This is a principal $S^1$ -bundle, and a generator of $\pi_3(S^2)$ . From the long exact sequence of the bundle: $$\cdots\pi_n(S^1)\to \pi_n(S^3)\to\pi_n(S^2)\to\cdots$$ we get that $\pi_n(S^3)\cong \pi_n(S^2)$ for all $n\geq 3$ . In particular, $\pi_3(S^2)\cong\pi_3(S^3)\cong\Z$ .