Homeomorphism between $k$-simplex and a product of $k$ unit intervals How to prove that there is a homeomorphism between the $k$-simplex $[v_0, v_1, \dots, v_k]$ and a product of $k$ unit intervals?
This question is motivated by 'Lecture Notes on Elementary Topology' by I. Singer, where it is claimed that this "is not difficult to prove using barycentric coordinates" (Chapter 4, Page 71).
The $k$-simplex is defined as
$$
[v_0, \dots, v_k]
= \left\{\sum_{i=0}^{k} a_i v_i
\text{ such that } a_i \geq 0 \text{ and }
\sum_{i=0}^{k} a_i = 1
\right\} .
$$
$a_i$ are called barycentric coordinates.
(EDIT: $\{v_0, \dots, v_k\}$ is a set of $C$-independent vectors, namely the set $\{v_1-v_0, \dots, v_k-v_0\}$ is linearly independent.)
For example, given three noncolinear vectors, $[v_0, v_1, v_2]$ is the triangle with vertices located at $v_0$, $v_1$ and $v_2$.
According to the theorem that I am asking about, there should be a homeomorphism between that triangle and the unit square $[0,1]\times[0,1]$.
I do not know how to find that homeomorphism. My only idea is that the barycentric coordinates $(a_0, a_1, \dots, a_k)$, due to the condition $\sum_{i=0}^{k} a_i = 1$, are homeomorphic to a subset of $\Pi_{i=1}^{k} [0,1]$ by the map $f:[v_0, \dots, v_k] \rightarrow \Pi_{i=0}^{k}[0, 1]$ defined as
$$f(a_0, a_1, \dots, a_k) = (a_1, a_2, \dots, a_k).$$
But that is not a homeomorphism to $\Pi_{i=1}^{k} [0, 1]$: for example, the point $(1,1,\dots,1) \in \Pi_{i=1}^{k} [0, 1]$ has no nonempty inverse image.
I am a beginner in elementary topology and would appreciate very much any help.
 A: You do not tell us what $v_0,\ldots, v_k$ are, but certainly they are noncolinear vectors in some $\mathbb R^N$.
Let $\Delta^k = [0,e_1,\ldots,e_k] \subset \mathbb R^k$ be the standard $k$-simplex, where the $e_i$ are the standard basis vectors of $\mathbb R^k$. It is the set $\{ (x_1,\ldots,x_k) \in \mathbb R^k \mid x_i \ge 0, \sum_{i=1}^k x_i \le 1 \}$. The $x_i$ are the barycentric coordinates of $x = (x_1,\ldots,x_k)$ associated to $e_i$, $x_0 = 1 - \sum_{i=1}^k x_i$ is the barycentric coordinate associated to $0$.
First note that each $k$-simplex is homeomorphic to $\Delta^k$. In fact
$$h : \Delta_k \to [v_0,\ldots, v_k], h(x_1,\ldots,x_n) = (1 - \sum_{i=1}^k x_i)v_0  + \sum x_i v_i$$
is a homeomorphism. Observe that $[v_0,\ldots, v_k]$ is contained in any ambient $\mathbb R^N$ wit $N \ge k$.
This means that it suffices to show that $\Delta^k$ is homeomorphic to the cube $I^k \subset \mathbb R^k$.
Le $Q = \{ (x_1,\ldots,x_k) \mid x_i \ge 0 \text{ for all } i \}$. Consider the norms $\lVert (x_1,\ldots, x_k) \rVert_1 = \sum_{i=1}^k \lvert x_i \rvert$ and $\lVert (x_1,\ldots, x_k) \rVert_\infty = \max_{i=1}^k \lvert x_i \rvert$. Both are continuous real-valued functions on $\mathbb R^k$. Let $B_1$ and $B_\infty$ denote the closed unit balls with respect to these norms, i.e. $B_1 = \{ x \in \mathbb R^k \mid \lVert (x_1,\ldots, x_k) \rVert_1 \le 1\}$, similarly $B_\infty$. Then $\Delta^k = B_1 \cap Q$ and $I^k = B_\infty \cap Q$. Define
$$\phi : \Delta^k \to I^k, \phi(x) = \begin{cases} \dfrac{\lVert x \rVert_1}{\lVert x \rVert_\infty}   x \ne 0 \\ 0 & x = 0 \end{cases}$$
$$\psi : I^k \to \Delta^k, \psi(x) = \begin{cases} \dfrac{\lVert x \rVert_\infty}{\lVert x \rVert_1}   x \ne 0 \\ 0 & x = 0 \end{cases}$$
It is readily checked that $\psi \circ \phi = id$ and $\phi \circ \psi = id$, thus $\phi$ and $\psi$ are bijections which are inverse to each other. Both maps are obviously continuous in all $x \ne 0$. But they are also continuous in $0$ since
$$\lVert \phi(x) - \phi(0) \rVert_\infty = \lVert \phi(x) \rVert_\infty = \lVert x \rVert_1 = \lVert x -  0 \rVert_1 ,$$
$$\lVert \psi(x) - \psi(0) \rVert_1 =  \lVert x -  0 \rVert_\infty .$$
This means that $\phi,\psi$ are homeomorphisms.
Edited:
Usually $\mathbb R^n$ is endowed with the Euclidean norm $\lVert (x_1,\ldots, x_k) \rVert_2 = \sqrt{\sum_{i=1}^k x_i^2 }$ which generates the standard Euclidean topology. It is well-known that all norms on $\mathbb R^n$ are equivalent, i.e. the topology generated by any norm is the Euclidean topology. Therefore, to show that a function $D  = \mathbb R^n \to R  = \mathbb R^n$ is continuous, we may take any norm $\lVert - \rVert_D$ on the domain $D$ and any norm $\lVert - \rVert_R$ on the range $R$, similarly for maps $\mathbb R^n \to \mathbb R$. However, in the context of this answer we do not need the general norm equivalence theorem. Just note that
$$\lVert (x_1,\ldots, x_k) \rVert_\infty \le \lVert (x_1,\ldots, x_k) \rVert_2 = \sqrt{\sum_{i=1}^k x_i^2 } \le \sqrt n \max_{i=1}^k \lvert x_i \rvert = \sqrt n \lVert (x_1,\ldots, x_k) \rVert_\infty$$
$$\lVert (x_1,\ldots, x_k) \rVert_\infty \le \lVert (x_1,\ldots, x_k) \rVert_1 \le n \lVert (x_1,\ldots, x_k) \rVert_\infty$$
