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For a simply connected simplicial complex, a theorem of Whitehead (Derived categories for the working mathematician, bottom of page 2) explains that the associated chain complexes with coefficients in $\mathbb{Z}$ $$K \textrm{ : } \rightarrow C_n(X) \rightarrow C_{n-1}(X) \cdots $$ contains more information than the singular homology/cohomology groups (two such simplicial complexes are homotopic iff there is a certain relation between their associated chain complexes involving the chain complex of another simplicial complex).

Question: Let $X$ be a compact simplicial complex. Can one recover the torsion in $H^i(X, \mathbb{Z})$ from knowing the complex of simplicial chains (with $\mathbb{Q}$-coefficents)? Is there a procedure to do this?

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I made a mistake (see comments below), so I modified the question. – LMN Jan 2 '13 at 19:53
Dear LMN, As Fernando Muro says, a cochain complex over $\mathbb Z$ is isomorphic to its cohomology (regarded as a complex with trivial differentials). This is even more evidently true when $\mathbb Z$ is replaced by a field such as $\mathbb Q$. Thus your assertion that "the associated chain complex contains more information ..." is false. Regards, – Emerton Jan 3 '13 at 3:22
LMN, do you mean, for instance, that the cup product on cochains contains more information than its cohomology groups? Otherwise it's not true that the cochain complex has more information than the cohomology groups (as others have said). – Hiro Lee Tanaka Jan 3 '13 at 6:02

1) is false, Whitehead's theorem doesn't say that. Actually, any complex over a hereditary ring, eg the integers, is quasi-isomorphic to its cohomology.

2) No, lens spaces have quasi-isomorphic singular (co)chains but different integral cohomology.

3) Yes, by the answer to 1)

Maybe you're interested in doing all this functorially. Since this is a very important point, if this is what you want you should specify all this explicitly, eg what would be the source category, the target, etc.

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I guess that (3) would be yes if the sentence preceding it in LMN's question were accurate, but it isn't. It is possible to algebraically recover singular cohomology with coefficients in $\hat{\mathbb{Z}}$ or $\mathbb{C}$, but not $\mathbb{Z}$. – Donu Arapura Jan 2 '13 at 19:24
@Donu you are right! – Fernando Muro Jan 2 '13 at 19:28
Fernando, Thanks for your answer. I had misquoted Whitehead's theorem, so I changed my question appropriately. – LMN Jan 2 '13 at 19:45
@LNM it's still unaccurare. As I've said in my answer, the chain complex does not contain more information than the homology, unless you consider it as a functor. – Fernando Muro Jan 2 '13 at 20:13
@LMN my answer is complete, however if you don't have enough background you'll have to make año effort to fully understand it. Concerning Whiteheads's thm, you haven't got it yet: the complex and the homology contain exactly the same information. A homotopy type is much more complicated than the singular or cellular complex, even in the simply conected case (a hypothesis that you have overlooked). – Fernando Muro Jan 2 '13 at 21:07

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