How to prove that it is a surjective homomorphism?? Ergodic Theory, Symbolic Dynamics and Hyperbolic Spaces, Chapter 1, Page 12.

Let $\mathcal{M}$ be the group of all Möbius transformations. It is well known that the map
$$A=\left(\begin{array}{cc} a &amp; b \\ c &amp; d\end{array}\right)\mapsto g_A,\ g_A(z)=\frac{az+b}{cz+d}$$
is a homomorphism of $SL(2,\mathbb{C})$ onto $\mathcal{M}$ [...]

How to prove that this homomorphism is surjective?
In $SL(2,\mathbb{C})$ we have $ad-bc=1$ where in $\mathcal{M}$ we have $ad-bc\not=0$.
So $\mathcal{M} = \{ g :\hat{\mathbb{C}} \to \hat{\mathbb{C}}, \exists(a,b,c,d) \in \mathbb{C}^4, ad-bc \ne0, g(z)=\frac{az+b}{cz+d}  \} $
So take any $(a,b,c,d) \in \mathbb{C}^4$ with $ad-bc \ne 0$, and let called $\mathcal{M} \ni g:z \mapsto\frac{az+b}{cz+d}$.
Let $t$ be one of the two square root of $ad-bc$ then
$$A = \begin{pmatrix}a/t &amp; b/t \\ c/t &amp; d/t
 \end{pmatrix} \in SL(2,\mathbb{C})$$
and $g_A(z)=\frac{za/t +b/t}{zc/t+d/t}=\frac{az+b}{cz+d}=g(z)$
So we find a preimage of $g$ and the map is surjective.
The inverse of a matrix with determinant $1$ also has determinant $1$.
So $g_{A^{-1}}$ is also in $SL(2,\mathbb{C})$, which shows that $g_A$ is both onto and one-to-one.
Indeed, $$g_{A^{-1}}\circ g_A=g_A\circ g_{A^{-1}}={\rm id}.$$