What are the Betti numbers of a double pinched torus?

Question

What are the betti numbers for a double pinched torus? A intuitive explanation for which holes these betti numbers correspond to would also be much appreciated.

Answer 1

The Betti numbers of a space are really the ranks of the integral homology groups of that space, so let's try to work those out. If you know about the Mayer-Vietoris homology long exact sequence, you should be able to extract the result fairly quickly.

Let $X$ denote the twice pinched torus. Notice that both "halves" of $X$ are homeomorphic to 2-spheres, and that we may decompose $X$ into two open subsets $U \simeq S^2$, $V \simeq S^2$, the union of whose interiors is all of $X$, by taking each open to be one half of $X$ extended a bit into the other half in such a way that it deformation retracts back to a 2-sphere. You should now be familiar with the homology of $U$, $V$ and $U \cap V$, the latter of which deformation retracts to the disjoint union of two points, and you may infer the homology of $X$, hence the Betti numbers of $X$, from the Mayer Vietoris homology long exact sequence associated to this situation, which looks like:

$... \to H_n(U \cap V) \to H_n(U) \oplus H_n(V) \to H_n(X) \xrightarrow{\delta} H_{n-1}(U \cap V) \to ...$

Note: From the mere fact that $X$ is connected, you know that $H_0(X) = \mathbb{Z}$, hence that the zeroth Betti number is 1; because $X$ admits a CW complex structure of dimension 2 composed of two 0-cells, two 1-cells and two 2-cells, you know that all the Betti numbers $\geq$ 3 must vanish, and that the first and second Betti numbers must be either 0, 1 or 2. The above method allows you to get to the end of the story.