how I must prove that the fundamental group of this shape with base point which is heavily shaded is $\left \langle a^{2},b^{2},aba^{-1},bab^{-1} \right \rangle$,and I want to use van kampen theorem.very naturally I think to pick the eight figure each side of the base point and do the van kampen theorem but it didn't go well,it will be great if you help me with some hint or explanation.

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  • $\begingroup$ The right side has fundamental group $\langle a^2, aba^{-1}\rangle$. You can see this by applying van Kampen to the figure eight with a base point in the middle, call it $c$. Then you can take the isomorphism $\beta_a:\pi_0(\infty,c)\cong\pi_0(\infty,s)$ (where $s$ is the shaded point) given by the upper edge $a$. $\endgroup$ Mar 27, 2014 at 14:39
  • $\begingroup$ my problem is exactly here :why it is $\langle a^2, aba^{-1}\rangle$? $\endgroup$
    – kpax
    Mar 27, 2014 at 15:15
  • 1
    $\begingroup$ The $\pi_0(\infty, c)$ is generated by $a^2$ and $b$. The isomorphism $\pi_0(\infty, c)\cong\pi_0(\infty, s)$ sends a loop $\gamma$ to $a\gamma a^{-1}$. $\endgroup$ Mar 27, 2014 at 15:21

1 Answer 1


Applying van Kampen to the open cover formed by small enough open neighborhoods around the two figure eight $\Large∞$ which meat at the heavily shaded dot $\bullet$, you see that $$π_0(\Large∞\hspace{-5pt}\small\bullet\hspace{-5pt}\Large∞, \small\bullet)\approx π_0(\Large∞\hspace{-3pt}\small\bullet,\bullet) *π_0(\small\bullet\hspace{-3pt}\Large∞,\small\bullet)$$ Now $π_0(\Large∞\hspace{-9pt}\small\bullet\hspace{6pt},\bullet)$ is generated by $a^2$ and $b$, and there is an isomorphism $$\beta_a:π_0(\Large∞\hspace{-9pt}\small\bullet\hspace{6pt},\bullet)\to π_0(\small\bullet\hspace{-3pt}\Large∞\small,\bullet)$$ given by sending the class of a loop $\gamma$ to the class of the loop $a\gamma a^{-1}$, which means that $π_0(\small\bullet\hspace{-3pt}\Large∞\small,\bullet)$ is generated by $a^2$ and $aba^{-1}$.


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