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?

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.)

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 ( Suspension of an excisive pair ), whence one doesn't even end up with a pair of spaces after suspension.

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.

  • $\begingroup$ The stable homotopy groups of the mapping cone of the map of spectra $\Sigma^{\infty}A\rightarrow\Sigma^{\infty}A$ induced by the inclusion $A\subset X$ $\endgroup$ – Fernando Muro Jan 15 '12 at 16:29
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    $\begingroup$ I don't think so $\endgroup$ – Fernando Muro Jan 15 '12 at 17:09
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    $\begingroup$ it replaces relative groups by absolute groups of the cone and then stabilises them instead. How is that not circumventing? $\endgroup$ – old account Jan 17 '12 at 10:10
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    $\begingroup$ Because classical relative homotopy groups are ordinary homotopy groups of the homotopy fiber of the inclusion $\endgroup$ – Fernando Muro Jan 17 '12 at 11:15
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    $\begingroup$ Whatever approach you like, what I say is not circumventing, it's simply true! :-) $\endgroup$ – Fernando Muro Jan 17 '12 at 14:08

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