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Is the following true: If two chain complexes of free abelian groups have isomorphic homology modules then they are chain homotopy equivalent.

flag	asked Jan 6 2010 at 22:05 Stephen Bigelow 235●6

	Is the question you are asking, if the following statement is true: "H(X) isomorphic to H(Y) implies X chain homotopic to Y for X,Y complexes of free abelian groups" (like your text suggests) or is it about the following statement: "X chain homotopic to Y implies H(X) isomorphic to H(Y) for X,Y complexes of free abelian groups" (like your title suggests)? – Konrad Voelkel Jan 6 2010 at 23:03
	@Konrad: I think it's the first question. The title should probably be parsed "Is homology enough to detect chain homotopy equivalence?". – Qiaochu Yuan Jan 6 2010 at 23:15

Tyler Lawson · Accepted Answer · 2010-01-06 23:22:57Z UTC

Yes, this is true. Suppose $C_*$ is such a chain complex of free abelian groups.

For each $n$, choose a splitting of the boundary map $C_n \to B_{n-1}$ , so that $C_n \cong Z_n \oplus B_{n-1}$ . (You can do this because $B_{n-1}$ , as a subgroup of a free group, is free.) For all $n$, you then have a sub-chain-complex $\cdots \to 0 \to B_n \to Z_n \to 0 \to \cdots$ concentrated in degrees $n$ and $n+1$, and $C_*$ is the direct sum of these chain complexes.

Given two such chain complexes $C_*$ and $D_*$ , you get a direct sum decomposition of each, and so it suffices to show that any two complexes $\cdots \to 0 \to R_i \to F_i \to 0 \to \cdots$ , concentrated in degrees $n$ and $n+1$, which are resolutions of the same module $M$ are chain homotopy equivalent; but this is some variant of the fundamental theorem of homological algebra.

This is special to abelian groups and is false for modules over a general ring.

Leonid Positselski · Answer 2 · 2010-01-06 23:50:39Z UTC

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Yes, this is true, and it does not matter whether the complexes are bounded from any side (nor of course does it matter whether the homology is finitely generated). This is so because:

The homotopy category of free abelian groups is equivalent to the derived category of abelian groups. This holds even for unbounded complexes, since the category of abelian groups has a finite homological dimension.
Any complex of abelian groups is quasi-isomorphic to its homology, since the category of abelian groups has homological dimension 1.

answered Jan 6 2010 at 23:50

Leonid Positselski
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	This answer seems to assume $H_(C_)$ is torsion-free. If $C_$ is a chain complex with torsion in its homology, then it is indeed quasi-isomorphic to its homology (considered as a chain complex $H_$ with all boundary maps zero), but it is certainly not chain homotopy equivalent to $H_$ (for example, if it were, then $H^(C)$ would be isomorphic to the cohomology of the complex $H_$ , which is wrong). So in general, I don't think it's really helpful to observe that $C_$ is quasi-isomorphic to $H_*$ . Mariano's very similar answer clarifies this point. – Dan Ramras Apr 30 2010 at 6:18
	This answer assumes nothing of the kind. You are right that a complex of free abelian groups is only quasi-isomorphic and not homotopy equivalent to its homology in general. The positive answer to the original question follows from my assertions 1. and 2. as follows: if two complexes have isomorphic homology groups, then by 2. they are isomorphic in the derived category, and if they are also complexes of free abelian groups, then by 1. they must be isomorphic in the homotopy category. – Leonid Positselski Apr 30 2010 at 14:24
	Ah, thank you. I misread the statement 1., I think, and was imagining you had said only that the homotopy category of free abelian groups is equivalent to the derived category of free abelian groups. But you're right, 1. is correct as stated and the result does follow. – Dan Ramras Apr 30 2010 at 16:07

Mariano Suárez-Alvarez · Answer 3 · 2010-01-06 23:31:07Z UTC

The natural functor $K^b(\mathbb Z\mathrm{-free})\to D^b(\mathbb Z)$ from the homotopy category of bounded complexes of finitely generated free abelian groups to the derived category of bounded complexes of finitely generated abelian groups is an equivalence. This means that a map of bounded complexes of finitely generated free abelian groups which induces an isomorphism in homology is an homotopy equivalence.

This and the fact that one can always lift a morphism $f:H_\bullet(X)\to H_\bullet(Y)$ between the homologies of two complexes of free abelian groups to a morphism $\tilde f:X\to Y$ of complexes which induces $f$ give an affirmative answer to your question.

Does homology detect chain homotopy equivalence?

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