How to show the standard $n$-simplex is homeomorphic to the $n$-ball

Solution 1:

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?)

Solution 2:

So, why are $g$ and $g^{-1}$ continuous in @Seirios answer?

Here are the main facts (all easily verifiable)

1. The barycenter $c$ has all its coordinates equal to $1/(n+1)$.
2. The standard simplex $\Delta^n$ is included in the hyperplane $H=\{x\mid\sum_ix_i=1\}$.
3. If $x_{(1)}$ denotes the smallest coordinate of vector $x$, then the application $x\mapsto x_{(1)}$ is continuous.
4. The projection $f\colon B[c,r]\cap H\setminus\{c\}\to\partial\Delta^n$ is $$ f(x) = c + \rho(x)(x-c), $$ where $$ \rho(x) = \frac{1}{1-x_{(1)}(n+1)}. $$
5. 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 $$ \frac{\Vert x-c\Vert}{\Vert f(x)-c\Vert} = 1 - x_{(1)}(n+1). $$
6. If $y=g(x)$ then $$ 1 - y_{(1)}(n+1) = \frac{r}{\Vert f(x)-c\Vert}(1 - x_{(1)}(n+1)). $$
7. If $y=g(x)$ then $f(y)=f(x)$.
8. The inverse of $g$ is $$ h(y) = c + \frac{\Vert f(y)-c\Vert}{r}(y-c) $$ (similarly to part 7, show that $z=h(y)\implies f(z)=f(y)$.)
9. (Bonus) $r=\sqrt{1 - 1/(n+1)}$ (not required to complete the proof.)

Solution 3:

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].

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

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

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

4. 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 $f(x)=f(z)$.

5. 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 $x\ne0$.)

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

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

8. Conclude that $g$ is an homeomorphism.

9. 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].