How do I show that $\phi(z) = i\frac{1-z}{1+z}$ is a bijection $\phi: D \to \mathbb{H}$ between the open disk $D = \{z \in \mathbb{C} : |z| < 1 \}$ and the upper half plane $\mathbb{H} = \{ z \in \mathbb{C} : \text{Im} z > 0 \}$?
If I understand the question correctly, showing $\phi$ is both surjective and injective tells me nothing about whether there is a bijection between the specified sets? And the same holds for finding an explicit inverse?
I think I am not understanding how to approach a question like this, it's the first of its kind for me. And I am also not understanding how to think about it. Perhaps I am missing some basic prior knowledge.
Best Answer
Note that if we solve $w=\phi(z) = i {1-z \over 1+z} $ formally then we get $z = {i-w \over i+w}$.
Note that $-1$ in the domain and $-i$ in the range (codomain) are special. If we try to solve $\phi(z) = -i$ we get $1=-1$ hence the range of $\phi$ does not include $-i$. In particular, it shows that $\phi$ is a bijection $\phi: \mathbb{C} \setminus \{ -1\} \to \mathbb{C} \setminus \{ -i\}$. (As an aside, as a map on the Riemann sphere, $\phi: C_\infty \to C_\infty$ is a bijection and $\phi(-1) = -i$.)
Hence if we can show that $\phi(D) = \mathbb{H}$ we are finished.
(Note that $-1 \notin D$ and $-i \notin \mathbb{H}$.)
We have $\phi(x+iy) = i{1-x-iy \over 1+x+iy} = i{(1-x-iy) (1+x-iy) \over (1+x)^2+y^2} = i{1-x^2-y^2 -2 iy \over (1+x)^2+y^2} = {-2y +i(1-x^2-y^2) \over (1+x)^2+y^2}$.
Hence we see that $\operatorname{im} \phi(x+iy)>0 $ iff $x^2+y^2 < 1$.