# Fundamental Group of Punctured Plane

What is the fundamental group of $(\mathbb{C} \setminus {\{0\}})~/~\{e,a\}$, where $e$ is the identity homeomorphism and $az = -\bar{z}$? Clearly this is homeomorphic to the half cylinder , which is homotopy equivalent to the half circle. But what is the fundamental group of the half circle?

• I do not understand what is $(\mathbb{C} \setminus {\{0\}})~/~\{e,a\}$. Are you factoring $\mathbb{C}\setminus {\{0\}}$ by those two homeomorphisms? Well if it is homotopic to half circle which is contractible, than the fundamental group has to be trivial. – tom Apr 13 '14 at 19:22
• @tom She is letting the group $\{e,a\}$ act on the space $\Bbb C\setminus\{0\}$ and then quotienting the space by that action (i.e. identifying $z$ with $-\bar{z}$). – anon Apr 13 '14 at 19:24
• Yes, anon is correct. So the fundamental group is trivial? Why is the half circle contractible while the full circle is not? Formally I understand why the full circle is not contractible, but intuitively I don't see the difference. – Jen Apr 13 '14 at 19:31
• A circle is wrapped around a gaping hole. Intuitively you can't contract the circle without deleting said hole. The same can't be said for part of a circle. – anon Apr 13 '14 at 19:34
• Take rubber band, put it around bottle. Now try to squash the rubber band into a one point while keeping it on the surface of bottle. You can't really do it. But now cut the rubber band, than you can do it. – tom Apr 13 '14 at 19:36

Your space is homotopic to half circle $\{ e^{i\theta} : \theta\in[0,\pi]\}$ which is contractible. We can write down the contraction $$H(\theta,t) = e^{i \min\{\theta,\pi(1-t)\}}.$$
There is another way of putting this question. The cyclic group $C_2$ of order $2$ acts on the circle $S^1$ by conjugation $z \mapsto \bar{z}$. The fundamental group of the circle at $1$ is $\mathbb Z$ and the induced action on $\mathbb Z$ is $n \mapsto -n$; the quotient of $\mathbb Z$ by this action is cyclic of order $2$, which is the wrong answer, as shown by the answer of tom.
The resolution of this is that the action has two fixed points, $1,-1$; to avoid making a choice we consider the fundamental groupoid $\pi_1(S^1, \{-1,1\})$ and the induced action on this groupoid. The quotient groupoid by this action is indeed the groupoid $\mathcal I$ with two objects $-1,1$ and one arrow $\iota: -1 \to 1$.
The general theory of groups acting on groupoids is due to Higgins and Taylor, and is covered in Chapter $11$ of Topology and Groupoids.