Explicit homeomorphism between $\mathbb{R}^n$ and $(0,1)^n$

I would like to prove that the map $f : \mathbb{R}^n\to (0,1)^n$ defined by

$f(x)=(1_{(0,+\infty)}f_1(x_1)+1_{(-\infty,0])}f_2(x_1),...,1_{(0,+\infty)}f_1(x_n)+ 1_{(-\infty,0]}f_2(x_n))$ where $f_1(x_i)=\frac{1+x_i}{2+x_i}$ and $f_2(x_i)=\frac{1}{2-x_i}$.

is an homeomorphism.

My idea is to prove that it is the case when $n=1$ and shows that a vector function where each component is an homeomorphism is still an homeomorphism.

The case $n=1$. We show that $f_1(x)=\frac{1+x}{2+x}$ is an homeomorphism from $(0,\infty)$ to $(\frac{1}{2},1)$. Let $f(x_1)=f(x_2)$ then we have $(1+x_1)(2+x_2)=(1+x_2)(2+x_1)\iff x_2+2x_1+x_1x_2=x_1+2x_2+x_1x_2\iff x_2-x_1=2(x_2-x_1)$ so $x_1=x_2$ and $f_1$ is injective.

Now let $y\in(\frac{1}{2},1)$. We have $y=\frac{1+x}{2+x}\iff x=\frac{2y-1}{1-y}$ which is in $(0,\infty)$ and gives also the expression of $f_{1}^{-1}$. The continuity of both $f_1$ and its inverse is well known so $f_1$ is an homeomorphism.

Now consider $f_2(x)=\frac{1}{2-x}$ from $(-\infty,0]$ to $(0,\frac{1}{2}]$, the continuity is clear, as for the injectivity. Let consider $y\in(0,\frac{1}{2}]$. Then $y=\frac{1}{2-x}\iff x=\frac{2y-1}{y}$ which proves the surjectivity and clearly this inverse is continuous so $f_2$ is an homeomorphism.

Now we observe that $\mathbb{R}=(-\infty,0]\cup(0,+\infty))$ and $(0,1)=(0,\frac{1}{2}]\cup(\frac{1}{2},1)$ and consider the map $f= 1_{(0,+\infty)}f_1(x)+1_{(-\infty,0])}f_2(x)$, it is the wanted homeomorphism between $\mathbb{R}\to(0,1)$.

Now consider $f : \mathbb{R}^n\to(0,1)^n$ defined by

$$f(x)=(f_1(x_1),...,f_n(x_n))$$where each $f_i$ is an homeomorphism from $\mathbb{R}\to(0,1)$. We infer that :

It is a bijection, continuous as each component function is continuous and its inverse is given by $f^{-1}:(0,1)^n \to\mathbb{R}^n$

$$f^{-1}(x)=(f_{1}^{-1}(x_1),...,f_{n}^{-1}(x_n))$$

which is continuous as each component function is.

Using this result, we conclude that

$$f(x)=(1_{(0,+\infty)}f_1(x_1)+1_{(-\infty,0])}f_2(x_1),...,1_{(0,+\infty)}f_1(x_n)+ 1_{(-\infty,0]}f_2(x_n))$$

is an homeomorphism between $\mathbb{R}^n$ and $(0,1)^n$.

I would like to know if the arguments used here are correct please. I think the idea should be this one but I may lack precision.

Thank you a lot !

I think also that such homeomorphism (if it is the case) is a semi-algebraic mapping. It can be seen by writing explicitly the graph of the mapping, each « component equality » correspond to a polynomial equality