The stable-homotopy-homology-theory - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T17:54:25Z http://mathoverflow.net/feeds/question/85747 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/85747/the-stable-homotopy-homology-theory The stable-homotopy-homology-theory Fabian Hebestreit 2012-01-15T16:23:38Z 2012-01-17T10:02:44Z <p>Hi</p> <p>Is there a way to stabilise relative homotopy groups into giving the stable-homotopy-homology-functor? The fact that the homotopy excision theorem holds for exactly the same kind of pair that occurs in the Eilenberg-Steenrod axioms seems to indicate, that this should be possible, doesn't it?</p> <p>For CW-pairs \$(X,A)\$ simply applying unreduced suspension repeatedly, taking homotopy groups accordingly (basepoints become unnecessary) and going to the limit works fine, I think. (Though I may have overlooked something once more.) </p> <p>For a while I thought this approach might just work for arbitrary pairs, however Tom Goodwillie (thankfully) set me straight by pointing out, that this is rubbish: The suspension of a subspace need not even be a subspace of the suspension ( <a href="http://mathoverflow.net/questions/84971/suspension-of-an-excisive-pair" rel="nofollow">http://mathoverflow.net/questions/84971/suspension-of-an-excisive-pair</a> ), whence one doesn't even end up with a pair of spaces after suspension.</p> <p>Has this approach been studied in the literature? My standard textbooks only construct reduced stable homotopy groups and I've been wondering why, ever since first learning about stable homotopy groups. So i would be very happy with a reference and somewhat content with an answer as to why this can't possibly work.</p>