Determine bijective conformal self maps of $\Bbb C \setminus \{0,1\}$ I'm asked how to determine the bijective conformal self maps of $\Bbb C \setminus \{0,1\}$.
My attempt is to use the fact that Moebius transformations are uniquely determined by where I send $3$ points (where I'm allowed to pick $\infty$) and are invertible and conformal. Hence by seeing $\{0,1\}$ to $\{0,1\}$ and $\infty$ to $\infty$ I have a family of conformal bijective self maps of $\Bbb C \setminus \{0,1\}$.
How to prove that they exhaust all the possible self maps of $\Bbb C \setminus \{0,1\}$?
 A: Let $f$ be an automorphism of of $\Bbb{C}\setminus\{0,1\}.$ Since $f$ is injective according to Picard theorems non of the points $0, 1$ can be an essential singularity. Therefore at these points we need to have either removable singularities or poles of order one. Moreover these are the automorphisms of Riemann sphere that fixes the set $\{0, 1, \infty\}$. Thus we have following cases for possible automorphisms $f$:


*

*Möbius transformations that fixes $0, 1$ and $\infty$ 
$$z$$

*Möbius transformations that fixes $0$; interchange $1$ and $\infty$
$$\dfrac{z}{z-1}$$


*Möbius transformations that fixes $1$; interchange $0$ and $\infty$


$$\dfrac{1}{z}$$


*Möbius transformations that fixes $\infty$; interchange $0$ and $1$


$$1-z$$


*Möbius transformations that sends $0\to1\to\infty\to 0$
$$\dfrac{1}{1-z}$$

*Möbius transformations that sends $0\to\infty\to 1\to 0$
$$\dfrac{z-1}{z}$$
Also any composition of these Möbius transformations is also a candidate for such an automorphism, but here we have computed all such possible compositions. The set of these Möbius transformations form a subgroup of PGL ($2, \Bbb{C}$) called anharmonic group and it is isomorphic to the Dihedral group of order 6.
