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$\newcommand{\R}{\mathbb{R}}$ I'm being asked to show there is no covering map $\R P^2\to T^2$ (the torus) by showing that every map $\R P^2\to T^2$ is null-homotopic, by lifting to the universal cover.

I know the universal cover of $T^2$ is $\R^2$ (call the map $p$), hence given any $f:(\R P^2,y_0)\to(T^2,x_0)$, it lifts to a unique $\tilde f:(\R P^2,y_0)\to(\R^2,\tilde x_0)$ for some $\tilde x_0\in p^{-1}(x_0)$.

I don't really see what is special about $\R P^2$ and $T^2$ in particular that will make any lift of this type null-homotopic, and I'm pretty stuck. I wish I could say I've tried more but I just don't know... does anybody have any hints/insight? I'd prefer a nudge in the right direction rather than just being given the answer.

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    $\begingroup$ First, how do you know the map $f$ lifts? (It does, but you didn't say why.) The lift $\tilde f$ is nullhomotopic since $\mathbb R^2$ is contractible (this is a standard exercise). If $H$ is the nulhomotopy for $\tilde f$, then $p \circ H$ is a homotopy from $f$ to a constant map. $\endgroup$ – manthanomen Mar 9 '16 at 6:28
  • $\begingroup$ @manthanomen I'm fairly certain the lift exists because the only homomorphism $\mathbb{Z}_2\to \mathbb{Z}^2$ is the trivial map, hence $f_*\pi_1(\R P^2,y_0)=\{0\}=p_*\pi_1(\R^2,\tilde x_0)$. I really don't have as nice of intuition as I'd like for all this stuff. $\endgroup$ – Alex Mathers Mar 9 '16 at 6:34
  • $\begingroup$ @manthanomen OH wow okay I see why that works now. It made sense that $\tilde f$ was null-homotopic because $\R^2$ is contractible, but it seemed like that argument would work for any map between any spaces, because any (nice enough) space has a universal cover, so I was wondering what's special about $\R P^2$ and $T^2$ in this case. However, typing that out made me see it only works here because the only valid homomorphism between the two fundamental groups is the trivial one. $\endgroup$ – Alex Mathers Mar 9 '16 at 6:37
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If $p: \mathbb{R}P^2 \to \mathbb{T}^2$ is a covering map then $p_*: \pi_1(\mathbb{R}P^2, x) \to \pi_1(\mathbb{T}^2,p(x))$ which is the induced homomorphism, is injective. In view of this and the fact that the corresponding fundamental groups are $\mathbb{Z}_2$ and $\mathbb{Z} \oplus \mathbb{Z}$ then $p_*$ is the trivial map and so therefore $p$ is null-homotopic. Since $p$ supposedly is a covering for the $2$-torus then this implies $\textrm{id}_{\mathbb{T}^2}$ is homotopic to a constant map i.e $\pi_1(\mathbb{T}^2)$ is trivial which is a contradiction.

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