Hatcher Exercise 3.1.5
Let be a 1-cochain in with coefficients in – i.e. . is a function from 1-chains in to . Since, in general, singular 1-chains are precisely paths in , we will consider as a function from paths in to . For paths and , we will use to refer to path concatenation (not to be confused with the group operation in ).
For all of the following, assume is a co-cycle – i.e. , where is the coboundary map.
Proof: Consider a singular two chain – that is, a map from the 2-simplex to – such that and . We then have that is homotopic to . To see this, just consider that there is a deformation retraction of the simplex to the edge (by simply crushing it) – compose this deformation retraction with to get a homotopy from the image of which is to the image of . Now, since is a cocycle, is the zero map and . Then,
Corollary: is zero on constant paths.
Proof: If is a constant path, then . By the last theorem, , which could only be true in the group if .
Theorem: If is homotopic to (fixing endpoints), then
Proof: Let in explicitly describe the vertices of a 2-simplex. We will explicitly describe two singular 2-chains . Let be the homotopy from to , so that and .
Let . Observe that and (the constant path). Then , where is the path . Let . Then, (the constant path) and . In particular, again. Then, we have:
But since is zero on constant paths, the last term is zero and we have . Applying the same argument to gives , and we have by transitivity.
Theorem: is a coboundary if and only if depends only on the endpoints of
Proof: Let be an arbitrary singular 1-chain. If is a coboundary, then for a 0-cochain . But then , so only depends on the endpoints.
Suppose that depends only on the endpoints of for any . Pick a basepoint and define a 0-cochain by where is any path from to . This is well-defined due to the supposition about . Then, for any path from to , pick some path from to , and notice that is some path from to . Then, we have:
using the first theorem. Thus, .