 Solution

Represent a Möbius transformation by a matrix, with nonzero scalar multiples representing the same map. Matrix multiplication reproduces composition, so $\theta$ is a group homomorphism. Every Möbius map has a representing matrix $A\in G{L}_2(\mathbb{C})$; multiplying $A$ by either square root of $(\det A{)}^{-1}$ gives a matrix in $S{L}_2(\mathbb{C})$ representing the same map. Thus $\theta$ is surjective. A matrix in its kernel represents the identity and is therefore scalar; determinant one leaves

If a nonidentity Möbius map $T$ has two distinct fixed points, conjugate those points to $0$ and $\infty$. The conjugated map fixes both and hence has the form $S(z)=\mu z$ with $\mu \ne 0,1$. If $T$ has one fixed point, conjugate it to $\infty$. The resulting affine map has the form $az+b$; having no finite fixed point forces $a=1$ and $b\ne 0$, and conjugating by a scaling makes it $S(z)=z+1$.

Directly, the finite fixed points satisfyIncluding $\infty$ when appropriate, a nonidentity Möbius map consequently has one or two distinct fixed points.

Finally suppose $T$ has the unique fixed point $z_0$. With the conjugacy $R$ above,Every finite point tends to $\infty$ on the Riemann sphere, while $\infty$ itself is fixed. Applying the continuous map $R^{-1}$ shows thatfor every $z\in \mathbb{C}\cup \{\infty \}$.