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

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?)

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.)