Is it there a nice way of visualising $\pi_3(\mathbb{S}^2)$? I was wandering if there is a nice geometric way of visualising $\pi_3(\mathbb{S}^2)$. Thanks in advance.
 A: A generator is given by the Hopf fibration $p :S^3 \to S^2$.
A: Let me make a stab at amplifying @NicolasHemelsoet's answer, using a bunch of unsubstantiated but plausible claims: 
Suppose $f : S^3 \to S^2$ is a continuous map. Replace $f$ by a new map (which I'll still call $f$) that's actually smooth, but is homotopic to the original. 
Most points of $S^2$ will be "nice" (by Sard's Theorem), so that for any two such nice points $p$ and $q$ in the 2-sphere, their preimages are both 1-manifolds. To keep things simple, let's assume that they're actually connected 1-manifolds, hence are closed curves in the 3-sphere.
The preimage $\gamma$ of $p$ can be oriented as follows: suppose $P$ is a point with $f(P) = p$. Then $df(P)$ maps the tangent $u$ to $\gamma$ at $P$ to the zero vector, but if you pick two vectors $v$ and $w$ that are perpendicular to $u$ and look at $df(P)[v]$ and $df(P)[w]$, you get a basis for the tangent space of $S^2$ at $p$. If it's an oriented basis, good. If not, swap $v$ and $w$. OK, so now you have two vectors, $v, w$ at $P \in S^3$, and their images, under $df(P)$, form an oriented basis for $S^2$. Extend these to $v, w, u$; this will either be an oriented basis for the tangent space to $S^3$ at $P$ or anti-oriented. If it's anti-oriented, reverse $u$. Now the vector $u$ gives you an orientation for the curve $\gamma$. 
Whew!
So you've got two oriented curves in the 3-sphere, one for $P$, one for $Q$. You compute their linking number $s$. Amazingly enough, the value of $s$ is independent of the choice of $P$ and $Q$, and indeed, of $p$ and $q$ (as long as you stick to "nice" points). That linking number $s$ is an integer which tells you which element of $\pi_3(S^2)$ the function $f$ belongs to. 
For the hopf map, the preimages of any two points are two linking circles, so its degree is $\pm 1$. 
