Abstract homotopy invariance of homology

Question

When topology is involved, we know (singular) homology is homotopy invariant. However, homology and homotopy can be discussed in much more general contexts.

Living in $\mathsf{Ch}_\bullet(R\mathsf{Mod})$ for instance, it seems reasonable to ask whether the formalism continues to behave as it did when we had topology, i.e that homology would remain homotopy invariant. Here I mean homotopy in the sense of cylinder objects, which are, in our context, given by tensor products with the interval-complex.

Suppose there's a homotopy equivalence between two chain complexes. Do they have the same homology objects?

If it is true, what is the essence of the proof? I assume it does not involve games with simplices as does the proof for singular homology in Hatcher, so what does "make it true"?

If it were true, we'd be able to get the homotopy invariance of singular homology for free without playing with simplices, no?

Answer 1

Homology is a representable functor on the homotopy category of chain complexes. Namely, in the homotopy category of chain complexes of $R$-modules $H_n$ is represented by $R[n]$, which you should think of as a chain complex incarnation of the $n$-sphere.

You don't get the homotopy invariance of singular homology for free this way since you still need to know that a homotopy equivalence of spaces induces a chain homotopy equivalence.

Answer 2

If you have a chain homotopy equivalence, then it induces isomorphisms in homology. There are various ways to see this – but here is the outline of a hands-on proof:

1. Show that nullhomotopic chain maps induce zero maps in homology.
2. Deduce that homotopic chain maps induce the same map in homology.
3. Deduce that chain homotopy equivalences induce isomorphisms in homology.

However, to go from homotopy of continuous maps to chain homotopy requires some geometric input – after all, $C_{\bullet} (X \times Y)$ is not literally the same as $C_{\bullet} (X) \otimes C_{\bullet} (Y)$. (That said, they are chain homotopy equivalent: this is the Eilenberg–Zilber theorem.)