Hatcher Exercise 3.1.5 Let \varphi be a 1-cochain in X with coefficients in G – i.e. \varphi \in C^1(X,G) = \Hom(C_1(X), G). \varphi is a function from 1-chains in X to G. Since, in general, singular 1-chains are precisely paths in X, we will consider \varphi as a function from paths in X to G. For paths f and g, we will use f \cdot g to refer to path concatenation (not to be confused with the group operation in C_1(X)). For all of the following, assume \varphi is a co-cycle – i.e. \varphi \in \ker \delta_1, where \delta_1 : C^1(X,G) \rightarrow C^2(X,G) is the coboundary map. Theorem: \varphi(f \cdot g) = \varphi(f) + \varphi(g). Proof: Consider a singular two chain F – that is, a map from the 2-simplex [abc] to X – such that F \vert_{[a b]} = f and F \vert_{[b c]} = g. We then have that F \vert_{[ac]} is homotopic to f \cdot g. To see this, just consider that there is a deformation retraction of the simplex [abc] to the edge [ac] (by simply crushing it) – compose this deformation retraction with F to get a homotopy from the image of F \vert_{[ab] \cdot [bc]} which is f \cdot g to the image of F \vert_{[ac]}. Now, since \varphi is a cocycle, \delta_1 \varphi is the zero map and (\delta_1 \varphi)(F) = 0. Then, % \mjqed Corollary: \varphi is zero on constant paths. Proof: If f is a constant path, then f \cdot f = f. By the last theorem, \varphi(f \cdot f) = \varphi(f) = \varphi(f) + \varphi(f), which could only be true in the group G if \varphi(f) = 0. \mjqed Theorem: If f is homotopic to g (fixing endpoints), then \varphi(f) = \varphi(g) Proof: Let a = (0, 0), b = (0, 1), c= (1,1) in \mathbb{R}^2 explicitly describe the vertices of a 2-simplex. We will explicitly describe two singular 2-chains F, G : [abc] \rightarrow X. Let H : I \times I \rightarrow X be the homotopy from f to g, so that H(0, \cdot) = f and H(1, \cdot) = g. Let F(x,y) = H(y,x). Observe that F\vert_{[ab]} = f and F\vert_{[bc]} = f(1) = g(1) (the constant path). Then F\vert_{[ac]} = z, where z is the path t \mapsto H(t,t). Let G(x,y) = H(x,y). Then, G\vert_{[ab]} = f(0) = g(0) (the constant path) and G\vert_{[bc]} = g. In particular, G\vert_{[ac]} = z again. Then, we have: (\delta_1 \varphi)(F) = \varphi(f) - \varphi(z) + \varphi(f(1)) = 0 But since \varphi is zero on constant paths, the last term is zero and we have \varphi(f) = \varphi(z). Applying the same argument to G gives \varphi(g) = \varphi(z), and we have \varphi(f) = \varphi(g) by transitivity. \mjqed Theorem: \varphi is a coboundary if and only if \varphi(f) depends only on the endpoints of f Proof: Let f : [ab] \rightarrow X be an arbitrary singular 1-chain. If \varphi is a coboundary, then \varphi = \delta_0 \gamma for a 0-cochain \gamma. But then \varphi(f) = \gamma(f\vert_{[b]}) - \gamma(f\vert_{[a]}), so \varphi only depends on the endpoints. Suppose that \varphi(f) depends only on the endpoints of f for any f. Pick a basepoint x_0 \in X and define a 0-cochain \gamma by \gamma(x) = \varphi(f_x) where f_x is any path from x_0 to x. This is well-defined due to the supposition about \varphi. Then, for any path g from a to b, pick some path f_a from x_0 to a, and notice that f_a \cdot g is some path from x_0 to b. Then, we have: (\delta_0 \gamma)(g) = \gamma(b) - \gamma(a) = \varphi(f_a \cdot g) - \varphi(f_a) = \varphi(f_a) + \varphi(g) - \varphi(f_a) = \varphi(g) using the first theorem. Thus, \delta_0 \gamma = \varphi. \mjqed