curve such that $S\setminus\gamma$ is still connected.
Let $\gamma_1$ and $\gamma_2$ be the two most popular curves on $S_1$ that cut the torus into a disk. Since $f$ is a homeomorphism, the image of these curves under $f$ (i.e. $f(\gamma_1)$ and $f(\gamma_2)$) must also cut $S_2$ into a disk. Now, I claim that $f(\gamma_1)$ and $f(\gamma_2)$ must also be non-separating in $S_2$. This is a good exercise to stew on by yourself!
Moreover, I claim that no pair of non-separating curves can cut $S_2$ into a disk. To prove this, you pick your favorite pair of non-separating curves on $S_2$, and cut along them. Show this is not a disk. Then, for any other pair of non-separating curves $\alpha_i$, there is a homeomorphism taking the $\alpha_i$ curves to your favorite curves. Thus, cutting along the $\alpha_i$ curves cannot yield a disk.
However, from earlier we know that $f(\gamma_i)$ must cut $S_2$ into a disk, thus obtaining our contradiction.