Let [ 0 , 1 ] n + 2 be the n + 2 )-dimensional unit cube. Consider the set A ⊂ [ 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 ∈ { 1 , . . . , n + 2 }. My question: is it true that the set A is homeomorphic to the n-dimensional sphere 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 S 1 .

Question

Let   [  0  ,  1      ]          n      +      2       be the   n  +  2  )-dimensional unit cube. Consider the set   A  ⊂  [  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  ∈  {  1  ,  .  .  .  ,  n  +  2  }. My question: is it true that the set   A is homeomorphic to the   n-dimensional sphere             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             S              1      .

Phillip Fletcher · Accepted Answer

The set   A is always homeomorphic to       S    n  . Specifically, let   P   be the plane       x    1    +  ⋯  +      x          n      −      2        =  0 in             R              n      +      2      , and let   π  :            R              n      +      2        →  P   be orthogonal projection. I claim that1.   π is injective on   A, and   π  (  A  ) is homeomorphic to   A.2.   π  (  A  ) is the boundary of   π            (        [  0  ,  1      ]          n      +      2                  )       in   P.For the first claim, observe that no two points of   A differ by a multiple of   (  1  ,  1  ,  …  ,  1  ), and therefore   π is injective on   A. Since   A is compact, it follows that   π  (  A  ) is homeomorphic to   A  .For the second claim, consider the subset of   P   defined by the inequalities      x    i    −      x    j      ≤    1for all   i  ,  j  ∈  {  1  ,  …  ,  n  +  2  }. A simple argument with coordinates shows that this set is precisely   π            (        [  0  ,  1      ]          n      +      2                  )      , and the boundary of this set is precisely   π  (  A  ).Incidentally, the polytope   π            (        [  0  ,  1      ]          n      +      2                  )       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.

Let [0,1]^(n+2) be the n+2)-dimensional unit cube. Consider the set A⊂[0,1]^(n+2) consisting of all points (x_1,...,x_(n+2)) such that xi=0, xj=1 for some i,j∈{1,...,n+2}. My question: is it true that the set A is homeomorphic to the n-dimensional sphere S^n and how to show this (if true)?

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