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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

Let’s prove it!

For the sake of getting some contradiction, let $f:S_1to S_2$ 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 $Ssetminusgamma$ 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.

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.

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.

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.