Homology, A Different Kind of Cohomology

A Different Kind of Cochain

Cher sang a "different kind of love song", though it has a lot in common with other popular songs. Similarly, this is a different kind of cochain, but all the theorems you know and love carry over.

Let C be a chain complex and let M be a reference module. We have already described the cochain of C with respect to M, denoted C′. Specifically, C′_i is the set of functions from C_i into M. There is another form of cochain in which C′_i is the set of functions from M into C_i, denoted hom(M,C_i). I'm going to call this C′_i as well. I know, I should probably come up with something else, but this form of cochain isn't used very often, and when it is, I'll try to be clear.

To map C′_i into C′_i+1, follow the functions from M into C_i with the boundary operator from C_i into C_i+1, giving functions from M into C_i+1, which are elements of C′_i+1. Notice that the boundary operator runs the same direction as the original chain complex. The arrows are not reversed.

Go from C′₆ to C′₇ to C′₈ and the original boundary operator is applied twice, from C₆ to C₇ and from C₇ to C₈. This is 0, hence C′ is a chain complex. It therefore has a homology, and I will call it the cohomology of C in this context.

You can probably guess what's coming up next. Let f be a chain map from C to E. Compose f with the functions from M into C_i to get functions from M into E_i. Thus f induces a homomorphism f′ from C′ into E′. Because bf = fb, bf′ = f′b, and f′ is a chain map. That's all we need to imply a map from the cohomology of C into the cohomology of E.

As usual, the sum, composition, and restriction of chain maps implies the same on the homomorphisms on the cohomology of C.

If f and g are homotopic chain maps, courtesy of d, compose d with the functions from M into C to get functions from M into E. This isn't a chain map, but it is a homomorphism from C′_i into E′_i+1, and it proves f′ and g′ are chain homotopic. Both functions induce the same map on the cohomology of C.

If C and E are homotopy equivalent, apply the above to show C′ and E′ are equivalent, hence C and E have the same cohomology.

If C and E have subchains, such that f maps the subchain of C into the subchain of E, f′ maps the relative cohomology of C into the relative cohomology of E. The sum, composition, and restriction of chain maps implies the same on the homomorphisms on the relative cohomology of C.

Each one of these paragraphs should be expanded into a page of corroborating algebra, but it's just like the theorems we've seen before, so I'll leave the details to you.