I recently saw a homeorphism $f:B(0,1) \to \mathbb{R^n}$
Which goes as following : $f(x) = \frac {x}{1-||x||}$, $f^{-1}(x) = \frac {x}{1+||x||}$.
I'm required to fill in the details of the proof that it is indeed a homeorphism, and to find a homeorphism between $(0,1)^n \to \mathbb{R}^n$, This topic is rather new to me(I only picked up the definition of homeorphism), so I`ll be glad if someone could take a look at my proof and point out if any mistakes were made, or let me know about a more elegant way to prove it.
1) Show that $f(f^{-1}(y))=y$ and $f^-1(f(y))=y$ :
$$f\left(\frac{y}{1+\|y\|}\right)=\frac{\frac{y}{1+\|y\|}}{1-\left\|\frac{y}{1+\|y\|}\right\|}=\frac{\frac{y}{1+\|y\|}}{1-\frac{\|y\|}{1+\|y\|}} = \frac{y(1+\|y\|)}{(1+\|y\|)^2-\|y\|(1+\|y\|)} = y$$
$$f^{-1}\left(\frac{y}{1-\|y\|}\right) = \frac{\frac{y}{1-\|y\|}}{1+\left\|\frac{y}{1-||y||}\right\|} = \frac{\frac{y}{1-\|y\|}}{1+\frac{\|y\|}{1-||y||}} = \frac {y(1-\|y\|)}{(1-\|y\|)^2+\|y\|(1-\|y\|)}=y$$
2)Showing $f,f^{-1}$ are continous. $f^{-1}$ is obviously continous as a composition of elementary functions and the norm function (which is continous), $f$ is continous for the same reason, and since $||x||<1$ for every $x\in B(0,1)$.
Now to prove the homeorphism $(0,1)^n \to \mathbb{R}^n$ we consider $R=(\mathbb{R}^n, ||.||_\infty)$ The maximum norm.
We have $R \supset B(0,1) = \{x\in \mathbb{R}^n \space s.t \space ||x||<1 \} = \{x\in \mathbb{R}^n \space s.t\space max(|x_1|,..|x_n|)<1\} = (-1,1)^n$
$f$ from above is a homeorphism between $B(0,1)\to \mathbb{R}^n$ - only this time we use $||.||_\infty$ as our norm.
We can now take $g(x) = 0.5x+(0.5,...0.5)$ as an homeorphism between $g:(-1,1)^n \to (0,1)^n$ . now $f\circ g^{-1}:(0,1)^n \to\mathbb{R}^n $ is a homeorphism.