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.
Best Answer
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.