How to show that Mobius transformation is surjective on the unit disk?

Question

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,...

Answer 1

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.