By Kaczynski , Mischaikow , Mrozek
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Additional info for Algebraic Topology A Computational Approach
We do this for two reasons. First, each interval can be represented by a graph and so using the types of arguments employed in the previous section we can compute the homology. Second, we can actually draw pictures of the functions. This latter point is to help us develop our intuition, in practice we will want to apply these ideas to problems where it is not feasible to visualize the maps, either because the map is too complicated or because the dimension is too high. With this in mind let X = ;p2 2] R, Y = ;2 4] R and let f : X !
MOTIVATING EXAMPLES 46 a linear operator, we can match the topological expression bd ( a b] b c]) = fag fcg with the algebraic expression @ ( a b] + b c]) = @ ( a b]) + @ ( b c]) = a+b+b+c = a + 2b + c = a + c: Continuing in this way we have that @ ( a b] + b c] + c d] + d e]) = a + b + b + c + c + d + d + e = a + e: As an indication that we are not too far o track observe that on the topological level bd 0 1] = fag feg. 2. 2: Topology and algebra of boundaries in ;1. e. algebraic objects whose boundaries add up to zero, are topologically nontrivial.
E. elements of the kernel of the boundary operator. So de ne Z0(G Z2) := ker @0 = fv 2 C0(G Z2) j @0 v = 0g Z1(G Z2) := ker @1 = fv 2 C1(G Z2) j @1 v = 0g Since @0 = 0 it is obvious that Z0(G Z2) = C0(G Z2). We also observed that cycles which are boundaries are not interesting. To formally state this, de ne the set of boundaries to be B0(G Z2 ) := im @1 = fv 2 C0(G Z2 ) j 9 e 2 C1(G Z2 ) such that @1 e = vg B1 (G Z2) := im @1 = f0 2 C0(G Z2 )g Observe that B0 (G Z2) C0(G Z2 ) = Z0(G Z2 ). Since we have not yet de ned @2 we shall for the moment declare B1(G Z2 ) = 0.
Algebraic Topology A Computational Approach by Kaczynski , Mischaikow , Mrozek