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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 $$A=\left(\begin{array}{cc} a & b \\ c & d\end{array}\right)\mapsto g_A,\ g_A(z)=\frac{az+b}{cz+d}$$ 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 $ad-bc\not=0$.

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  • $\begingroup$ Maybe you should give the definition of Möbius transformations from your book. Indeed sometimes function $z \mapsto (az+b)/(cz+d)$ is the definition ( en.wikipedia.org/wiki/M%C3%B6bius_transformation ) $\endgroup$ – EtienneBfx Jan 28 at 14:20
  • $\begingroup$ @EtienneBfx In $SL(2,\mathbb{C})$ we have $ad-bc=1$ where in $\mathcal{M}$ we have $ad-bc\not=0$. $\endgroup$ – Neil hawking Jan 28 at 16:36
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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 $$A = \begin{pmatrix}a/t & b/t \\ c/t & d/t \end{pmatrix} \in SL(2,\mathbb{C})$$ 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.

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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, $$g_{A^{-1}}\circ g_A=g_A\circ g_{A^{-1}}={\rm id}.$$

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  • $\begingroup$ In $SL(2,\mathbb{C})$ we have $ad-bc=1$ where in $\mathcal{M}$ we have $ad-bc\not=0$. $\endgroup$ – Neil hawking Jan 28 at 18:09
  • $\begingroup$ Yes, we know that. So? $\endgroup$ – John B Jan 28 at 18:23
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    $\begingroup$ @JohnB The question was not to show that $g_A$ is onto. $\endgroup$ – Marktmeister Jan 28 at 18:44
  • $\begingroup$ @Marktmeister The question was only to show that $g_A$ is onto. And everybody knows that the action is factored by the determinant... $\endgroup$ – John B Jan 28 at 18:45
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    $\begingroup$ @JohnB No. The goal was to show that the mapping $SL(2,\mathbb C) \to \mathcal M$, $A \mapsto g_A$ is onto. $\endgroup$ – Marktmeister Jan 28 at 18:47

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