Unreduced Suspension Isomorphism - MathOverflow

Unreduced Suspension Isomorphism

2011-11-09T10:42:56Z

Tending to a lecture on homotopy theory, the following question occured to me (is that a correct sentence?):

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?

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) $(x,0)$.

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.

Answer by Neil Strickland for Unreduced Suspension Isomorphism

2011-11-09T11:00:20Z

On the linguistic question: I would say "While attending a lecture on homotopy theory, the following question occurred to me."

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.

Answer by Tom Goodwillie for Unreduced Suspension Isomorphism

2011-11-09T14:45:33Z

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