= Solution
The <universal cover> of $S^1\times S^2$ is
$$
p:\mathbb R\times S^2\longrightarrow S^1\times S^2,
\qquad p(t,z)=(e^{2\pi it},z).
$$
Because $S^3$ is simply connected, any <continuous> map $f:S^3\to S^1\times S^2$ lifts to $\tilde f:S^3\to\mathbb R\times S^2$ after choosing a lift of one base point. The <lifting criterion for a covering space> applies because $f_*\pi_1(S^3)=0$. Equivalently, path lifting defines the lift and simple connectedness makes its value independent of the chosen path.
The cover deformation retracts onto $\{0\}\times S^2$, so $H_3(\mathbb R\times S^2;\mathbb Z)=0$. Since $f=p\circ\tilde f$, functoriality of <homology (mathematics)> gives
$$
f_*:H_3(S^3;\mathbb Z)\xrightarrow{\tilde f_*}
0\xrightarrow{p_*}H_3(S^1\times S^2;\mathbb Z).
$$
Therefore $\boxed{f_*=0\text{ and }\deg f=0}$, with the last equality using the oriented three-dimensional fundamental classes.
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