Proving a homeomorphism between two closed disks in $\mathbb{C}$ 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.
 A: No, what you have done is not correct. $a \leq b$ and $c \leq d$ does not give you $\frac a c \leq \frac b d$. 
Can you show that $f$ is continuous? The best way to complete the proof is to verify (by direct calculation) that $f(g(z))=g(f(z))=z$ where $g(z)=\frac {z+a} {1+\overline {a}z}$. Since $g$ is same as $f$ with $a$ changed to $-a$ it follows that $g$ is also continuous. Also, $g$ is the inverse of $f$ so $f$ is a homeomorphism. 
Continuity of $f$ follows from the fact that it is a ratio of two continuous functions and the denominator $1-\overline {a} z$ cannot be $0$: $|\overline {a} z| <1$ whenever $|z| <1$. 
A: The first fact follows from the observation that if $r,s\leq 1$ then $r^2 +s^2\leq 1+r^2 s^2 $, hence we pick the square of the norm and spanned it using the formula for the norm in the complex plane and make a comparison between denominator and numerator.
The metric, I think is just the normal Euclidean metric, and for your proof, we find that if $1-\bar{a} z\neq 0$, i.e. if $z\neq a/\vert a\vert^2 $(It's bound to be true since the norm of $\frac{a}{\vert a\vert^2 }$ is $\frac{1}{\vert a\vert }$, which is larger than $1$, but $\vert z\vert\leq 1$), then there is a lower bound for $\vert 1-\bar{a} z\vert $ for $z$ sufficiently closed to $z_1 $, hence the function is continuous.
