By the uniformization theorem every connected and simply connected surface $M$ is conformally equivalent to one of the following three surfaces: open disk $D$, complex plane $\mathbb{C}$, or $2$-sphere $S^2 = \mathbb{C}\cup \{\infty\}$.

Let $H(M)$ be the group of conformal automorphisms of $M$ which is the same as the group of its biholomorphisms.

If $M=S^2$, then it directly follows from Schwartz lemma that $H(M)$ consists of Möbius transformations which leave $D$ invariant. This group is the same as the group of Möbius transformations of the upper half-plane (having thus real coefficients). Hence, $H(D)$ can be identified with the group of Möbius transformations of the real line. In other words, $H(D) = \mathrm{PSL}(2,\mathbb{R})$.

Further, it is written in many sources, including Wikipedia, that

**$H(S^2)$ is exactly the group of Möbius transformations**, and so it is isomorphic with the group $\mathrm{PSL}(2,\mathbb{C})$.It will then follow from 2) that $H(\mathbb{C})$ consists of affine maps $z\mapsto az+b$, i.e. it is the subgroup of $\mathrm{PSL}(2,\mathbb{C})$ fixing $\infty$, so $H(\mathbb{C})=\mathrm{Aff}(\mathbb{C})$.

My question is where one can find an exact proof of 2) that $H(S^2) = \mathrm{PSL}(2,\mathbb{C})$. Thank you in advance. I would very appreciate any information about that.