I am given the following problem set:
Show that the cross-ratio is invariant under Moebius transformation, meaning that $$D \left(L_g(z_1),L_g(z_1),L_g(z_1),L_g(z_1)\right) = D(z_1,z_2,z_3,z_4) \qquad \forall z_i \in \mathbb{P}_1\mathbb{C} \quad i= 1, \ldots,4 \qquad \quad \text{where}\quad z_1 \neq z_4 \ , \ z_2 \neq z_3 \quad \text{and} \quad \forall g \in \text{GL}(2,\mathbb{C})$$ without using the generating matrices of GL$(2,\mathbb{C})$
The way they don't want to see:
Since we know that GL$(2,\mathbb{C})$ is generated by the matrices: $$\begin{pmatrix} 1 & b \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} a & 0 \\ 0 & d \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$$ we can show that the cross-ratio is invariant under each of these matrices. Henceforth this way is not the one they are looking for so I am puzzled how to prove it.
Maybe by showing that $L_g$ is bijective ?
Let $S(z)= (z, z_2, z_3, z_4)$ and $T(z)$ be a Mobius Transformation, then $ST^{-1}(z)$ takes $T(z_2)$ to $1$, $T(z_3)$ to $0$, and $T(z_4)$ to $\infty$. In other words, $ST^{-1}(z)=(z, T(z_2),T(z_3),T(z_4))$. Therefore $(z_1,z_2,z_3,z_4)=S(z_1)=ST^{-1}(T(z_1))=(T(z_1),T(z_2),T(z_3),T(z_4))$.