Show that the function
$$\phi (z) = \frac{z}{1 – |z|} $$
defines a homeomorphism between the unit disc $D_{1}(0)$ and $\mathbb{C}$.
Now this is obviously bijective, as
$$ \phi^{-1} (z) = \frac{z}{1 + |z|}$$
gives an inverse function, as
$$ \phi^{-1} \circ \phi (z) = \frac{\frac{z}{1 – |z|}}{1 + \left| \frac{z}{1 – |z|}\right|} = \frac{z}{1 – |z| + |z|} = z$$
Noting that $| \ 1 – |z| \ | = 1 – |z|$ as $|z| < 1 $ always.
Similarly, we have
$$ \phi \circ \phi^{-1} = \frac{\frac{z}{1 + |z|}}{1 – \left| \frac{z}{1 + |z|} \right|} = \frac{z}{1 + |z| – |z|} = z$$
Noting that $| \ 1 + |z| \ | = 1 + |z|$ as $1 + |z| > 0$ always.
I am assuming continuity follows from the fact that both $\phi$ and $\phi^{-1}$ are compositions of continuous functions?
Furthermore, how do I know that the range of the first is all of the complex plane, and the range of the second is just the unit disc? That is,
$$ \phi (D_{1}(0)) = \mathbb{C} \quad \text{and} \quad \phi^{-1}(\mathbb{C}) = D_{1}(0)$$
Best Answer
That's not the inverse function of $\phi$. You just took the multiplicative inverse which is not the same. What you need is some continuous function $\psi:\mathbb{C}\to D_1(0)$ such that $\phi\circ \psi=Id_{\mathbb{C}}$ and $\psi\circ\phi =Id_{D_1(0)}$.
More information: https://en.wikipedia.org/wiki/Inverse_function
You should try $\psi(z)=\dfrac{z}{|z|+1}$.