Showing that an open unit disk is homeomorphic to the surrounding space I was wondering that can I show that an open unit disk in $\mathbb{R}^n$ is homeomorphic to the surrounding space $\mathbb{R}^n$ by arguing that all of the half-axes of the space are homeomorphic to the half-axes of the unit disk. By half-axes of the space I refer to the two parts, $(-\infty, 0], [0, \infty)$ of each coordinate axis, and by half-axes of the open unit disk I refer to the intervals $(-1, 0], [0, 1)$ along all of the coordinate axis of the surrounding space.
So essentially I would argue that by some univariate functions $f$ and $-f$ all $[0, 1) \approx [0, \infty)$ and $(-1, 0] \approx (-\infty, 0]$. Then the mapping $g(x_1, x_2,\dots,x_n)$ from the $n$-dimensional open disk to the $\mathbb{R}^{n}$ could be constructed by considering the sign of the arguments, so that if $x_i \in [0, 1)$ then the $i$th coordinate of the image of $g$ is determined by $f$, and if $x_i \in (-1, 0]$ then it is determined by $-f$.
 A: Map $x \in \Bbb R^n$ to $f(x)=\frac{1}{1+\|v\|}\cdot v$, where $\cdot$ is scalar multiplication and the norm is the standard Euclidean one. So every vector is scaled down (in the same direction) so that its norm becomes
$$\left\| \frac{v}{1+\|v\|} \cdot v \right\| = \left(\frac{1}{1+\|v\|}\right)\|v\| = \frac{\|v\|}{1+\|v\|} < 1$$
and so $f$ maps $\Bbb R^n$ into the open unit ball $B(0,1)$.
$f$ is clearly continuous as the norm as a function is continuous and division in $\Bbb R$ and scalar multiplication are too.
$f$ is invertible as we can solve $f(tv)=v$ for $v \in B(0,1)$ uniquely and it's easy to see we get $f^{-1}(v)=\frac{1}{1-\|v\|}\cdot v$ which is also continuous (and well-defined on $B(0,1)$).
A: You say "if $x_i\in[0,1)$". But not all such coordinates combined give a point in an open disk (assuming the standard Euclidean norm). For example for $n=2$ the point $(0.9,0.9)$ does not belong to the unit disk. Your approach would be ok for $(-1,1)^n$, but not for unit disk.
The simpliest method for $B\to\mathbb{R}^n$ homeomorphism is to map $v\mapsto f(\lVert v\rVert)\cdot v$ where $f:[0,1)\to[0,\infty)$ is a homeomorphism, say $f(x)=\frac{\pi}{2}\cdot\tan(x)$. That's because it has a simple inverse: $v\mapsto f^{-1}(\lVert v\rVert)\cdot v$. As a nice bonus: this approach works with any norm.
