 Solution

The universal cover of $S^1\times S^2$ isBecause $S^3$ 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 $f_{\ast }{\pi }_1(S^3)=0$. 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 givesTherefore $\boxed{f_{\ast }=0\text{ and }\mathrm{deg}f=0}$, with the last equality using the oriented three-dimensional fundamental classes.