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Mar 18, 2015 at 20:29 comment added Jeff Strom Good point, and not one I have resolved yet. I sort of answered the question I thought you asked, not the one you did ask. Still thinking.
Mar 18, 2015 at 19:12 comment added user1437 Great! This is explicit enough for me. However, does this actually imply that the two resulting groups are not isomorphic? Perhaps the isomorphisms to the same abelian group are different on the underlying sets?
Mar 17, 2015 at 18:55 comment added Eric Wofsey This group is computed for the standard $H$-space structure here.
Mar 17, 2015 at 13:54 comment added Mark Grant Nice! And you might hope to compute the groups $[S^3\times S^3,S^3]$, using the cofibration sequence $S^3\vee S^3\to S^3\times S^3 \to S^3\wedge S^3 = S^6$.
Mar 17, 2015 at 13:36 history answered Jeff Strom CC BY-SA 3.0