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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.

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    $\begingroup$ Think about fundamental groups or cohomology rings for two different approaches. $\endgroup$ – user98602 Jul 23 '16 at 2:14
  • $\begingroup$ @MikeMiller I'm sorry that I'm not familiar with cohomology rings. Is it impossible to prove without cohomology? $\endgroup$ – wooa0923 Jul 23 '16 at 2:32
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    $\begingroup$ See the clause of the sentence before "or". $\endgroup$ – user98602 Jul 23 '16 at 2:35
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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.

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    $\begingroup$ It's worth noting that not every $f$ needs to be null-homotopic, though they always induce the zero map on homology. The Hopf map $\eta: S^3 \to \Bbb{CP}^1$, given by $f(z,w) = z/w$, is not null-homotopic, and factors through $\Bbb{RP}^3$. Indeed every map $\Bbb{RP}^3 \to S^2 \times S^1$ must be of the form $\iota n \eta$, where $\eta$ is the Hopf map, $n$ denotes the unique (up to homotopy) degree $n$ map $S^2 \to S^2$, and $\iota$ is the inclusion $S^2 \to S^2 \times S^1$; none of these are homotopic. $\endgroup$ – user98602 Jul 23 '16 at 4:36
  • $\begingroup$ Eh, the above comment isn't quite right. I'm missing at least one map, and probably an infinite family of them. Oh well. $\endgroup$ – user98602 Jul 23 '16 at 4:49

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