Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
add comment

2 Answers

up vote 7 down vote accepted

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

share|improve this answer
    
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
add comment

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.

share|improve this answer
    
Again on the linguistic part: I'm not attending the lecture, I'm the guy in charge of the exercise session and so on, but I don't know the english word for that, in german (I'm in münster, with the lecture given by christian ausoni) it's Übungsleiter. On the mathematics: oh right, I tried it for the universal example but used a single sphere, not a wedge. Stupid, not realizing there are two inputs... –  old account Nov 9 '11 at 13:32
    
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
2  
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
    
yes, but it is the reduction to the universal case, that i did not fully understand, when going through it this afternoon. For the reduction step on needs a map SX∨SX→S(X∨X) and this map will not be a homotopyequivalence in general. I overlooked however that the argument does not need it. So by now I'm quite convinced this works as well. But i do not know how to accept multiple answers. –  old account Nov 9 '11 at 17:38
    
@Fabian: Maybe the word you are looking for is teaching assistant: en.wikipedia.org/wiki/Teaching_assistant . –  Rasmus Bentmann Jan 5 '12 at 18:55
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.