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The purpose of this question is to understand the computations to get the expression of the fundamental group in a simple case using the Van Kampen theorem.

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Let $X$ be the three holes object obtained from the union of two "figure eight" $A$ and $B$. The fundamental groups $\pi_1(A)$ and $\pi_1(B)$ (computed at $O$) are both isomorphic to $\mathbb{Z} * \mathbb{Z}$ and by extension I suppose that $\pi_1(X)$ will be isomorphic to $\mathbb{Z} * \mathbb{Z} * \mathbb{Z}$.

Using the notations from http://mathworld.wolfram.com/vanKampensTheorem.html, $N$ is the free subgroup of $\pi_1(A) * \pi_1(B)$ generated by $a_2 b_1^{-1}$.

The Van Kampen theorem indicates that:

\begin{equation} \pi_1(X) \approx (\pi_1(A) * \pi_1(B)) / N \end{equation}

As I understand, each coset of this quotient group is generated by applying one element of $\pi_1(A) * \pi_1(B)$ to each element of $N$. So for example if a representative element is $a_1a_2b_2a_1$, the corresponding coset will contain elements such as $a_1 a_2 b_2 a_1 a_2 b_1^{-1}$, $a_1 a_2 b_2 a_1 2 a_2 2 b_1^{-1}$, etc... Since the elements of $N$ are only present at the right end of those resulting elements, I fail to understand how $\mathbb{Z} * \mathbb{Z} * \mathbb{Z}$ might emerge as the fundamental group of $X$.

Can you help me with understand, from the expression of some of the cosets of $(\pi_1(A) * \pi_1(B)) / N$, why does $\pi_1(X) \approx \mathbb{Z} * \mathbb{Z} * \mathbb{Z}$ ?

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$N$ is not the subgroup generated by some elements; it is the normal subgroup generated by some elements, and so contains all of their conjugates. Practically what this means is that when you work $\bmod N$ you can replace $a_2$ with $b_1$ wherever it appears in a word, and vice versa (which should make sense since these are the same loop in $X$).

So you can describe $\pi_1(X)$ equivalently as either the free group on generators $a_1, a_2, b_2$ or $a_1, b_1, b_2$.

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  • $\begingroup$ So for example the words $a_2 a_1 b_2 a_2$ and $b_1 a_1 b_2 b_1$ belong to the same quotient group. What is the element $n$ of $N$ so that $a_2 a_1 b_2 a_2 + n = b_1 a_1 b_2 b_1$ ? $\endgroup$ – vkubicki Jan 11 '16 at 20:20
  • $\begingroup$ @vkubicki: it's uniquely determined by that condition: it must be $(a_2 a_1 b_2 a_2)^{-1} b_1 a_1 b_2 b_1 = a_2^{-1} b_2^{-1} a_1^{-1} a_2^{-1} b_1 a_1 b_2 b_1$. This is, again, not in the subgroup generated by $a_2 b_1^{-1}$, but it is in the normal subgroup generated. $\endgroup$ – Qiaochu Yuan Jan 11 '16 at 21:39
  • $\begingroup$ OK, thanks this gets clearer now ! $\endgroup$ – vkubicki Jan 12 '16 at 22:53

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