Homology, Chain Equivalence

Chain Equivalence

A chain map f from C into D is a chain equivalence if there exists an inverse chain map g such that the composition map in either order is chain homotopic to the identity map from C onto C or from D onto D. In this case we say C and D are homotopy equivalent, or they have the same homotopy type.

If f is the identity map on C, set g = f to demonstrate chain equivalence. Thus C is homotopy equivalent to itself.

By definition, C is homotopy equivalent to D iff D is homotopy equivalent to C.

The composition of two chain equivalences gives another chain equivalence. Verify this in steps. If f and g are the successive chain equivalences, let f′ and g′ be their inverses. With gg′ homotopic to I, fgg′ is homotopic to fI. Put f′ on the right and fgg′f′ is homotopic to ff′, which is homotopic to I. The same holds in the reverse direction, hence fg is a chain equivalence.

Being homotopy equivalent is, well, an equivalence relation, as its name suggests. Thus each chain complex represents a specific homotopy type.

Isomorphic Homologies

If C and D are homotopy equivalent via f and g, then fg produces a map that is homotopic to the identity. By the previous theorem, fg induces the identity map on the homology of C. This is the map on the homology of C, induced by f, followed by the map on homology induced by g. In other words, the composition of two homomorphisms yields the identity map. The same is true in the reverse direction, hence the homomorphisms are isomorphisms, and the homologies of C and D are isomorphic.

Tensoring with a Common Module

Tensor the chain complex C with a common module S. Each Ci becomes Ci×S, and each boundary homomorphism is tensored with the identity on S. apply the boundary twice and the C component is 0, making the tensor product 0. Thus C×S is a chain complex.

It's homology groups may not be the same as those of C. In fact, each Ci×S could drop to 0, giving a 0 homology.

Assume C and E are homotopy equivalent, as demonstrated by the chain maps f and g. Tensor C, E, f, and g with S. Apply f×S and g×S to get an endomorphism on C×S. Let d be the homotopy that equates fg with the identity map on C. Tensor d with S to build the new homotopy that equates (f×S)(g×S) with the identity map on C×S. In other words, C×S and E×S are homotopy equivalent, and have the same homology.