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!

Note that
$$f(z)=\frac{z-a}{1-\overline{a}z}$$
has inverse
$$g(z)=\frac{z+a}{1+\overline{a}z}.$$
