Show that the homology induced homomorphism $f_*:H_3(RP^3)\rightarrow H_3(S^2\times S^1)$ is a zero map. Let $f:\mathbb{RP}^3\rightarrow S^2\times S^1$ be a continuous map.
Prove that induced map $f_*:H_3(\mathbb{RP}^3)\rightarrow H_3(S^2\times S^1)$ is a zero map.
I found that the third homology of both spaces are $\mathbb{Z}$ using cellular homology. But I can't show that the induced map is a zero map.

Following Mike's nice hint, note that $\pi_{1}(\mathbb{R}\mathbb{P}^{3}) = \mathbb{Z}/2\mathbb{Z}$, and $\pi_{1}(S^{2} \times S^{1}) \cong \pi_{1}(S^{2}) \times \pi_{1}(S^{1}) \cong \mathbb{Z}$, so the induced map $\pi_{1}(\mathbb{R}\mathbb{P}^{3}) \to \pi_{1}(S^{2} \times S^{1})$ must be the zero map. Recalling that $p \colon S^{2} \times \mathbb{R} \to S^{2} \times S^{1}$ is the universal cover of $S^{2} \times S^{1}$, Proposition 1.33 of Hatcher
implies that $f$ lifts to a map $\tilde{f} \colon \mathbb{R}\mathbb{P}^{3} \to S^{2} \times \mathbb{R}$ satisfying $f = p \circ \tilde{f}$. By functoriality, we must also have $f_{\ast} = p_{\ast} \circ \tilde{f}_{\ast}$ for the induced maps on $H_{3}$. But $H_{3}(S^{2} \times \mathbb{R}) \cong H_{3}(S^{2}) = 0$, so $f_{\ast}$ must be the zero map.   
