homeomorphism between a boundary and a sphere I started another question related to this. Consider $|\Delta^n|:=\{(x_0,..,x_n)\in\mathbb{R}^{n+1}:\sum_{i=0}^n x_i=1, x_i\in[0,1]\;\forall i\}$ the geometric realization of the standard n-simplex $\Delta^n$ endowed with the subspace topology of the standard topology on $\mathbb{R}^{n+1}$. Let $U=\{x\in \mathbb{R}^{n+1}:\sum_{k=0}^nx_i=1\}$ and $S^{n-1}=\{x=(x_0,..,x_{n-1})\in\mathbb{R}^{n}: \|x\|_2=1\}$. 
Consider $\partial_{U}|\Delta^n|=\{x\in|\Delta^n|: x_i=0\; \text{ for one of the i}\}$ the boundary of $|\Delta^n|$ in U, whereby U is homeomorphic to $\mathbb{R}^{n}$. 
I am searching for a homeomorphism between $\partial_{U}|\Delta^n|$ and $S^{n-1}$.  
I have  translate either $\partial_{U}|\Delta^n|$ or $S^{n-1}$ because if $x\in \partial_{U}|\Delta^n|$, the $x_i$ are nonnegative. Moreover the homeomorphism has to send the middlepoint of $\partial_{U}|\Delta^n|$ to the middlepoint of $S^{n-1}$ and 0 is the middlepoint of $S^{n-1}$ but on the boundary of $\partial_{U}|\Delta^n|$. I tried it with a map $\partial_{U}|\Delta^n|\to 1+S^{n-1}$, $1+S^{n-1}=\{x\in \mathbb{R}^{n}: (x_0-1)^2+..+(x_{n-1}-1)^2=1\}$ $x\mapsto 1+\frac{x-1}{\|x-1\|}$ but it doen't fit. I tried it a few days to find out the homeomorphism but i have no idea what is correct and i need the map for a longer proof...
Could you tell me the right homeomorphism? Regards.
 A: I'm going to simplify notation and write $\partial\Delta^n$ instead of $\ \partial_U |\Delta^n|$.
The space $U$ is not only homeomorphic to $\mathbb{R}^n$, it is isometric to $\mathbb{R}^n$. Pick an isometry $f \colon U \to \mathbb{R}^n$. Next, pick any point $x$ in the interior of $f(\Delta^n)$, and pick a number $r>0$ such that the closed ball $\overline B_r(x)$ around $x$ of radius $r$ is still in the interior of $f(\Delta^n)$. The choices of $f,x,r$ can all be made quite concretely if one so desires. Let $S_r(x)$ be the boundary of $\overline B_r(x)$, so $S_r(x)$ is homeomorphic to $S^{n-1}$, in fact there is a similarity homeomorphism of $\mathbb{R}^n$ which restricts to a homeomorphism between $S_r(x)$ and $S^{n-1}$; again this can be made quite concrete.
To define a homeomorphism $h : S_r(x) \to f(\partial\Delta^n)$: for each $y \in S_r(x)$, let $\vec{xy}$ denote the ray based at $x$ passing through $y$, and let
$$h(y) = \vec{xy} \cap f(\partial\Delta^n)
$$
This is well-defined, one-to-one, and onto, by convexity. Continuity is geometrically obvious, can also be proved abstractly using convexity, and can also be proved by writing down a formula. 
