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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?

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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.

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  • $\begingroup$ For homotopy invariance, is it already enough that the homology functor is defined on the homotopy category? How should I think of its representability? Lastly, where can I find this result? $\endgroup$ – user153312 Feb 5 '15 at 0:53
  • $\begingroup$ @Exterior: yes. The values of any representable functor on a category are automatically invariant under isomorphisms in that category (which in this case is chain homotopy equivalences). This is probably an exercise in various homological algebra textbooks. $\endgroup$ – Qiaochu Yuan Feb 5 '15 at 1:40
  • $\begingroup$ When you say homotopy category of chain complexes, do you formally mean quotienting Hom-sets by chain-homotopic arrows or "abstract-homotopic" arrows? These are the same whatever the case though, right? $\endgroup$ – user153312 Feb 5 '15 at 1:48
  • $\begingroup$ @Exterior: if they're not the same, then "abstract-homotopic" is the wrong notion. $\endgroup$ – Qiaochu Yuan Feb 5 '15 at 1:52
  • $\begingroup$ By "abstract-homotopic" here, I mean there exists an arrow $X\times I\rightarrow Y$ where $X,Y,I$ are chain complexes, $I$ being the inverval complex as here, such that its compositions on the inclusions $e_0,e_1$ are respectively $f,g$. The cylinder $X\times I$ can be taken either grading-wise as $X_n\oplus X_n \oplus X_{n-1}$ or as the tensor product $X\otimes I$. This seems like the right notion, I just want to make sure the homotopy categories are the same. $\endgroup$ – user153312 Feb 5 '15 at 1:59
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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.)

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  • $\begingroup$ I think perhaps my question wasn't clear enough (or I'm missing something, which is also likely). I'm asking about the abstract machines without any geometry. Just a homotopy equivalence in the abstract cylinder-object sense, and whether chain complexes homotopy-equivalent in this sense have the same homology objects. So no topological spaces :) $\endgroup$ – user153312 Feb 5 '15 at 0:45
  • $\begingroup$ A homotopy equivalence in the cylinder sense is exactly a chain homotopy equivalence. $\endgroup$ – Zhen Lin Feb 5 '15 at 7:57

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