By Andreescu T., Feng Z.

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**Extra info for 102 Combinatorial problems from the training of USA IMO team**

**Example text**

Since I satisfies NE(2,1 ,1), there is a natural transformation A : G(2 ,1,l)QI ~ G 2QI so that for all objects X, Y, A(X, Y) is a filler function . If X is an object of C, then 'x given by ,6 = ,f = eo(X)rJ(X), ,6 = I dx x 1 28 is a (2,1,1)-box in QI(X, X x 1). A(X,X x Ihx)el(X x 1) to get the required natural transformation i : ( ) x I -----t ( ) xI. 8)). We thus see how extra structure in the shape of fillers in the cubical sets QI(X, Y) can be used to obtain a more richly structured homotopy theory.

We say I satisfies the Kan condition NE(n,v,k) if there is a natural transformation ). : G(n,v,k)QI ---t GnQI, such that for any X, Y, objects in C, is a filler map. If, in addition , each )'(X, Y) is compatible with degeneracies, I is 26 said to satisfy the Kan condition DNE(n,lI,k). If for some n, I satisfies E(n, II, k) (resp. NE(n, II, k), resp. DNE( n, II, k)) for all II = 0,1, k = 1"", n, we say that I satisfies the Kan condition E(n) (resp. NE(n), resp. DNE(n)) . Examples 1. 24)). 2. For any category C the trivial cylinder Id on C satisfies the Kan condition DNE( n) for all n.

Moreover 7J;( i x I) = ( i x 1). The proof of the lemma will be complete when we have proved g'g ~ldx. To do this we first consider the (2,1,1 )-box "I = ("(6, -, "15, "In given by "If = , "15 = This has a filler p : X X 7J;(g x 1), 12 - "16 = (69 = ga(X). X. We set F = l~(p), so F: X x [ - X and F: g'g ~ 1dx . ) by 03 og p(i x [2), = oJ = ia(X)a(X x 1), or = 0; = (i x 1)(a(A) x I) = 7J;(i x 1)(a(A) x 1). Thus 0 can be illustrated by the following figure. 38 iu(A) cj>(i x 1) cj>( i iu(A) X X cj>( i x 1) 1) cj>( i {12 0 cj>(i 813 X 1) 812 iu(X)u(X X I) 1) cj>(i X iu(A) 1) II 1/;( 9 X 1)(i X 1) fL (i cj>(i X 12) F( i X X 1) 1) One verifies easily that 8 is a (3,1,1)-box in QI(A,X) .

### 102 Combinatorial problems from the training of USA IMO team by Andreescu T., Feng Z.

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