Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The theorem states:

The suspension map $\pi_{i}( S^{n})\rightarrow \pi_{i}(S^{n+1})$ is an isomorphism when $i<2n-1$ and a surjection when $i=2n-1$. In the case where $X$ is an $(n-1)$-connected CW complex, this holds for the suspension map $\pi_{i}(X)\rightarrow \pi_{i}(SX)$.

But what explicitly is this suspension map?

share|cite|improve this question
I think that you've lost a $+1$ somewhere. The map is from $\pi_i(S^n)$ to $\pi_{i+1} (S^{n+1})$, or at least it used to be back when I learned about it... – John Hughes Apr 23 '14 at 11:56
You should remove the tag "fiber bundles"... They really don't appear in your question. Also, John is right (see my answer). – Bruno Stonek Apr 23 '14 at 12:02

I am writing this answer as a supplement to Bruno's excellent (and sufficient) answer to provide an alternative perspective.

Let $X$ be a topological space and let $SX$ denote the suspension of $X$. Note that $X$ is the union of the upper cone $C_{+}X$ and the lower cone $C_{-}X$ (I assume that you have seen this before so I won't be explicit with the notation but I can be so if you like).

The suspension map is defined as follows:

$\pi_{i}(X)\cong \pi_i(C_{+}X,X)\to \pi_i(SX,C_{-}X)\to \pi_{i+1}(SX)$

where the isomorphisms can be derived from the appropriate long exact sequences of homotopy groups of a pair (using the contractibility of the cones $C_{+}X$ and $C_{-}X$) and the middle map is induced by inclusion. (Of course, this is assuming you are familiar with the notion of relative homotopy groups.)

Hope this helps!

share|cite|improve this answer
+1! I've expanded my answer to include some stuff that in particular explains how to obtain the suspension from two cones. – Bruno Stonek Apr 23 '14 at 12:25
Dear @Bruno, thank you! Your answer is great and the picture certainly adds a lot of intuition now. It is one thing to be able to write down a definition and another thing to really understand what has been written down so I really like your answer. – Amitesh Datta Apr 23 '14 at 12:31

Suspension is a functor $\Sigma$ defined on the homotopy category of pointed topological spaces. It maps $[X,Y]$ to $[\Sigma X,\Sigma Y]$, by mapping an arrow $f:X\to Y$ to an arrow $\Sigma f:\Sigma X\to \Sigma Y$ (see below) and being homotopy invariant. Here the brackets denote homotopy classes of based maps.

Therefore it maps $\pi_n(X)=[S^n,X]$ to $[\Sigma S^n,\Sigma X]\cong [S^{n+1},\Sigma X] = \pi_{n+1}(\Sigma X)$.

You seem to ask for clarification concerning the explicit construction of the suspension functor, namely at the level of maps. I believe Arkowitz's explanation in "Introduction to Homotopy Theory" is very explicit, and in turn, complements Amitesh's answer (you can ignore the definitions of $c$ and $j$ as they are not relevant for our purposes):

enter image description here

Geometrically, you should picture $\Sigma f$ going from $\Sigma A$ to $\Sigma A'$, and being just $f$ at every horizontal slice, which is a time $t$.

share|cite|improve this answer
Does this answer your question or is there something that you'd like me to expand? – Bruno Stonek Apr 23 '14 at 11:59
Is it some how possible to write it similarly to how we can write a normal suspension in the form of a quotient space? – Rhoswyn Apr 23 '14 at 12:09
Dear @Rhoswyn, I am not sure if this directly answers your question but the definition of the suspension of $X$ as a certain quotient space is formally equivalent to writing the suspension as a union of cones with common base $X$. And it is possible to define the suspension map using these cones - e.g., please see my answer below. – Amitesh Datta Apr 23 '14 at 12:14
@Rhoswyn: I think you are asking how to explicitly define $\Sigma f$ for an arrow $f$. I've included this in my answer. – Bruno Stonek Apr 23 '14 at 12:37

Your Answer


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.