Let $B_1(0,0) \subseteq \mathbb{R}^2$ denote the open ball of radius $1$ about the origin. One of the first things you learn while studying (general) topology is that $\mathbb{R}^2 \approx B_1(0,0)$. Explicitly, there is a homeomorphism $f: \mathbb{R}^2 \to B_1(0,0)$ given by $f(\mathbf{x}) = \frac{\mathbf{x}}{1 + |\mathbf{x}|}$, with inverse $\mathbf{x} \mapsto \frac{\mathbf{x}}{1 - |\mathbf{x}|}$. In coordinates, one has $f(x, y) = \left(\frac{x}{1 + \sqrt{x^2 + y^2}},\frac{y}{1 + \sqrt{x^2 + y^2}} \right)$.
My question: are there rational functions (defined on all of $\mathbb{R}^2$) $g,h: \mathbb{R}^2 \to \mathbb{R}$ so that the function $(x,y) \mapsto (g(x,y), h(x,y))$ is a homeomorphism $\mathbb{R}^2 \to B_1(0,0)$, with inverse also given by well-defined rational functions in each component?
Naively, if you simply drop the square root signs in the function $f$ above, the resulting function $(x, y) \mapsto \left(\frac{x}{1 + x^2 + y^2},\frac{y}{1 + x^2 + y^2} \right)$ is no longer a homeomorphism. For instance, the image of the parabola $\mathcal{P} := \{(x,x^2): x \in \mathbb{R}\}$ folds in on itself at the origin in $B_1(0,0)$ (this roughly follows from observing that the $y$-degree of the denominator in $\frac{y}{1 + x^2 + y^2}$ is larger than that of the numerator).
Maybe a simpler question: is the above result true if $\mathbb{R}^2$ is replaced by $\mathbb{R}^1$ (with $B_1(0,0)$ now an open interval of radius $1$ about the origin)?
If the answers to these questions are negative, does this work on perhaps a Zariski-open subset of $\mathbb{R}^2$? If somehow the answers are positive (!), does this generalize from $\mathbb{R}^2$ to $\mathbb{R}^n$?
Thanks for reading.
For $B_1(0,0)\to\Bbb R^2$, you can take $$ (x,y)\mapsto\left(\frac x{1-x^2-y^2},\frac y{1-x^2-y^2}\right)$$ but its inverse is not rational. Do you see why we cannot get rid of the denominator $1-x^2-y^2$ and it is an obtacle?