# What is the suspension used in the Freudenthal suspension theorem?

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?

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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!

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+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):

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 $t$.

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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