I have the following function $f(z)=\frac{z-a}{1-\bar{a}\,z}$, where $z\in\mathbb{C}$ and $|a|<1$ from the closed disk $|z|\leq 1$ to the closed disk $f(z)\leq 1$. I need to show that the disks are homeomorphic. I have shown that the function is one-to-one. It is easy to show that it is onto but I am having trouble showing that for any $w\in \{f(z)\in\mathbb{C}:|f(z)|\leq 1\}$, $\frac{w+a}{1+\bar{a}\,w}\in \{z\in\mathbb{C}:z\leq 1\}$ (to show that $f(\frac{w+a}{1+\bar{a}\,w})=w$). I have done the following
\begin{align\*}
\left|\frac{w+a}{1+\bar{a}\,w}\right|&\leq\frac{|w|+|a|}{1+|\bar{a}||\,w|}\\
&\leq 1.
\end{align\*}
Is this correct? The other issue I am facing is that the question does not mention the metric to use when proving the function is continuous. In any case (assuming the standard metric) I need to show for $\epsilon>0$ and $z\_{1}\in\{z\in\mathbb{C}:z\leq 1\}$, $\exists\,\delta(\epsilon)>0$: $$|z-z\_{1}|<\delta\Rightarrow|f(z)-f(z\_1)|<\epsilon.$$ To find $\delta$ I have done the following: $$|f(z)-f(z\_1)|=\frac{|z-z\_1|}{|1-\bar{a}\,z||1-\bar{a}\,z\_1|}.$$
(Don't know what to do with the denominator (tried simplifying)). Need some assistance proceeding further.