By Kaczynski , Mischaikow , Mrozek

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**Example text**

R. Observe that F (a s) = (1 ; s)f (a) + sg(a) > 0 and F (b s) = (1 ; s)f (b) + sg(b) < 0 Thus, we can apply the intermediate value theorem to F ( s) for any s 2 0 1]. This family of functions is a special case of what is known as a homotopy. 2 Let X and Y be topological spaces. Let f g : X ! Y be continuous functions. f is homotopic to g if there exists a continuous map F : X 0 1] ! Y such that F (x 0) = f (x) and F (x 1) = g(x) for each x 2 X . The map F is called a homotopy between f and g. f homotopic to g is denoted by f g.

To answer this lets repeat what we have said. c is a boundary so we can write c = @b for some chain b. Thus f#(c) = f#(@b). But we want f#(c) to be the boundary of some chain. What chain? The only one we have at our disposal is b, so the easiest constraint is to ask that f#(c) = @f# (b).

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.