Whitehead Bracket lies in the kernel of suspension

Question

On Wikipedia https://en.wikipedia.org/wiki/Whitehead_product it is stated that for $f\in \pi_k(X), g\in \pi_l(X)$ the reduced suspension of their Whitehead bracket (Whitehead product) is nullhomotopic. I am looking for a proof of this fact $(k,l>1)$.

I have tried the following approach. Let $\alpha: S^{k+l-1}\to S^k \vee S^l$ be the attaching map of the $(k+l)$-cell of $S^k\times S^l$ with the usual cell structure. It suffices to show that its reduced suspenison $\Sigma \alpha: S^{k+l}\to S^{k+1}\vee S^{l+1}$ is nulhomotopic. Why I believe this could be the case? $\Sigma \alpha$ is the attaching map of the $(k+l+1)$-cell of $\Sigma (S^{k}\times S^l)$, where $\Sigma (S^{k}\times S^l)$ has the suspended cell structure. In this stack exchange question Suspension of a product - tricky homotopy equivalence they produce a homotopy equivalence $\Sigma(S^k\times S^l) \simeq S^{k+1}\vee S^{l+1} \vee S^{k+l+1} $. Hence using $\Sigma \alpha$ or a constant map $S^{k+l}\to S^{k+1}\vee S^{l+1}$ gives homotopy equivalent spaces (their mapping cones are homotopy equivalent), but I don't see how this could implie that $\Sigma \alpha$ is nullhomotopic.

Answer 1

Notation: let $\bar\alpha:D^{k+l}\to S^{k}\times S^l$ denote the characteristic map associated with attaching the top cell in $S^k\times S^l$. Let $i_{S^k}, i_{S^l}:S^k,S^l\to S^k\vee S^l$ and $j:S^k\vee S^l\to S^k\times S^l$ be standard inclusions, and let $\pi_{S^k},\pi_{S^l}:S^k\times S^l\to S^k, S^l$ be projections.

Using the notation of the first answer in the post you linked to, we know that the map $$\psi:=\Sigma(i_{S^k}\pi_{S^k})+\Sigma(i_{S^l}\pi_{S^l}):\Sigma(S^k\times S^l)\to \Sigma(S^k\vee S^l)$$ satisfies $$\psi\circ \Sigma j\simeq \text{id}_{\Sigma(S^k\vee S^l)}$$ by cogroup structures of reduced suspensions.

The relationship between the attaching map and the characteristic map suggests that $$\Sigma\bar\alpha\circ \Sigma i=\Sigma j\circ\Sigma\alpha$$ where $i:S^{k+l-1}\to D^{k+l}$ is the inclusion of boundary. Composing with $\psi$, we get $$\ast\simeq \ast\circ\Sigma i\simeq(\psi\circ\Sigma\bar\alpha)\circ \Sigma i\simeq (\psi\circ\Sigma j)\circ \Sigma \alpha\simeq \Sigma\alpha$$ since $\psi\circ\Sigma\bar\alpha$ is a map from $D^{k+l}$ which is contractible. This shows that $\Sigma\alpha$ is null-homotopic. (and of course the top cell of $S^k\times S^l$ splits off after suspension)