# The fundamental group of Klein bottle/{3pt}

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

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 $$1+n$$.

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