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By Kaczynski , Mischaikow , Mrozek

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Numerical Analysis and Its Applications: Second InternationalConference, NAA 2000 Rousse, Bulgaria, June 11–15, 2000 Revised Papers

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7 Let ;n := fx 2 Rn+1 j jjxjj = 1g: There is no deformation retraction of ;n to a point. We include this example at this point to try to indicate that this is a nontrivial problem. In particular, we encourage you to try to nd a proof of this fact. As motivation for the study of this subject we assure you that once you know homology theory, this example will become a triviality. 1 Prove that homotopy is an equivalence relation. 2 Let f g : X ! Y be continuous maps. Under the following assumptions on X and Y prove that f g.

Since T has no free vertices, there is an edge e2 with vertices v2 such that v1+ = v2;. Continuing in this manner we can label the edges by ei and the vertices by vi where vi; = vi+;1. Note since there are only a nite number of vertices, at some point vi+ = vj; for some i > j 1. Then fej ej+1 : : : eig forms a loop. This is a contradition. 11 Every edge is homotopic to a point. Proof: Let e be an edge with vertices v ; and v+. Since e is a line segment it is homeomorphic to 0 1]. Let h : 0 1] !

But what about f#1 ( 0 1]) where f#1(f0g) = f#1(f1g) = f0g? e. that f#1 ( 0 1]) = 0. 2. APPROXIMATION OF MAPS 65 rules to each of the intervals we obtain the following matrix 20 0 0 03 66 0 0 0 0 77 66 0 1 0 1 77 f#1 = 66 1 0 0 1 77 66 7 4 1 0 0 0 75 1 0 0 0 In guring out how to de ne f#1 we used the phrase \it seems reasonable to de ne" but this does not mean we should not de ne it a di erent way. Given our choice for f#0 are there any restrictions on the way we de ne f#1? The answer is an emphatic yes.

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