The map $f(z) = \frac{z-a}{1-\bar{a}z}$ preserves unit circle and open unit ball. Given $f(z) = \dfrac{z-a}{1-\bar{a}z}$, with $|a|<1$.
I showed that if $|z|=1$, then $|f(z)|=1$; if $|z|<1$, then $|f(z)|<1$. 
However, I am stuck at showing that the map $f$ is "onto". 
Is there any elementary way of showing that this map is onto? 
I looked at similar questions here, but they are only showing that $f$ is "into".
Thank you very much!

The algebra used in proving that $f$ is one-to-one (or "into") can also be used to show that $f$ is "onto".
\begin{align}
w & = \frac{z-a}{1 - \bar a z} \\[10pt]
z & = \frac{w+a}{1+\bar a w} = \frac{w-b}{1-\bar b w}
\end{align}
The same argument that shows that if $z$ is on the unit circle then so is $w$ will show that if $w$ is on the unit circle, then so is $z$, because $b$, like $a$, is within the closed unit disk.  Hence for every $w$, there is a suitable $z$, so $f$ is onto.
