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There are informative and easily accessible images and videos that illustrate the Hopf fibration $S^3\to S^2$ by describing what happens to the fibers in the unit cube $(0,1)^3\approx S^3\backslash \ast$, e.g. those by Niles Johnson. Is there a similar way to visualize the map $S^3\to S^2\vee S^2$, which is the attaching map of the $4$-cell of $S^2\times S^2$ and allows one to the construct the first interesting higher whitehead product?

I understand that this map is not a fibration - it's the restriction of the product of the characteristic maps of the $2$-cells of $S^2\vee S^2$ but, if possible, I think it'd be interesting to be able to understand this visually in a $3$-cube as some kind of higher analogue of conjugation.

Note: this was originally a post on MSE that received no answers.

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    $\begingroup$ Take the standard Heegaard splitting of $S^3$ along a torus. In the first term $S^1 \times D^2$, first collapse the $S^1$ factor then compose with the map $D^2 \to D^2/S^1 \cong S^2$, mapping to the first sphere in $S^2 \vee S^2$, with the boundary $S^1 \times S^1$ all mapping to the basepoint. Similarly with the second term $D^2 \times S^1$ of the splitting, where you collapse the circle and then the boundary of the disc, using this to map to the second sphere. You can see this fairly explicitly thinking of $S^2 = \Bbb R^2 \cup \infty$. $\endgroup$
    – mme
    Commented Jun 15, 2019 at 16:41
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    $\begingroup$ The Pontriagin argument that homotopy groups of spheres is equivalent to embedded framed cobordism, this argument generalizes. Rather than making your map $S^n \to S^k$ transverse to the antipodal point of the base-point, you make your map $S^n \to S^{k_1} \vee S^{k_2}$ transverse to the antipodal points of the wedge/basepoint. Now rather than framed cobordism of manifolds, you've got framed cobordism of disjoint (i.e. labelled) manifolds. So how do you measure the Whitehead product? Linking number of the corresponding pair of framed manifolds. $\endgroup$
    – Ryan Budney
    Commented Jun 15, 2019 at 16:55

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The natural generality is as follows. For any finite-dimensional inner-product spaces $U$ we have a unit sphere $S(U)\subset U$ and a one-point compactification $S^U\simeq S(U\oplus\mathbb{R})$. If $V$ is another finite-dimensional inner-product space, then we can ask about the attaching map of the top cell in the product $S^U\times S^V$. This naturally arises as a map $$ w\colon S(U\oplus V)\to S^U \vee S^V = S(U\oplus\mathbb{R}) \vee S(V\oplus\mathbb{R}) $$
which can be described like this: $$ w(u,v) = \begin{cases} v^{-2}(2u\sqrt{v^2-u^2},2u^2-v^2) \in S(U\oplus\mathbb{R}) & \text{ if } \|u\| \leq \|v\| \\ u^{-2}(2v\sqrt{u^2-v^2},2v^2-u^2) \in S(V\oplus\mathbb{R}) & \text{ if } \|v\| \leq \|u\|. \end{cases} $$ (Here $u^2$ means $\|u\|^2=\langle u,u\rangle$.)

The fibre of $w$ over the basepoint is homeomorphic to $S(U)\times S(V)$. The fibre over all other points of $S(U\oplus\mathbb{R})$ is $S(V)$, and the fibre over all other points of $S(V\oplus\mathbb{R})$ is $S(U)$.

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