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This is in the same vein as my previous question on the representability of the cohomology ring. Why are the homology groups not corepresentable in the homotopy category of spaces?

flag	asked Oct 20 2009 at 15:09 Dinakar Muthiah 3,387●9●31

Tyler Lawson · Accepted Answer · 2009-10-21 12:30:05Z UTC

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Corepresentable functors preserve products; homology does not.

One replacement is the following. Let X be a CW-complex with basepoint. Then the spaces {K(Z,n)} represent reduced integral homology in the sense that for sufficiently large n, the reduced homology H_k(X) coincides with the homotopy groups of the smash product:

pi_n+k(X ^ K(Z,n)) = [S^n+k, X ^ K(Z,n)]

This is some kind of "stabilization", and it factors through taking the n-fold suspension of X. Taking suspensions makes wedges more and more closely related to products. This doesn't make homology representable, but provides some alternative description that's more workable than simply an abstract functor.

edited Oct 21 2009 at 12:30

answered Oct 20 2009 at 16:03

Tyler Lawson
18.7k●2●44●90

	That's still not what "representable" means for a functor. Representable means that there is an object D such that Hom(D,X) is the functor. Homology is not representable, even in the stable case. That's why no-one ever talks about homology operations. Also, I'm not sure about your formula for the stabilisation. The equivalence between a spectrum and its Omega-spectrum holds in the category of spectra, not in the category of spaces. So you ought to have Sigma^n X wedge K(Z,k) and while this maps to X wedge K(Z,k+n) it isn't an equivalence. – Andrew Stacey Oct 20 2009 at 18:40
	Yes. It's not representable; I was just trying to give the closest thing I know. Also, the formula only works for n roughly larger than twice k, not all k simultaneously. – Tyler Lawson Oct 20 2009 at 20:53
	I see. But your last sentence is slightly misleading as it seems to imply that stably, homology is representable (say, for spectra). This isn't the case so, in the interest of clarity, I'd prefer it if it were edited. Technically, a generalised homology theory is defined for spaces by E_k(X) = {S,Sigma^infty X wedge E}^{-k}. Due to Freudental's suspension theorem, we can compute that by taking a sufficiently high limit, as you say (though I'm still not sure about your forumla - I'll need to write it out in full to be sure and these comment boxes are just that little bit too shor – Andrew Stacey Oct 21 2009 at 7:46
	I've edited it slightly and I hope it's more palatable. For the formula you use Freudenthal suspension twice to pass from X ^ K(Z,n) to X ^ S^1 ^ K(Z,n) to X ^ K(Z,n+1). – Tyler Lawson Oct 21 2009 at 12:41

Eric Peterson · Answer 2 · 2010-09-04 17:05:25Z UTC

While it's true that there are lots of internal things that a corepresentable homology functor wouldn't support, I think it's also enlightening to see that you wouldn't get the nice sorts of dualities that homology and cohomology theories have. After all, we've already agreed that cohomology theories ought to be somehow representable, so maybe we should start there. Instead of using stable maps $X \to E_n$ to produce $n$-degree $E$-cohomology classes of $X$, you can think of these instead as elements in the stable homotopy groups $\pi_{-*}^S F(X, E)$ , where $F(X, E)$ denotes the function spectrum of maps $X \to E$. This presentation makes the right choice for defining $E$-homology somehow much more obvious: the functor $F(X, -)$ has an adjoint, called the smash product (this is the whole point of the smash product -- it plays the role of "tensor product" for spaces!), and so for homology we think about maps $S^n \to E \wedge X$ instead. That homology and cohomology are not (usually) exact duals in a linear algebraic sense is somehow measuring the twist introduced by this adjunction. This does actually turn out to be the right definition for homology; (extraordinary) homology theories in the traditional sense are in fact modeled by functors of the form $\pi_*^S (E \wedge -)$.

This construction has a number of attractive features -- for instance, it means that we can (under some flatness and ringy conditions) think about "homology cooperations" associated to a spectrum, and they look like $E_* E$, a pleasant mirror of cohomology operations living in $E^* E$. We also always get a pairing $E^* X \times E_* X \to E_* E$ of cohomology and homology classes that lands in homology operations, by composing as $S^n \to E \wedge X \to E \wedge E$. (This pairing even gets used occasionally, though I'd be hard-pressed to come up with an obvious citation.)

For the most familiar homology theory, singular homology with $\mathbb{Z}/p$-coefficients, this flatness business does hold, the operations and cooperations even turn out to be $\mathbb{Z}/p$-vector space duals, and the coaction and action line up in the way you'd expect from Milnor's work.

(This belongs as a comment on Lawson's answer, I think, but it looks like I'm too new here to make that happen.)

Urs Schreiber · Answer 3 · 2009-10-22 09:19:04Z UTC

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The dual to cohomology in the functorial sense turns out to be not homology, but homotopy (as described here).

For instance ordinary homotopy groups are the homotopy theory co-represented by spheres.

answered Oct 22 2009 at 9:19

Urs Schreiber
6,458●13●21

The duality to which you refer is called Eckmann-Hilton duality, and is also intimately tied in with Koszul duality of operads. – Dev Sinha Mar 30 2010 at 21:44

Why is homology not (co)representable?

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