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Algebraic topologists like to cook up algebraic invariants on topological spaces in order to answer questions, so they are often concerned with how strong those invariants are. Currently, I am concerned with just how much information is lost when moving from a space to `the' chain complex associated to that space.

Now, I should be a bit more specific here. There are many homology theories in which one takes a space, cooks up a chain complex, and takes its homology. I am mainly interested in just singular homology for this question, but if you can only think of an answer using sheaf cohomology or some other homology theory then that's alright. Now, the actual question is:

Do there exist two spaces that have chain homotopic associated chain complexes, but are not, themselves, homotopic?

I imagine that, unless the answer to the question is "no," it would be a bit difficult to show that the two spaces are not homotopic, since we have taken away the tool we usually use to prove such facts. My first thought was that one might be able to cook up a counterexample by looking at two spaces whose compactifications give different homologies... but I wasn't able to come up with anything immediately. (Another method may be looking at homotopy groups... but they're so hard to compute, I didn't even try this approach). I probably didn't give this as much thought as I should have, so if the answer to this question is somewhat trivial, then go ahead and scold me and I'll go put some more effort into thinking about it.

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    $\begingroup$ Any chain complex is homotopy equivalent to its homology, so your question reduces to asking whether there are non-homotpic space with isomorphic homologies, and this is classical. $\endgroup$
    – domenico fiorenza
    Aug 12, 2010 at 20:07
  • $\begingroup$ What does this mean? That a chain complex is chain homotopic to the homology groups regarded as a complex? Any complex that is not acyclic can't have that property. $\endgroup$
    – Josh
    Aug 12, 2010 at 20:45
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    $\begingroup$ You've heard of cup products? They reflect more structure in the singular chains than just that of a chain complex. The chain complex of a space does indeed forget a lot about the space, but if it is regarded as some sort of more refined algebraic object then it forgets less. $\endgroup$
    – Tom Goodwillie
    Aug 12, 2010 at 20:49
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    $\begingroup$ One has to be careful. Complexes are in general not homotopy equivalent to their homology complexes (if they are the homology would be a direct factor of the complex which is not true if the complex is free and the homology has torsion). I guess everyone was thinking of quasi-isomorphic. $\endgroup$
    – Torsten Ekedahl
    Aug 12, 2010 at 22:15
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    $\begingroup$ Yes, I was thinkng of quasi-isomorphisms. Sorry for having called it homotopy equivalence: working too much with dglas one gets the awful habit of thinking of complexes of vector spaces rather than of arbitrary modules.. Thanks for having fixed that. $\endgroup$
    – domenico fiorenza
    Aug 13, 2010 at 6:47

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A bounded below complex of free $R$-modules is acyclic if and only if it is contractible (in the sense that the identity map is chain homotopic to zero). Since the singular chain complex of a space is constructed out of free $\mathbb{Z}$-modules, any space with no homology would have to contract to a point, which is not the case (check out the wikipedia page for some examples/references).

Thus the chain homotopy exhibiting contractibility of the identity does not come from a topological homotopy.

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  • $\begingroup$ And more generally, a complex is split exact if and only if the identity map is null-homotopic... I'm surprised I didn't think of this! Thanks for the answer! $\endgroup$
    – Dylan Wilson
    Aug 12, 2010 at 21:07

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