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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 $\pi_1(T)$.

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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.)

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    I meant exactly what I said: see Ringo's definition of $T$.2010-08-15
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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.

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    @Michael Lugo: I agree with you.2010-08-15
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This is a response to Robin Chapman's answer. (For some reason I am not able to ask this directly under his question.)

Why do we get that formula from van Kampen? The double torus is the union of the two open subsets that are homeomorphic to $T$ and whose intersection is $S^1$. So by van Kampen this should equal the colimit of $\pi_1(W)$ with $W \in {T,T,S^1}$. I thought the colimit in the category of groups is just the direct sum, hence the result should be $\pi_1(T) \oplus \pi_1(T) \oplus \pi_1(S^1)$.

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    Ringo, you could ask the moderators to merge your two accounts (and so add your rep points together).2010-08-19
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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.