Why is homology not (co)representable? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T09:54:01Z http://mathoverflow.net/feeds/question/1438 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/1438/why-is-homology-not-corepresentable Why is homology not (co)representable? Dinakar Muthiah 2009-10-20T15:09:52Z 2010-09-04T17:05:25Z <p>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? </p> http://mathoverflow.net/questions/1438/why-is-homology-not-corepresentable/1444#1444 Answer by Tyler Lawson for Why is homology not (co)representable? Tyler Lawson 2009-10-20T16:03:50Z 2009-10-21T12:30:05Z <p>Corepresentable functors preserve products; homology does not.</p> <p>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<sub>k</sub>(X) coincides with the homotopy groups of the smash product:</p> <pre>pi<sub>n+k</sub>(X ^ K(Z,n)) = [S<sup>n+k</sup>, X ^ K(Z,n)]</pre> <p>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.</p> http://mathoverflow.net/questions/1438/why-is-homology-not-corepresentable/1853#1853 Answer by Urs Schreiber for Why is homology not (co)representable? Urs Schreiber 2009-10-22T09:19:04Z 2009-10-22T09:19:04Z <p>The dual to cohomology in the functorial sense turns out to be not homology, but <a href="http://ncatlab.org/nlab/show/homotopy+(as+an+operation)" rel="nofollow">homotopy (as described here)</a>. </p> <p>For instance ordinary homotopy groups are the homotopy theory co-represented by spheres.</p> http://mathoverflow.net/questions/1438/why-is-homology-not-corepresentable/2299#2299 Answer by Eric Peterson for Why is homology not (co)representable? Eric Peterson 2009-10-24T15:40:38Z 2010-09-04T17:05:25Z <p>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 <code>$\pi_{-*}^S F(X, E)$</code>, 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 $\pi_*^S (E \wedge -)$.</p> <p>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.)</p> <p>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.</p> <p>(This belongs as a comment on Lawson's answer, I think, but it looks like I'm too new here to make that happen.)</p>