T. =-?-·-·-·-?-for-its-source, . :=-+-·-·-·-+-for-its, and . Target, We can now state the main theorem of this section: the n-dimensional standard chromatic subdivision

, Before we prove it, let us show a helper lemma about what partial cube chains in the cube look like: Lemma 4.25

, Since a partial cube chain can always be extended to a total one (by filling the holes with the paths that are required to exist by definition), the result follows c = (c 1

, If c is a cube in n+1 , we write proc c ? [n] for the set of indexes where c has symbol '0', and view c ? [n] for the set of indexes where c has symbol '0' or

, c ) be a partial cube chain in n+1 . If k ? k , then view c k ? view c k . f : PCh( n+1 ) t s ? ChSub(? n ) as follows, Let c =, issue.1

, {(i, view c k ) | i ? proc c k }

, First, i ? view c k since i ? proc c k ? view c k . The views in f (c) can be totally ordered by inclusion thanks to Lemma 4.28. Finally, suppose (i, view c k ) and (j, view cm ) are two elements of f (c) such that i ? view cm . Then c m has a '0' or '+' symbol at position i, and by Lemma 4.25, all the subsequent cubes must have a '+' at position i. Since i ? proc c k

, The sets proc c i are disjoint by, c ) be of type

, The picture below illustrates the map f for n = 2: partial cube chains in the 3-dimensional cube correspond to the simplexes of the 2-dimensional chromatic subdivision

, We now define g : ChSub(? n ) ? PCh( n+1 ) t s . Let X = {(i 0 , X i 0 ), . . . , (i m , X im )} be a simplex in ChSub(? n )

. ?-x-im, Let (I k ) 1?k? be the (unique) partition of {i 0 , . . . , i m } such that: -For j, j ? I k , X j = X j . We write this common view X I k . -The inclusion X I k X

, where c k has symbol '0' at all positions j ? I k ; symbol '+' at all positions j ? (X I k \ I k ); and symbol '?' at the remaining positions

, The only effect a path in the cube can have on vertices is to turn '?' symbols into '+' symbols, so we need to check that whenever ? + (c k ) has a '+' symbol at some position, then ? ? (c k+1 ) also does. ? + (c k ) has a '+' symbol exactly at the positions j ? X I k . Since X I k ? X I k+1 , c k+1 has either a '0' or a '+' at those positions

, It is straightforward to check that f ? g(X) = X. We take the notations from Definition 4.31, with g(X) = (c 1

. {(i,-x-i-k-)-|-i-?-i-k-},

, Since the sets (I k ) 1?k? form a partition of {i 0 , . . . , i m }, and X I k = X i for i ? I k , we get f ? g(X) = X. Bibliography

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