Let $S_{g,m}$ be a surface of genus $g$ with $m$ punctured, we know the fundamental group of $S_{g,0}$ is $$ \pi_1(S_{g,0}) = \left\langle a_1, b_1, \dots, a_g, b_g {~\large\mid~} [a_1, b_1] \dots [a_g,b_g] = 1 \right\rangle, $$ then what is the fundamental group of $S_{g,m}$? Thanks in advance.


1 Answer 1


Hint: Show that $S_{g,m} \simeq \bigvee_{i=1}^{m+2g-1}S^1$ for $m>0$. To do this, use the fundamental polygon of $S_g$; this should get easier as you add more holes!

  • 1
    $\begingroup$ In particular, you probably showed this for $m=1$ when originally calculating the fundamental groups of $S_{g,0}$ via Van Kampen's theorem. $\endgroup$
    – Dan Rust
    Jun 10, 2014 at 13:27
  • $\begingroup$ So $\pi_1(S_{g,m})=\mathbb{Z}^{m+2g-1}$, $m>0$? $\endgroup$
    – user151938
    Jun 11, 2014 at 5:40
  • $\begingroup$ @user151938 Careful: what you wrote generally refers to the free abelian group. $\pi_1(S_{g,m})$ is the free group on $m+2g-1$ generators. $\endgroup$
    – user98602
    Jun 11, 2014 at 5:47

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