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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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4 Answers 4

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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"$T$ deformation retracts to a bouquet of two circles" -- you mean, $T$ minus a point(or disc) deformation retracts to a bouquet of two circles? –  user1119 Aug 14 '10 at 19:10
I meant exactly what I said: see Ringo's definition of $T$. –  Robin Chapman Aug 15 '10 at 7:17

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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Doesn't this answer your question? –  anonymous Aug 14 '10 at 17:51
-1: A merely related link is not an answer to a specific question. You could as well point to wikipedia. –  Rasmus Aug 14 '10 at 17:54
However, if the original answer had included the magic words "see Exercise 0.2", this would be a perfectly fine answer. (Note for those who don't follow the link: Baez gives the solution there. I don't think it makes sense to just give a link to where somebody has asked a question.) –  Michael Lugo Aug 14 '10 at 21:29
@Michael Lugo: I agree with you. –  Rasmus Aug 15 '10 at 10:57

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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No, the colimit in the category of groups is the amalgamated product as I described. In the category of Abelian groups the colimit is more or less the direct sum, but fundamental groups are not Abelian in general. You should consult a text dealing with van Kampen, e.g., Hatcher's. –  Robin Chapman Aug 18 '10 at 12:02
PS you can't post a comment in this section because you have two user IDs! –  Robin Chapman Aug 18 '10 at 12:02
Thank you! I now know why I was wrong. It should have been that "coproducts" (not colimits in general) are direct sums in the category of groups. (My old user ID was before I connected my account to my google account and after clearing the cookies I could not longer access it). –  Ringo Starr Aug 18 '10 at 19:35
Ringo, you could ask the moderators to merge your two accounts (and so add your rep points together). –  Robin Chapman Aug 19 '10 at 6:17

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.

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