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This question is similar to Find the smallest triangulation of the n-dimensional but easier :

How to show the n-dimensional cube can be triangulated into exactly n! simplices?

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Let me point out that the triangulation described by achille and by user8262 are the same (well, in achille's there is the choice of what vertex you pick, both in the $n$-cube and in the recursion process over faces; to get the triangulation of user8262 you just choose the "smallest" vertex in each face, where smallest means "least number of coordinates equal to 1") –  Francisco Santos Sep 27 '13 at 18:24

3 Answers 3

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If the cube has coordinates $x_i$ ($1\leq i\leq n$, $0\leq x_i\leq 1)$ then for every permutation $\sigma\in S_n$ you have the simplex given by $0\leq x_{\sigma(1)}\leq x_{\sigma(2)}\leq\dots\leq x_{\sigma(n)}\leq 1$.

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Very nice. For each point we can determine the simplex it is in (or if some coordinates are equal: the simplices) by looking at the permutation needed to sort the coordinates into ascending order. –  Hagen von Eitzen Aug 24 '13 at 9:16
    
Good, thanks. This is the best answer so now. Any better solution would be appreciate. –  Zia Aug 24 '13 at 10:47

You can triangulate $Δ^n × I$ into $(n + 1)$ $(n + 1)$-simplices ($Δ^n$ is $n$-simplex, $I$ is unit interval $[0, 1]$). See [Hatcher, Proof of 2.10] where it is used to prove homotopy invariance of singular homology. The idea is following: Let $[v_0, …, v_n]$ be the simplex $Δ^n × \{0\}$ and $[w_0, …, w_n] = Δ^n × \{1\}$. Then the triangulation is $\{[v_0, …, v_i, w_i, …, w_n]: i ≤ n\}$.

Now you can triangulate the cube inductively: $I^n = I × \bigcup_{i < (n - 1)!} Δ_i^{n - 1} = \bigcup_{i < (n - 1)!} \bigcup_{j < n} Δ_{ij}^n$.

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Nice of you to express "Proof of Hatcher", for those guys don't have access to ref. –  Zia Aug 24 '13 at 8:42
    
@Zia: That was aditional information for someone who has the access and it also expreses that the core argument is not my idea. But I have rewritten the decomposition so I don't know what is the problem. If you need more explanation, just ask in comment what is not clear. You can draw a picture of the decomposition for low dimensions to get the idea. –  user87690 Aug 24 '13 at 8:53
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@Zia Here is Hatcher’s “Algebraic Topology”. Proof of 2.10 starts from page 119 of the file = page 111 of the book. –  Alex Ravsky Aug 24 '13 at 11:49

Pick one vertex from your $n$-cube. This vertex has $n$ hyper-faces opposite to it. Each hyper-face itself is a $(n-1)$-cube and be triangulated into $(n-1)!$ copies of $(n-1)$-simplex. Forming convex hulls with the original vertex gives you $n! = n \times (n-1)!$ copies of $n$-simplex.

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