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The fundamental group of Klein bottle has been well discussed in many materials, I'm think if it is requested to calculate the fundamental group of Klein bottle/{3pt} or even more generally Klein bottle/{n points}. I found it might has deformation retraction onto Mobius band with $n$ circles but not very sure. I'm looking forward to idea on it

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The Euler characteristic of the Klein bottle $K$ is equal to $0$.

From this, you can conclude that the Euler characteristic of $K$ minus $n$ points is $-n$.

It is a general fact that for any compact connected surface $S$ and any finite nonempty subset $P$, the complement $S-P$ deformation retracts onto a finite 1-complex $\Sigma$. So if we let $\Sigma_n$ denote a finite 1-complex onto which $K$ minus $n$ points deformation retracts, it follows that the Euler characteristic of $\Sigma_n$ equals $-n$ (and $\Sigma_n$ is connected).

The fundamental group of every finite connected graph $G$ is free of some rank $r \ge 1$, and the Euler characteristic of $G$ equals $1-r$.

Setting $-n = 1-r$, it follows that $r = 1+n$. So, the fundamental group of $K$ minus $n$ points is a free group of rank $1+n$.

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  • $\begingroup$ If for K#K/{3pt} will the outcome be different ? $\endgroup$ – Xin Hu Oct 18 '18 at 10:22
  • $\begingroup$ It will be different, but that's another question. $\endgroup$ – Lee Mosher Oct 18 '18 at 16:43
  • $\begingroup$ Is the question about spaces with 3 points removed or spaces with 3 points identified to a single point? $\endgroup$ – Ronnie Brown Oct 21 '18 at 14:55

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