Is the torus with one hole homeomorphic to the torus with two holes?

Question

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

Answer 1

One of these has $H^1(\Sigma_1,\Bbb Z)\cong \Bbb Z^2$ and the other $H^1(\Sigma_2,\Bbb Z)\cong \Bbb Z^4$.

In general the $g$-holed torus $\Sigma_g$ has $H^1(\Sigma_g,\Bbb Z)\cong \Bbb Z^{2g}$.

One can see this by considering a representative loop for each of the $2g$-homotopy classes on the $g$-holed torus, and cutting along these, and by homeomorphism taking any pair of such cut surfaces to one another.

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Informally you can simply fix a base-point for loops to start and end at. On the $1$-holed torus $\Sigma_1$ there are two topologically distinct types of loops. Those that go 'through' the hole, and those that go around the torus. Similarly on the $2$-holed torus, there are $4$ different types of generating loops, either going 'around' each hole, or through it.

These generating loops all contribute $1$ rank to $H^1(\Sigma_g,\Bbb Z)$. So really there is nothing scary about this.

Answer 2

You can get a figure $8$ as the intersection of a plane with the torus with two holes but not with the one with only one hole.

Answer 3

Note: The following argument should be taken loosely, and not as a proof. It assumes the classification of surfaces (which is silly).

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.

Answer 4

I like your basic idea, but I would make it more precise as follows: in a two-holed torus, you can cut along two disjoint circles and leave the space still connected; in a one-holed torus, you cannot do this.

The first part is pretty easy and I think you have already found such circles.

So why can't we do this in a one-holed torus? Here I am going to wave my hands a little.

There are basically two kinds of circles in a torus. The trivial kind is just a little circle that cuts out a disk. To see the other kind, look at the torus as a unit square with the left and right edges glued together, and the top and bottom edges glued together (look up "flat torus" for more). Now draw a line starting at the lower left corner and going up and to the right with a slope of (say) $1/2$. This will end up going twice around the horizontal axis and once around the vertical axis before it comes back to the lower left, giving us a circle. You can do this for any slope $p/q$, including the special slope $\infty$ that goes straight up. The "favorite" choices of circle correspond to $\infty$ (vertical) and $0$ (horizontal); try drawing $3/5$ for a more interesting one.

The key thing here is that, up to a homeomorphism of the torus, those circles I just described are all the circles. This is not easy to prove (algebraic topology was invented to answer questions like this) but it should seem plausible if you experiment.

Assuming that's true, how would we find two disjoint circles in the torus which do not disconnect it? Clearly neither of the circles can be trivial, so they both have that $p/q$ form. In fact they must have the same slope: the first one cuts the torus into a cylinder, and the second one cuts that cylinder into two cylinders.

This shows that the two circles must always disconnect the torus, so you're done.