Is there a homeomorphism between the $n$-dimensional cube to the $n$-dimensional simplex that doesn't create hard corners? Is there a homeomorphism between the $n$-dimensional cube to the $n$-dimensional simplex that doesn't create hard corners?  This is related to a similar question found here.  However, the mappings in response to that question create corners.  Specifically, I've been playing with schemes where


*

*Set a direction to squeeze the point $x\in[0,1]^n$, $v$

*Project $x$ into the nullspace of $v$, $y = x - \frac{v^Tx}{v^Tv}v$

*Draw a line between $x$ and $y$, $l(\alpha) =  \alpha(y-x)+x$

*$l(0)=x$, so we know it's in the hypercube.  Find
$$
  \alpha_l = \arg\max\{\alpha\geq 0:l(\alpha)\in[0,1]^n\}
  $$
$$
  \alpha_r = \arg\min\{\alpha\leq 0:l(\alpha)\in[0,1]^n\}
  $$
Basically, how far left and right we can travel and stay inside the cube

*Define the squeeze factor
$$
  \alpha = \frac{\alpha_l}{\alpha_l-\alpha_r}
  $$
Basically, how far between the leftmost and rightmost points we are along the direction $v$.

*Determine a new leftmost and rightmost point on the simplex,
$$
  z_l = l(\alpha_l)
  $$
$$
  z_r = \frac{1-\sum x_i}{\sum (y_i-x_i)}
  $$
It turns out the left most point is the same for the cube and the simplex.

*Use the squeeze factor to determine the new mapping for $x$
$$
  w=\alpha(z_r-z_l)+z_l
  $$


Anyway, that was a little circuitous, but it works.  It turns out if $v=(1,\dots,1)$, we get the first answer to the linked question above.  If we set $v = x$, we get the second scheme.
Alright, so what's the problem?  This mapping makes something that kind of looks like a reentrant corner on smooth areas mapped to the simplex.  For example, if we set $v=(1,\dots,1)$ and map a circle centered at $(0.5,0.5)$ of radius $0.25$ to the simplex, we go from

to

Basically, we get a heart shaped object with these hard corners.  I've tried a bunch of different schemes for choosing $v$ and I keep getting corners, which is undesirable.  I really want something that's mapped smoothly to the simplex.
Anyway, is there a smooth map from the hypercube to the simplex where where something like a circle mapped from the hypercube to the simplex won't have these corners?

Edit 1
I think @Del has a good answer and here's what it looks like.  Given a rectangular grid

The scheme transforms this into

which shows how the grid is squished to fit into the simplex.  It turns out this is smooth for the use case I described above.  We can see this with the deformed circle

Anyway, thanks for the help!
 A: I have one $C^\infty$ in the interiors: first of all the cube I'm considering is $C=(0,1)^n$, while the simplex $S$ is the interior of the convex hull of $\{0,e_1,\ldots,e_n\}$, where $\{e_i\}$ is the standard orthonormal basis of $\mathbb R^n$. If $O$ denotes the open positive orthant then $C=O\cap \{x:\|x\|_\infty< 1\}$ and $S=\{x:\|x\|_1< 1\}$.
The diffeomorphism is $f:C\to S$ defined by
$$\left(f(x)\right)_i=x_i\prod_{i<j\leq n}(1-x_j)= x_i(1-x_n)\cdots (1-x_{i+1}).$$
For instance in $\mathbb R^3$ it's
$$ \begin{pmatrix}x\\y\\z\end{pmatrix}\mapsto \begin{pmatrix} x'\\y'\\z'\end{pmatrix}=\begin{pmatrix}x(1-y)(1-z)\\y(1-z)\\z\end{pmatrix}.$$
You can check by backwards substitution that the inverse is
$$ \begin{pmatrix}x'\\y'\\z'\end{pmatrix}\mapsto \begin{pmatrix} x\\y\\z\end{pmatrix}=\begin{pmatrix}\frac{x'}{1-y'-z'}\\\frac{y'}{1-z}\\z'\end{pmatrix}$$
or in general
$$(f^{-1}(x'))_i=\frac{x'_i}{1-\sum\limits_{i<j\leq n}x'_j}.$$
It is obviously smooth. Let us check it sends $C$ to $S$ by induction on $n$. The case $n=2$:
$$x_1(1-x_2)+x_2=(1-x_2)(x_1-1)+1\leq 1$$
and assuming the result true for $n-1$
$$\sum_{i=1}^n x_i\prod_{i<j\leq n}(1-x_j)=x_n+(1-x_n)\sum_{i=1}^{n-1}\prod_{i<j\leq n-1}(1-x_j)\leq x_n +(1-x_n)=1.$$
Moreover $f^{-1}$ sends $S$ to $C$:
$$\frac{x'_i}{1-\sum_{i<j\leq n}x'_j}\leq \frac{1-\sum_{j\neq i}x'_j}{1-\sum_{i<j\leq n}x'_j}\leq 1.$$
What it essentially does is collapsing some faces into other lower dimensional ones (as suggested by Pedro Tamaroff in the comments) and interpolating sort of linearly in the middle.
A: See my answer to the duplicate question here.

