How does the Hopf map generate $\pi_3(S^2)$? I have been studying the Hopf fibration which is an example of a map from $S^3$ to $S^2$. It is a member of $\pi_3(S^2)$ and shows that this group is non-trivial. It can be shown using a long exact sequence applied to the Hopf fibration that $\pi_3(S^2) \cong \mathbb{Z}$. 
However, the set of notes I am using to learn this material (here) then claims that the Hopf map is a generator of this group. Furthermore, the Hopf invariant, which is a sort of linking number of the preimages of distinct points can be used to define this isomorphism with $(\mathbb{Z},+)$. 
I fail to see how the hopf map generates $\pi_3(S^2)$. Specifically, I would like to know how you can compose the hopf map to generate maps with higher hopf invariant that make up this integer-like group structure. It would be really helpful to explicitly see this in action (i.e. pointers on how to writ this out using elementary algebra).
Note: I have a Chemistry background so my understanding of Mathematics is quite basic though I am willing to learn.
 A: Suppose we write $S^3$ as the set of all pairs $(z, w)$ of complex numbers with $|z|^2 + |w|^2 = 1$, and write $S^2$ as the complex plane plus a point at infinity. The the Hopf map is just
$$
(z, w) \mapsto \frac{z}{w}.
$$
(The formula you gave is, I believe, pretty much the same as this, but after stereographic projection from the plane to 3-space. 
To get a degree-two map, we can simply take a degree two map from $S^2$ to $S^2$ and compose. In particular, we can define
$$
H_2(z, w)= \left(\frac{z}{w}\right)^2
$$
and more generally, define
$$
H_k(z, w)= \left(\frac{z}{w}\right)^k
$$
for every integer $k$. The map $k \mapsto H_k$ defines an injective (use the Hopf invariant!) map from $\Bbb Z$ to $\pi_3(S^2)$. And $H_1 = H$ is a generator, because the Hopf invariant for that map is a $1$, which is a unit. 
A: The Hopf map is the projection map of a bundle $ S^1 \rightarrow S^3 \rightarrow S^2$. From this fiber bundle we get a long exact sequence. This sequence splits up into exact sequences $0 \rightarrow \pi_n (S^3) \rightarrow \pi_n (S^2) \rightarrow 0$ for $n>2$ because the higher homotopy groups of a circle vanish. In particular, this isomorphism is coming from post composition with the Hopf map. Specializing to $n=3$, this isomorphism must take any generator of $\pi_3 (S^3)$ to a generator of $\pi_3 (S^2)$. Since a generator of the first is the identity, the Hopf map must generate $\pi_3 (S^2)$.
