[Math] Explicit Homeomorphism from $\mathbb{R}^2$ to open disk

general-topology

I already asked a question here homeomorphism from $\mathbb{R}^2 $ to open unit disk.
I am trying to prove that $\mathbb{R}^2 \cong \{(x,y)\in \mathbb{R}^2|x^2+y^2<1\}$ The function that apparently solves this is $$f(z)=\frac{z}{||z||+1}$$
for $z\in \mathbb{R}^2$. In order to show this, I need to prove that $f$ is a continuous bijection. Continuity is obvious, but I see no reason why this is a bijection– I also see no reason why it should have a continuous inverse. Can someone explain this to me/point me in the right direction?
Thanks

Best Answer

Solve it directly. That is, solve $w = {z \over 1+ \|z\|}$ for $z$.

We see that $w=0$ iff $z=0$.

It is clear that $w$ and $z$ lie on the same line, so we can look for $z$ of the form $tw$ such that $w = {tw \over 1+ \|tw\|}$. It should be clear that if $w \neq 0$ then $t \ge 0$.

Then we have $1+t \|w\| = t$ and so $t={1 \over 1-\|w\|}$, from which we get $g(w) = {w \over 1-\|w\|}$.

We have $g(0) = 0$, $g$ is continuous, and we have $g(f(z)) = z$. It follows that $f$ is bijective.

Related Question