The Freudenthal suspension theorem states in particular that the map
$$
\pi_{n+k}(S^n)\to\pi_{n+k+1}(S^{n+1})
$$
is an isomorphism for $n\geq k+2$.
My question is: What is the intuition behind the proof of the Freudenthal suspension theorem?
[Math] the intuition behind the Freudenthal suspension theorem
at.algebraic-topologygt.geometric-topologyintuition
Related Solutions
One can think of the five lemma in terms of the two four lemmas. I think this makes it clearer... for instance drop the $A_1$ and $B_1$ from your diagram. If the maps from $A_2$ and $A_4$ to $B_2$ and $B_4$ are epimorphisms and the morphism $A_5 \to B_5$ is monic then the cokernel of $A_3 \to A_4$ is a subobject of the cokernel of $B_3 \to B_4$. So morally $B_3$ is an "extension" of quotients of $A_2$ and $A_4$ and we have not "killed less stuff" in the bottom row so $A_3 \to B_3$ should also be an epimorphism.
See
J. Hyam Rubinstein and Robert Sinclair. "Visualizing Ricci Flow of Manifolds of Revolution", Experimental Mathematics v. 14 n. 3, pp. 257–384. (Journal link)
(Image from that paper, via Wikipedia)
Best Answer
There are two proofs I particularly like:
A Morse-theoretic proof, probably due to Bott, can be found in Milnors book. Idea: consider the space of all paths on $S^n$ from the north to the south pole, which is homotopy equivalent to $\Omega s^n$. There is the energy function on this space. One does Morse theory: critical points correspond to geodesics (they are not non-degenerate, but Bott proved that a good deal of Morse theory works nevertheless). The set of absolute minima is the space of minimal geodesics connecting the poles. This is homeomorphic to $S^{n-1}$. All other critical points have index at least (rouhgly) $2n$. Therefore, the inclusion of the set of absolute minima into the whole space is $2n$-connected.
A spectral sequence proof (see Kirk, Davis, Lectures on Algebraic topology). Consider the homology Leray-Serre spectral sequence of the path-loop fibration $\Omega S^n \to PS^n \to S^n$. The total space $P S^n$ is contractible. Look at the shape of the spectral sequence; you'll see that for $k \leq 2n$ (again, only a rough estimate), all differentials out of the $E_{k,0}$-slot are zero, except of the last one, which gives an isomorphism $H_k (S^n)=E_{k,0}^{2} \to E_{0,k-1}^{2} = H_{k-1} (\Omega S^n)$. It is credible (but nontrivial to prove, this uses the transgression theorem) that this isomorphism is the same as the composition $H_k (S^n) \cong H_{k-1} (S^{n-1}) \to H_{k-1} (\Omega S^n)$ of the suspension isomorphism and the natural map $S^{n-1} \to \Omega S^n$. Thus the natural map is a homology isomorphism in a range of degrees, and by the Hurewicz theorem, this holds for homotopy groups as well.