Let $\text{SU}(1,1)=\left\{\left( \begin{array}{ccc} \alpha & \beta \\\overline\beta & \overline\alpha \end{array} \right)\mid \alpha,\beta\in \mathbb C,|\alpha|^2-|\beta|^2=1\right\}$. I need to prove that $\text{SU}(1,1)$ acts transitively on $\mathbb D=\{z\in\mathbb C\mid |z|\lt 1\}.$
First I defined a map $\text{SU}(1,1)\times \mathbb D\to \mathbb D$ such that $\biggr(\left( \begin{array}{ccc}\alpha & \beta \\\overline\beta & \overline\alpha \end{array} \right), z\biggr)\mapsto\dfrac{\alpha z+\beta}{\overline\beta z+\overline\alpha}$, which I hope is the required group action.
Let $w=\dfrac{\alpha z+\beta}{\overline\beta z+\overline\alpha}$. Then $z=\dfrac{\overline\alpha w-\beta}{\alpha -\overline\beta w}$. Since $|z|\lt 1$, we have $\left|\dfrac{\overline\alpha w-\beta}{\alpha -\overline\beta w}\right|\lt 1$. It follows that $|w|\lt 1$, that is, $w\in \mathbb D$. So the map is well-defined. Also it is easy to verify that the given map is indeed a group action.
Now to prove the transitivity, my idea is to show that $0$ can be sent to any element in $\mathbb D$. $$\left( \begin{array}{ccc}\alpha & \beta \\\overline\beta & \overline\alpha \end{array} \right)\cdot 0=z\iff \frac{\beta}{\overline\alpha}=z\iff \beta=\overline\alpha z.$$ Since we require $|\alpha|^2-|\beta|^2=1$, we have $|\alpha|^2=\dfrac1{1-|z|^2}$. So $\alpha=\dfrac{e^{i\theta}}{\sqrt{1-|z|^2}}$ and $\beta=\dfrac{ze^{-i\theta}}{\sqrt{1-|z|^2}}$. Hence the action is transitive.
I think my proof also shows that the action of $\text{SU}(1,1)$ on $\overline{\mathbb D}=\{z\in\mathbb C\mid |z|\leq 1\}$ is not transitive as $0$ cannot be sent to ,say $i$.
Are my arguments correct?
Suggestions/other solutions are welcome.