Unreduced Suspension Isomorphism - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T00:06:45Z http://mathoverflow.net/feeds/question/80463 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/80463/unreduced-suspension-isomorphism Unreduced Suspension Isomorphism Fabian Hebestreit 2011-11-09T10:42:56Z 2012-01-05T16:23:21Z <p>Tending to a lecture on homotopy theory, the following question occured to me (is that a correct sentence?): </p> <p>Given a pointed space $(X,x)$, is the UNREDUCED suspension map $S:\pi_k(X,x) \rightarrow \pi_{k+1}(SX, \ast)$ a group homomorphism? </p> <p>Here unreduced suspension refers to $SX = X \times D^1 / \sim$, where $\sim$ collapses $X \times \{1\}$ and $X \times \{-1\}$ respectively, and the basepoint $\ast$ is (the one point set with element) $(x,0)$. </p> <p>The statement is contained in every book on homotopy theory and almost trivial for the REDUCED suspension. For wellpointed spaces this of course surfices to answer my question, but it has resisted several similar 'general nonsense'-arguments in the general case.</p> http://mathoverflow.net/questions/80463/unreduced-suspension-isomorphism/80465#80465 Answer by Neil Strickland for Unreduced Suspension Isomorphism Neil Strickland 2011-11-09T11:00:20Z 2011-11-09T11:00:20Z <p>On the linguistic question: I would say "While attending a lecture on homotopy theory, the following question occurred to me."</p> <p>On the mathematical question: I think you need only prove that $S(u+v)=S(u)+S(v)$ when $(X,u,v)$ is the universal example of a based space with two based maps $u,v:S^k\to X$, ie $(X,u,v)=(S^k\vee S^k,i_1,i_2)$. Here $X$ is well-pointed so there is no problem. </p> http://mathoverflow.net/questions/80463/unreduced-suspension-isomorphism/80490#80490 Answer by Tom Goodwillie for Unreduced Suspension Isomorphism Tom Goodwillie 2011-11-09T14:45:33Z 2011-11-09T14:45:33Z <p>You can reduce to the well pointed case by observing that every space $X$ admits a weak equivalence $X'\to X$ from a well pointed space. Now the composition $$\pi_kX'\to \pi_kX\to \pi_{k+1}SX$$ (composition of your map with an isomorphism) is the same as the composition $$\pi_kX'\to \pi_{k+1}SX'\to \pi_{k+1}SX$$ in which the first map is a homomorphism because $X'$ is well pointed and the second is a homomorphism because it is induced by a map of spaces</p>