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

$\begingroup$ 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)? $\endgroup$ – Konrad Voelkel Jan 6 '10 at 23:03

$\begingroup$ @Konrad: I think it's the first question. The title should probably be parsed "Is homology enough to detect chain homotopy equivalence?". $\endgroup$ – Qiaochu Yuan Jan 6 '10 at 23:15
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 $C_n \to B_{n1}$, so that $C_n \cong Z_n \oplus B_{n1}$. (You can do this because $B_{n1}$, as a subgroup of a free group, is free.) For all $n$, you then have a subchaincomplex $\cdots \to 0 \to B_n \to Z_n \to 0 \to \cdots$ 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.

$\begingroup$ Yes, subgroup of a free group is free. But how does it give the splitting? For instance how will you split the multiplication by 2 map from Z to Z ? $\endgroup$ – Anweshi Jan 6 '10 at 23:34

1$\begingroup$ Assuming you're describing a twoterm complex, the first copy of Z splits as 0 + 2Z, and the second copy of Z splits as Z + 0. The only information you need is that the kernel and image are free abelian (see Leonid's reason number 2). $\endgroup$ – S. Carnahan♦ Jan 7 '10 at 0:02

2$\begingroup$ @Anweshi The fact is this: If $0\toA\toB\toC\to0$ is a short exact sequence of $R$modules for some ring $R$, and $C$ is projective, then the sequence splits. The quotient group $C$ is free, hence projective. $\endgroup$ – Sammy Black Jan 7 '10 at 0:05

$\begingroup$ There's a thing called "acyclic models" mentioned in Spanier's book. Is it relevant to such questions.? $\endgroup$ – Anweshi Jan 7 '10 at 17:40

1$\begingroup$ Acyclic models are generally relevant to understanding something like the singular chain functor itself. I don't think they're of direct application here since this is a question purely in homological algebra. $\endgroup$ – Tyler Lawson Jan 7 '10 at 17:51
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 quasiisomorphic to its homology, since the category of abelian groups has homological dimension 1.

$\begingroup$ This answer seems to assume
$H_*(C_*)$
is torsionfree. If$C_*$
is a chain complex with torsion in its homology, then it is indeed quasiisomorphic 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 quasiisomorphic to$H_*$
. Mariano's very similar answer clarifies this point. $\endgroup$ – Dan Ramras Apr 30 '10 at 6:18 
$\begingroup$ This answer assumes nothing of the kind. You are right that a complex of free abelian groups is only quasiisomorphic 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. $\endgroup$ – Leonid Positselski Apr 30 '10 at 14:24

$\begingroup$ 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. $\endgroup$ – Dan Ramras Apr 30 '10 at 16:07
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.