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Prove that every continuous map $f:P^2\to S^1$, where $P^2$ is the projective plane, is nullhomotopic.

I think I need to use the fact that $\pi_1(P^2) = \mathbb{Z}/2\mathbb{Z}$ and covering space theory. The map $f$ induces a map $f_* : \pi_(P^2) = \mathbb{Z}/2\mathbb{Z} \to \pi_1(S^1) = \mathbb{Z}$, but I don't see why this map is trivial.

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    What have you tried so far? Is $P^2$ the projective plane? In that case, $\pi_1(P^2)=\mathbb{Z}/2$. HINT: What are the maps $\mathbb{Z}/2 \to \mathbb{Z}$?2012-05-09
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    @FredrikMeyer's suggestion is what you need. See also a [related post](http://math.stackexchange.com/questions/77891/null-homotopy-of-a-map-to-the-circle).2012-05-09
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    I'm learning covering space and this is an exercise for that.I think I should use things related to covering space but $\pi_1(\mathbb P^2)=\mathbb Z_2$...ah,maybe I see how to prove it.2012-05-09
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    @bgins Thank you but $f_*(\pi_1(\mathbb P^2))$ is not trivial,so I can't find a lift of f from $\mathbb R \to S^1$2012-05-09
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    @Fredrik Meyer Is there a contractible covering space of $S^1$ whose fundamental group is $\mathbb Z$?2012-05-09
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    @JiangnanYu No, because all contractible spaces have trivial fundamental group.2012-05-09
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    @JiangnanYu: Why would you say $f_*(\pi_1(\mathbb{P}^2))$ is not trivial?2012-05-09
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    @JoeJohnson126 Sorry,I mean we can't see directly that $f_*(\pi_1(\mathbb P^2))$ is trivial2012-05-09
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    @Fredrik Meyer Hi,I can't figure out the proof,could you tell me how the maps $\mathbb Z/2 \to \mathbb Z$ work?2012-05-09
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    If you are having trouble characterizing $\operatorname{Hom}_\mathbb{Z}(\mathbb{Z}_2,\mathbb{Z})$, you should probably go back and brush up on commutative algebra before you continue studying algebraic topology ...2012-05-09

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