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Nov 19, 2009 at 22:36 comment added Tyler Lawson That map should instead stably be equal to the map $S^{-m} \wedge S^0 \to S^{-n} \wedge S^0$. I think that this material should be covered in J. F. Adams' Chicago notes on stable homotopy theory - roughly the wedge is both product and coproduct in the stable homotopy category, and so is preserved by this dualization.
Nov 19, 2009 at 21:23 comment added Igor Belegradek Also you seem to be using that "dual of the wedge is wedge of the duals". Where can I find this stated or proved? Thanks!
Nov 19, 2009 at 21:14 comment added Igor Belegradek Sorry I misspoke about self-maps. What I do not understand is that you seems to be saying that $\Sigma^{n+s}(S^m\vee S^0)\to \Sigma^{m+s}(S^n\vee S^0)$ stably equals to $S^m\vee S^0\to S^n\vee S^0$. Is this correct?
Nov 19, 2009 at 20:55 comment added Tyler Lawson I'm confused, I don't see any maps which are self-maps - the closest is the map $\Sigma^{s+n} S^m_+ \to \Sigma^{s+m} S^n_+$, which have the same dimension but the basepoints suspend to cells of different dimension.
Nov 19, 2009 at 20:36 comment added Igor Belegradek I am puzzled that you are getting a self-map of a wedge of spheres while in my question it is a map of wedges of spheres of different dimensions. In other words, stably the dual is a map from the suspension of $S^m_+$ to $S^n_+$, not from $S^m_+$ to $S^n_+$.
Nov 19, 2009 at 20:22 history answered Tyler Lawson CC BY-SA 2.5