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Let $p:S^2\to P^2$ be the double covering of the real projective plane. Let $g:P^2 \to P^2$ be a map such that its induced homomorphism on fundamental group is not trivial.

I'd like to show that there exists $T:S^2\to S^2$ covering $g$ with $T(-x)=-T(x)$.

Given $x\in S^2$ take $\bar x=p(x)\in P^2$ and $\bar y=g(\bar x)\in P^2$. It is clear that the fiber over $\bar y$ has two points $y_1,y_2\in S^2$.

I guess that $T(x)$ should be one of them. Also I guess that the action of $\pi_1(P^2)$ should be relevant to choose between $y_1,y_2$. But I'm not sure how.

I tried to take $\alpha\in\pi_1(P^2,\bar x)$ a generator and considered its image $\beta=g_*(\alpha)\in\pi_1(P^2,\bar y)$. Then I tried to take a lifting. But what should be the criteria to choose the necessary points?

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  • Sorry, what does it mean for a map of the sphere to 'cover' a map of a quotient of the sphere? – Dan Rust Nov 14 '13 at 23:24
  • @DanielRust, $p(T(x))=g(p(x))$, for any $x\in S^2$. – Sigur Nov 14 '13 at 23:27
  • When you say the induced homomorphism $p_*$ is not trivial, you mean that it is the identity, right? (Since $\pi_1(P^2)=\mathbb{Z}/2\mathbb{Z}$...) – Ben Blum-Smith Nov 14 '13 at 23:30
  • @BenBlum-Smith, yes, $g_*$ is iso. – Sigur Nov 14 '13 at 23:31
  • @DanielRust, $p$ is the covering map. It is not injective since $p(x)=p(-x)$. Also the non trivial homomorphism is $g_*$. Furthermore, if $p$ would be homeomorphism the fundamental groups of $S^2$ and $P^2$ would be isomorphic. This is not the case. – Sigur Nov 14 '13 at 23:37
  • @Daniel: Inducing an isomorphism on $\pi_1$ doesn't tell you the map is a homeomorphism (but the converse is true). – Jason DeVito - on hiatus Nov 14 '13 at 23:38
  • @Daniel: Thinking of the hemisphere model of $\mathbb{R}P^2$, map each line of lattitude to twice the latitude, and map everything above $45^o$ to the north pole. This induces an iso on $\pi_1$ but is far from injective. – Jason DeVito - on hiatus Nov 14 '13 at 23:43
  • @DanielRust - I'm not following either. A map on a closed manifold that is homotopic to the identity definitely does not have to be a homeomorphism. – Ben Blum-Smith Nov 14 '13 at 23:43
  • Ah sorry, ignore my comment then. – Dan Rust Nov 14 '13 at 23:44

1 Answers1

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You're right that there exists such a $T$.

The map $(g\circ p)_*:\pi_1(S^2)\rightarrow \pi_1(P^2)$ has trivial image because $S^2$ is simply connected. It follows that $g\circ p$ lifts to a map $S^2\rightarrow S^2$ because $S^2$ is $P^2$'s universal cover. (This is an application of the "lifting theorem" of covering space theory. See the accepted answer to this question.) This is the desired $T$.

To check that $T(-x)=-T(x)$, take a path from $x$ to $-x$ in $S^2$, and push it down to $P^2$ via $p$; it becomes a generator for $\pi_1(P^2,\bar x)$ where $\bar x = p(x)$. Push it forward to $P^2$ via $g$; it remains non-nullhomotopic because of your assumption about $g_*$.

On the other hand, by first pushing the path forward to $S^2$ along $T$ and then down to $P^2$ via $p$, we see that $T(x)$ and $T(-x)$ must be distinct; otherwise the loop would end up nullhomotopic. But if they are distinct, they are opposites, because they both map to $g(\bar x)$ via $p$.

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