How to show the standard $n$-simplex is homeomorphic to the $n$-ball I am trying to show the standard $n$-simplex is homeomorphic to the $n$-ball.
Here, the standard $n$-simplex is given by $$\Delta^n=\left\{(x_0,x_1,\cdots,x_n)\in\mathbb{R}^{n+1}:\sum x_i=1,x_i\geq0\right\}$$ and the $n$-ball is given by
$$B^n=\{x\in\mathbb{R}^n:||x||\leq 1\}$$
Any help will be appreciated.
 A: So, why are $g$ and $g^{-1}$ continuous in @Seirios answer?
Here are the main facts (all easily verifiable)


*

*The barycenter $c$ has all its coordinates equal to $1/(n+1)$.

*The standard simplex $\Delta^n$ is included in the hyperplane $H=\{x\mid\sum_ix_i=1\}$. 

*If $x_{(1)}$ denotes the smallest coordinate of vector $x$, then the application $x\mapsto x_{(1)}$ is continuous.

*The projection $f\colon B[c,r]\cap H\setminus\{c\}\to\partial\Delta^n$ is
$$
     f(x) = c + \rho(x)(x-c),
   $$
where
$$
    \rho(x) = \frac{1}{1-x_{(1)}(n+1)}.
   $$

*The homeomorphism $g\colon\Delta^n\to B[c,r]\cap H$, defined as
$$
     g(x) = \begin{cases}
                c       &{\rm if\ }x=c,\\
                c + \frac{r}{\Vert f(x) - c\Vert}(x-c) &\text{otherwise},
            \end{cases}
   $$
is continuous at $c$ because
$$
     \frac{\Vert x-c\Vert}{\Vert f(x)-c\Vert} = 1 - x_{(1)}(n+1).
   $$

*If $y=g(x)$ then
$$
     1 - y_{(1)}(n+1) = \frac{r}{\Vert f(x)-c\Vert}(1 - x_{(1)}(n+1)).
   $$

*If $y=g(x)$ then $f(y)=f(x)$.

*The inverse of $g$ is
$$
     h(y) = c + \frac{\Vert f(y)-c\Vert}{r}(y-c)
   $$
(similarly to part 7, show that $z=h(y)\implies f(z)=f(y)$.)

*(Bonus) $r=\sqrt{1 - 1/(n+1)}$ (not required to complete the proof.)

A: More generally, if $X$ is star-shaped, the center of $X$ is the set $Z$ of all $c\in X$ such that, for all $x\in X$, the segment $\{(1-\theta)c + \theta x \mid 0\le\theta\le 1\}$ is included in $X$. Since the $n$-simplex is convex, hence star-shaped, and its center is open, the Theorem below implies that the $n$-simplex is homeomorphic to the $n$-ball.
Theorem. If $X\subseteq\mathbb R^n$ is compact, star-shaped and its center $Z$ has a non-empty interior, then $X$ is homeomorphic to the $n$-ball $B[0,1]\subseteq\mathbb R^n$.
Proof [sketch].


*

*After a possible translation, we can assume that $0\in \operatorname{int}(Z)$. In what follows, let $X^* = X\setminus\{0\}$.

*For every $x\in X^*$ define $\ell_x = \{tx \mid t\ge0\}$.

*Put $\bar t=\sup\{t\ge0 \mid tx\in X\}$. Since $X$ is compact, the sup is attained and we can define
$$
     f(x)=\bar{t}x.
 $$

*The following properties hold
a. $f(x) \in \operatorname{cl}(X)$.
b. The segment from $0$ to $f(x)$ is included in $X$.
c. $\Vert f(x)\Vert\ge\delta$, where $\delta>0$ satisfies $B[0,\delta]\subseteq Z$ [cf. 1].
d. If $z\in X^*$ is such that $f(x)$ and $f(z)$ define the same ray, then $f(x)=f(z)$.

*Assume momentarily that $f\colon X^*\to\operatorname{cl}(X)$ is continuous. Then, the function $g\colon X\to B[0,1]$ defined as
$$
  g(x) = \begin{cases}
          \displaystyle\frac{x}{\Vert f(x)\Vert} &\text{if } x\in X^*,\\[0.1 in]
          0 &\rm otherwise
        \end{cases}
$$
is continuous. (Hint: Assume $(x_i)_{i\ge1}\subseteq X$ converges to $x\in X$. Show that $g(x_i) \to g(x)$ by studying separately the cases $x=0$ and $x\ne0$.)

*Show that $g$ is injective. (Hint: Assume $x\ne y$ and analyze two cases $\ell_x=\ell_y$ and $\ell_x\ne\ell_y$.)

*Show that $g$ is surjective. (Hint: If $z\in B[0,1]$, put $y=\delta z$ and  $y=\Vert f(x)\Vert z$. Then $g(y)=z$.)

*Conclude that $g$ is an homeomorphism.

*Now prove that $f$ is continuous, as follows:
a. Let $(x_i)_{i\ge1}\subseteq X^*$ converging to $x\in X^*$.
b. Since $X$ is compact we can assume $f(x_i)\to z\in\partial X$.
c. If $f(z)\ne z$, let $H$ be the hyperplane orthogonal to $z$ and $B = H\cap B[0,\delta]$.
d. Let $K$ be the cone with vertex  $f(z)$ and base $B$. Then $K\subseteq X$ is a closed neighborhood of $z$. Contradiction.
e. Use that $x_i$ and $f(x_i)$ belong in $\ell_{x_i}$ and $x$ and $f(x)$ in $\ell_x$ to show that $f(x)$ and $f(z)$ belong in the same ray $x/\Vert x\Vert$.
f. Conclude that $f(x)=f(z)$ [cf. 4. d].
A: Hint: $\Delta^n$ is convex, so you may may project $\Delta^n$ onto a ball $B^n \supset \Delta^n$ with respect to its barycentric center $c$.
The projection $f$ can be described as follow: First, notice that without loss of generality $B^n$ may be supposed to be centered at $c$; let $r$ denote its radius. For every $p \in \Delta^n \backslash \{c\}$, the ray from $c$ to $p$ meets $\partial \Delta^n$ at only one point $f(p)$. Now, we may define the projection $$g(p)= c+\frac{r}{\|f(p)-c\|} \cdot (p-c).$$ 
(Another related question: Proof that convex open sets in $\mathbb{R}^n$ are homeomorphic?)
