TITLE: Visualize Fourth Homotopy Group of $S^2$
QUESTION [16 upvotes]: I know $\pi_4(S^2)$ is $\mathbb{Z}_2$. However, I don't know how to visualize it. For example, it is well known that $\pi_3(S^2)=\mathbb{Z}$ can be understood by Hopf Fibration. Elements in $\pi_3(S^2)=\mathbb{Z}$ can be understood as describing the number of links of the $U(1)$ fibers in $S^2$. 
So, do we have similar picture for $\pi_4(S^2)$? And, do we have similar topological invariant as links in previous case?

REPLY [2 votes]: The most explicit and geometric generator for $\pi_4(S^2)$ is probably the achiral genus-1 Lefschetz fibration $f : S^4 \to S^2$ with two Lefschetz singularies (one positive and one negative, with parallel vanishing cycles). In my knowledge, this Lefschetz fibration was first considered by Matsumoto ("On 4-manifolds fibered by tori", Proc. Japan Acad. 58 (1982), 298-301), and it is also described in Example 8.4.7 in the Gompf-Stipsicz book "4-Manifolds and Kirby Calculus".
I don't have now a proof of the homotopic non-triviality of this map, but I give this as a challenge!