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By Kaczynski T., Mischaikow K., Mrozek M.

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Then there exists a tree T with 0 free vertices. Let n be the number of edges in T . Let e1 be an edge in T . Label its vertices by v1; and v1+. 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.

As we will see later for more complicated spaces this need not be the case. We now need to formally de ne the boundary operators that were alluded to earlier. Let @0 : C0(G Z2 ) ! C;1(G Z2) @1 : C1(G Z2 ) ! C0(G Z2) @2 : C2(G Z2 ) ! C1(G Z2) be linear maps. Since we have chosen bases for these vector spaces, we can think of @0 , @1 and @2 as matrices. Since C;1(G Z2) = 0, it is clear that @0 must be the matrix with all zeros. Similarly, @2 is the zero matrix. e. the edges ei. In line with the previous discussion we make the following de nition.

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. 1. TOPOLOGY 41 X = Y = 0 1] X = ;1 and Y = 0 1] X is any topological space and y 2 Y is a deformation retract of Y .

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Algebraic topology: A computational approach by Kaczynski T., Mischaikow K., Mrozek M.


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