# Functors that are the homology of a chain complex

Is there an a priori reason why the singular homology and cohomology groups of a space should be computable as the homology of chain complexes? Certainly you can express any family of functors (say, indexed by $\Bbb Z$) from spaces into abelian groups as the homology of a funcotrial chain complex: just take the graded abelian group that is the sum of the groups with zero differential. As you'd expect, I don't find that satisfying. I could (possibly) rule such silliness out by adding some freeness hypothesis on the chain complex functors, but I don't necessarily wish to do that.

Similarly, is there any a priori reason one may expect to be able to calculate the homology of a loop space $\Omega X$ given the (singular or cellular) chain complex of the space $X$ (with some connectedness hypothesis)? In a way this is a silly question : as I understand it, one can completely reconstruct the original space $X$ (up to weak homotopy equivalence) from its singular simplicial set $(\mathrm{map}(\Delta^n,X))_{n\in\Bbb N}$, so surely its singular chain complex carries much more information than is expressed in its homology. Still I'm baffled that an algebraic procedure, the cobar construction, can under some mild hypothesis of connectedness, produce a chain complex model for the homology of the loop space, which is a topological construction.

Maybe I should phrase my question thusly : why is it that there are good functorial chain complexes that allow one to calculate various groups associated to a space and perform topological constructions algebraically, and should we expect it?

Later addition. The original motivation behind this question comes from rational homotopy theory which I'm only beginning to learn. The rational homotopy groups of a space suitably shifted in index form a (sign-)graded Lie algebra. They are all expressible as the homology of a differential graded Lie algebra. I hope this example makes my question clearer : I never expected the rational homotopy groups of a space to be homology groups of chain complexes in an interesting way.

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How do you define singular homology/cohomology if not as the homology/cohomology of the singular complex? –  Zhen Lin Nov 5 '13 at 16:30
@ZhenLin The cohomology of a space $X$ with coefficients in $G$ is naturally isomorphic to the the set of homotopy classes of maps $X\to K(G,n)$. Homology can also be defined using only homotopy theory, but I don't know how. It's done in Jeff Strom's book Modern Classical Homotopy Theory''. He defines all of homology and cohomology in terms of homotopy. My question is this : the homology groups are functors from spaces to abelian groups. Is there some a priori way to recognize those functors that arise in some interesting way (I know that's vague) as homology groups of chain complexes? –  Olivier Bégassat Nov 5 '13 at 16:34
Hmmm. You could do it that way. But then it's not obvious that the set of homotopy classes of maps $X \to K (G, n)$ has a group structure, let alone an abelian group structure when $G$ is abelian. And it's not obvious that any $K (G, n)$ spaces exist at all: after all, we know it's impossible to have a $K (G, 2)$ space if $G$ is non-abelian. All in all, it seems like a bad definition! –  Zhen Lin Nov 5 '13 at 17:33
There's more to those chains than meets the eye. There's a famous paper of Mandell "Cochains and Homotopy Type" which shows that under certain hypotheses, the hommotopy type of $X$ is determined by the homotopy type of the cochain complex $S^*(X)$, considered as a certain kind of algebra called an $E_\infty$ algebra. This encodes things like Steenrod operations, and many other things, all of which is more than sufficient for a loop space construction. In fact, for that all that is needed is the dg algebra structure of $S^*(X)$. –  Justin Young Nov 5 '13 at 20:06
@ZhenLin $K(A,n)$ can be defined as $B^nA$ (or using Dold–Puppe). In this construction it's a topological abelian group, so $[X,K(A,n)]$ clearly is an abelian group. Now the question is why this homological functor can be lifted to a triangulated functor to $D(Ab)$, something like this. –  Grigory M Dec 18 '13 at 18:31