Let $n$ be a positive integer. Clearly $\mathbb{R}^{n-1}$ and the interior of the $n$-simplex $
\delta\_n := \{x \in [0,1]^n:\,\Sigma\_k x\_n =1, (\forall i)\,x\_i>0\}
$ are homeomorphic. What I'm looking for is much weaker, however. Namely, does anyone know of an explicit continuous surjection:
$$
f:\mathbb{R}^{n-1}\rightarrow \delta\_n,
$$
which has a *continuous right-inverse*?


**Update:**
Consider the maps
$$
\begin{aligned}
f\_1:\mathbb{R}^{n-1} & \rightarrow (0,1)^{n-1}\\
x & \mapsto \frac1{2}\cdot \left(1+ \frac{x\_k}{1+|x\_k|}\right)\_{k=1}^{n-1}\\
f\_2:(0,1)^{n-1} & \rightarrow \delta\_n\\
x & \to \frac{\|x\|\_{\infty}}{\|x\|\_1} x I\_{x\neq 0};
\end{aligned}
$$
then, isn't $f\_2\circ f\_1$ a homomorphism; so in particular setting $f=f\_2\circ f\_1$ satisfies the conditions we are looking for.



Continuous surjection of $\mathbb{R}^{n-1}$ onto the interior of the $n$-simplex with continuous right inverse
Let $\Delta\subset \mathbb R^{n-1}$ be an isometric copy of $\delta\_n$ with barycenter in the origin. Let $j:\mathbb R^{n-1}\to[0,+\infty)$ be the associated Minkowski functional, defined by $x\in j(x) \partial\Delta $ for all $x\in \mathbb R^{n-1}$. Then $x\mapsto \frac {j(x)}{j(x)+1}x$ is a homeo $\mathbb R^{n-1}\to \Delta$.


