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.

• Are you looking for an idea or the details? Oct 24, 2013 at 15:30
• @copper.hat I am looking for the details, since I have actually thought about this problem for days. Thank you!
– YYF
Oct 24, 2013 at 15:35
• Unfortunately I don't have time for the details now. Oct 24, 2013 at 15:43

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?)

• What do you mean by "project"? I did not get it. Actually, I am self-learning algebraic topology by reading Rotman's book, and it is my first time learning simplices.
– YYF
Oct 24, 2013 at 15:54
• I added a description of the projection. Oct 24, 2013 at 16:42
• It is pretty clear on a figure, but you can use the following argument: $\partial \Delta^n$ consists in the intersections of $\Delta^n$ with the hyperplanes $x_i=0$. Because $c$ does not belong to any of these hyperplanes, any ray starting from $c$ can meet $\partial \Delta^n$ at most once, and in fact exactly once since $c \in \Delta^n$. Dec 6, 2014 at 19:27
• There is a generalization of this in Bredon's Topoloy and geometry, Proposition 16.4, page 56. The proposition says: ''A compact convex body $C$ in $\mathbf{R}^n$ with nonempty interior is homeomorphic to the closed $n$-ball''. Feb 4, 2015 at 20:49
• Why is $g$ continuous, why is $g$ continuous at $c$ and why is $g^{-1}$ continuous? May 17, 2017 at 7:12

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.)

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].