What is $\pi_3(S^2\times S^1)$? What is the third  homotopy group of $S^2\times S^1$? Here $S^n$ means the $n$-dimensional sphere.
I am a physicist and did not take courses on too mathematical topics. I just need a fact of this in a recent research project. I thought some mathematicians or students from the mathematical department may know the result. Details on derivation could be skipped (for me).
 A: We may identify: $$\pi_3(S^2\times S^1)\cong \pi_3(S^2)\oplus \pi_3(S^1),$$
via projection onto the two factors.
We know $\pi_3(S^1)\cong \pi_3(\mathbb{R})$ as this is unchanged by passing to the universal cover. As $\mathbb{R}$ is contractible, we get $\pi_3(\mathbb{R})=0$.
For the other factor, $\pi_3(S^2)\cong \pi_3(S^3)$, via the long exact sequence associated to the Hopf fibration, and using the fact that the second and third homotopy groups of the fiber $S^1$ are trivial (for the reason given above):
$$
\begin{array}{cccccc}
\to&\pi_3(S^1)\cong 0& \to& \pi_3(S^3)&\to&\pi_3(S^2)\\
\to&\pi_2(S^1)\cong0&\to
\end{array}
$$
We then have $\pi_3(S^3)\cong H_3(S^3;\mathbb{Z})\cong\mathbb{Z}$, via the Hurewicz homomorphism.
The net result is $\pi_3(S^2\times S^1)\cong\mathbb{Z}$, with the generator being the Hopf map composed with inclusion: $$S^3\stackrel{h}\to S^2 \hookrightarrow S^2\times S^1.$$
In simple terms, the Hopf map is the map from $$S^3=\{(z,w)|\,\,z\bar{z}+w\bar{w}=1\}\subseteq \mathbb{C}^2,$$ to $$S^2=\{[z:w]|\,\,(z,w)\in \mathbb{C}^2 \backslash \{(0,0)\}\},$$
which maps $(z,w)\mapsto [z:w]$.
