Kernel of inverse of the Möbius transformation

Question

Given $f(z)=\frac{az+b}{cz+d}$ the Möbius transformation. Calculate $ker(f^{-1}(id))$.

$$f(z)=\frac{az+b}{cz+d}=w\implies w(cz+d)=az+b\implies z(wc-a)+wd-b=0$$$$\implies z=\frac{-wd+b}{wc-a}=f^{-1}(z)$$ The parameter matrix is then $$P=\begin{bmatrix}-d & b \\ c & -a \end{bmatrix}.$$ What is $f^{-1}(id)$?

Answer 1

It seems the actual question is the following.

What is the kernel $${\rm ker}(\Phi)~=~\left\{M\in SL(2,\mathbb{C})\mid \Phi(M) = {\rm id}_{\mathbb{P}^1}\right\}, \qquad {\rm id}_{\mathbb{P}^1}(z)~:=~z, \qquad z~\in~\mathbb{P}^1, \tag{1}$$ of the group homomorphism $$SL(2,\mathbb{C})~\ni ~M~=~\begin{pmatrix} a & b \cr c & d \end{pmatrix} \quad\stackrel{\Phi}{\mapsto}\quad \Phi(M)~\in~{\rm Aut}(\mathbb{P}^1),\tag{2}$$ into the group ${\rm Aut}(\mathbb{P}^1)$ of Möbius transformations $$ \Phi(M)(z)~=~\frac{az+b}{cz+d} \quad?\tag{3}$$

Answer: $${\rm ker}(\Phi)~=~\{\pm {\bf 1}_{2\times 2}\}, \qquad {\bf 1}_{2\times 2}~:=~ \begin{pmatrix} 1 & 0 \cr 0 & 1 \end{pmatrix} . \tag{4}$$