Fundamental group of the double torus

Question

In May's "A Concise Course in Algebraic Topology" I am supposed to calculate the fundamental group of the double torus. Can this be done using van Kampen's theorem and the fact that for (based) spaces $X, Y$: $\pi_1(X\times Y) = \pi_1(X)\times\pi_1(Y)$? Or do I need other theorems to prove this?

I believe that this should equal $\pi_1(T)\times\pi_1(T)\times\pi_1(S^1)$ where $T$ is the torus minus a closed disc on the surface, but I do not know how to calculate $\pi_1(T)$.

Answer 1

Hi: Please see this link, exercise 0.2 in the pdf file written by Christopher Walker in March 2, 2007 for Math 205B - Topology class. This has a nice explanation as well as some more information.

Answer 2

By van Kampen's theorem, what you get is actually $$\pi_1(T)\ast_{\pi_i(S^1)}\pi_1(T)$$ which is an amalgamated product (a pushout in the category of groups). Roughly speaking if you have two groups $G_1$ and $G_2$ and embeddings $i_1$ and $i_2$ of a group $H$ in both then $G_1\ast_H\ast G_2$ is the group freely generated by the elements of $G_1$ and $G_2$ but identifying elements $i_1(h)$ and $i_2(h)$ for $h\in H$.

Now $\pi_1(T)$ can be computed using the fact that $T$ deformation retracts to a bouquet of two circles. (Think about the standard torus; fix a point and look at the circles through it going round the torus in the two natural ways.)

Answer 3

As a futher hint: the fundamental group of the torus $\mathbb{T}$ is generated by $a, b$ and has the relation $abAB$, where capitals denote inverses. If you remove an open disk then you get a once-holed torus $T$. Now the fundamental group is free (why?) and the boundary is the homotopic to the element $abAB$ (why?).

So you can take another copy of the torus, say $\mathbb{S}$, with fundamental group generated by $c, d$ and having relation $cdCD$. Again remove a disk to get a once holed torus $S$. Now carefully follow the answer already given, gluing $T$ and $S$, and so on.