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? 
 A: Your space is homotopic to half circle $\{ e^{i\theta} : \theta\in[0,\pi]\}$ which is contractible. We can write down the contraction
$$
H(\theta,t) = e^{i \min\{\theta,\pi(1-t)\}}.
$$
Fundamental group of contractible space is trivial.
A: 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. 
See also my mathoverflow answer to the use of many base points. 
