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 $\mathbb{S}^{1}$.