I am given an action of $\Bbb{Z}_{2}$ on $\Bbb{C}\setminus\{0\}$ by $x+iy\mapsto -x+iy$ and $\text{id}:\Bbb{C}^{*}\to\Bbb{C}^{*}$ .
Then what is the orbit space homeomorphic to?.
The question here seems to be the same as my action when defined on cartesian coordinates, i.e. $(x+iy)\sim (-x+iy)$ and it says that the space is homeomorphic to $\Bbb{R}P^{1}\times \Bbb{R}_{>0}$ . I can see that why this is intuitively true. That is we have $\Bbb{C}^{*}=S^{1}\times \Bbb{R}_{>0}$ and the action of $\Bbb{Z}_{2}$ makes the quotient $\Bbb{C}^{*}/\Bbb{Z_{2}}\cong \frac{S^{1}}{\Bbb{Z}_{2}}\times \Bbb{R}_{>0}$ . But I cannot figure out how explicitly is $\Bbb{Z}_{2}$ acting on the set.
EDIT:- I think I have it. The quotient space of $S^{1}$ under the action is just $S^{1}$ with $0\leq \theta\leq\pi$ with $\theta\sim \pi -\theta$ and for $\pi\leq \theta\leq 2\pi$ we have $\theta \sim 3\pi - \theta $ and for the circle we must have $0\sim 2\pi$ . This would be homeomorphic to $\partial([0,1]\times [0,1])$ with $(x,1)\sim (1,x)$ and $(x,0)\sim (0,x)$ for all $x\in [0,1]$. Thus if we define $f:\partial(I^{2})\to I$ by $f(x,0)=\frac{x}{2}=f(0,x)$ and $f(x,1)=f(1,x)=\frac{1+x}{2}$ , then it descends to a homeomorphism from the quotient $\partial I^{2}/\sim $ to $[0,1]$ . Can you verify if all of these are correct?
The "question here" you mention is about the antipodal action $z\mapsto-z$, different from yours.
Your reduction to $\Bbb{C}^{*}/\Bbb{Z_{2}}\cong \frac{S^{1}}{\Bbb{Z}_{2}}\times \Bbb{R}_{>0}$ is correct, but your $\frac{S^{1}}{\Bbb{Z}_{2}}$ is not $\Bbb{R}P^1$. It is $\left[-\frac\pi2,\frac\pi2\right]$ (each of the two bounds of this segment is alone in its orbit, while each of the other $\theta$'s has $\pi-\theta$ in its orbit).