about a subset of the multidimensional cube Let $[0,1]^{n+2}$ be the ($n+2)$-dimensional unit cube. Consider the set
$A\subset[0,1]^{n+2}$ consisting of all points $(x_{1},...,x_{n+2})$
such that  $x_{i}=0$, $x_{j}=1$ for some $i,j\in\{1,...,n+2\}$. My question:
is it true that the set $A$ is homeomorphic to the $n$-dimensional sphere
$\mathbb{S}^{n}$ and how to show this (if true)? If the answer is negative,
then what is the homotopy type of $A$?
Note that $A$ is the union of $(n+1)(n+2)$ $n$-dimensional faces of
$[0,1]^{n+2}$ and is symmetric with respect to the center of the cube. For
example, in case $n=1$, $A$ is the union of 6 edges of $[0,1]^{3}$ forming a (topological)
circle $\mathbb{S}^{1}$.
 A: The set $A$ is always homeomorphic to $S^n$.  Specifically, let $P\;$ be the plane $x_1 + \cdots + x_{n-2} = 0$ in $\mathbb{R}^{n+2}$, and let $\pi\colon \mathbb{R}^{n+2} \to P\;$ be orthogonal projection.  I claim that


*

*$\pi$ is injective on $A$, and $\pi(A)$ is homeomorphic to $A$.

*$\pi(A)$ is the boundary of $\pi\bigl([0,1]^{n+2}\bigr)$ in $P$.
For the first claim, observe that no two points of $A$ differ by a multiple of $(1,1,\ldots,1)$, and therefore $\pi$ is injective on $A$.  Since $A$ is compact, it follows that $\pi(A)$ is homeomorphic to $A$.
For the second claim, consider the subset of $P\;$ defined by the inequalities
$$
x_i - x_j \;\leq\; 1
$$
for all $i,j\in \{1,\ldots,n+2\}$. A simple argument with coordinates shows that this set is precisely $\pi\bigl([0,1]^{n+2}\bigr)$, and the boundary of this set is precisely $\pi(A)$.
Incidentally, the polytope $\pi\bigl([0,1]^{n+2}\bigr)$ that appears in this problem is a simple example of a zonotope.  It is a regular hexagon when $n=1$ and a rhombic dodecahedron when $n=2$.  In general, it is an $(n+1)$-polytope with $(n+2)(n+1)$ parallelotope faces.
