# Unreduced Suspension Isomorphism

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.

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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

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oooh embarassing. Yupp, that does it. You can even find a pointed map that is a (non-pointed) homotopy equivalence, right? $\left(X \times {0} \cup {x} \times I, (x,1)\right) \rightarrow (X,x)$ the projection, correct? –  old account Nov 9 '11 at 15:59
That is correct. –  Tom Goodwillie Nov 9 '11 at 17:10
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.
damn, for the argument to go through i seem to need $S$ to be coproduct preserving, which it is not. could you maybe elaborate on that answer? –  old account Nov 9 '11 at 14:06
I think Neil's argument is this: consider the universal sum $f: S^k\to S^k \vee S^k$, and think about $Sf: S(S^k)\to S(S^k\vee S^k)$. Using a homeomorphism $S(S^k)=S^{k+1}$, think of this as an element of $\pi_{k+1}S(S^k\vee S^k)$; you want to show this is the same as the element $S(i_1)+S(i_2)$, i.e., that two maps into $S(S^k\vee S^k)$ are homotopic. Use the homotopy equivalence $S(S^k\vee S^k)\to S^{k+1}\vee S^{k+1}$. –  Charles Rezk Nov 9 '11 at 15:52