Visualizing a Whitehead product: the attaching map $S^3\to S^2\vee S^2$ 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.
 A: 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 $S(U)$.
