How to show that Mobius transformation is surjective on the unit disk? For $z\in \hat{\mathbb{C}}$, let:
$$T(z)=\frac{az+b}{cz+d}~~~~~~~~~~~~~~~~ad-bc=1$$
be the mobius transformation on the Riemann sphere. 
1) Show that  $T:\hat{\mathbb{C}}\longrightarrow \hat{\mathbb{C}}$ is onto.
2) Show that $T:\mathbb{D} \longrightarrow \mathbb{D}$ is onto.
My purpose is to show that in both cases $T$ is bijective. It is easy to prove that $T$ is one-to-one on the Riemann and hence on $\mathbb{D}$. But for surjectivity,...

For (1), note that $T$ has inverse
$$T^{-1}(z)=\frac{dz-b}{-cz+a}.$$
But (2) isn't true in general. Whether or not $f$ maps into or onto $D$
depends strongly on what $a$, $b$, $c$ and $d$ are.
