What is the quotient of $\mathbb{C}^*$ by the inverse action? Let $f:\mathbb{C}^*\longrightarrow\mathbb{C}^*$ be the inverse map $z\mapsto z^{-1}$. What is the quotient. I think it is $\mathbb{C}$.
Any point $re^{i\theta}$ is identified with $\frac{1}{r}e^{-i\theta}$. So the complement of the closed unit disc is identified with the interior of the unit disc. Further the upper semi circular portion of the unit circle is mapped to the lower semi circular region. Hence we get a sphere with a puncture, which is $\mathbb{C}$. Is this correct?
 A: That looks right to me. If you're willing to multiply by $i$, then you can say that the left and right half-circles are identified via $ x + iy \mapsto -x + iy$. You can probably even make the correspondence explicit, via something like this map defined on the unit disk:
$$
r e^{it} \mapsto \frac{r^2 - 1}{r} e^{it}
$$
which takes the interval $[-1, 1]$ (aside from $0$) to the real line, sending both $-1$ and $1$ to $0$. 
A: You have a group of order two acting on the (complex projective) line $\Bbb P^1(\Bbb C)$, and you can think of it as $z\mapsto z+\frac1z$. This latter quantity $z+\frac1z$ is a good generator of the fixed field of your two-element group. Under the map $z\mapsto z+\frac1z$, zero and $\infty$ go to $\infty$, and $\pm2$ are the only values $c$ such that the inverse image of $c$ is not a pair of distinct points. That is, the map is ramified only at $1\mapsto2$ and $-1\mapsto-2$.
The upshot is that the quotient is again the complex projective line, or in other language, the Riemann Sphere.
