# How to prove that it is a surjective homomorphism?? [closed]

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 $$ad-bc\not=0$$.

• 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 ) – EtienneBfx Jan 28 at 14:20
• @EtienneBfx In $SL(2,\mathbb{C})$ we have $ad-bc=1$ where in $\mathcal{M}$ we have $ad-bc\not=0$. – Neil hawking Jan 28 at 16:36

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.

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

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