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Prove that there is no map of degree two from $S^2$ to the torus $T^2$.

I'm struggling with this problem. I've tried lifting the map to the covering space but I'm not sure what to do from there. I keep getting results that I know are wrong. Most examples I can find of problems like this involve maps from $T^2$ to $S^2$, and I think those call for different techniques than what is needed here.

topman12435
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1 Answers1

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Let $f : S^2 \to T^2$ be a continuous map and let $p : \mathbb{R}^2 \to T^2$ be the universal covering map. As $S^2$ is simply connected, $f$ lifts to a map $\hat{f} : S^2 \to \mathbb{R}^2$ such that $f = p\circ\hat{f}$. As $\mathbb{R}^2$ is contractible, both $p$ and $\hat{f}$ are homotopic to constant maps, and therefore $f$ is homotopic to a constant map. So the degree of $f$ is the degree of a constant map, which is zero.

Michael Albanese
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  • So the degree 2 part of this question is irrelevant? It's a map of any degree greater than 0? – topman12435 Jun 28 '16 at 21:39
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    That's correct, the only thing special about $2$ is that it is not $0$. – Michael Albanese Jun 28 '16 at 21:41
  • This result is also known as $\pi_2(T^2)=0$, and it can alternatively be shown by way of $T^2=S^1\times S^1$ and $\pi_2(S^1)=0$ (which, by the way, is also easiest done via universal covering). – Arthur Jun 28 '16 at 21:51
  • @Arthur I haven't really learned much about $\pi_2$, can you explain that? – topman12435 Jun 28 '16 at 22:02
  • It's true in general that $\pi_n(X\times Y)=\pi_n(X)\times \pi_n(Y)$. Other than that, I don't think there much to say; you just apply that result in order to reduce to the simpler calculation of $\pi_2(S^1)$. Although, as I said, the proof is more or less exactly the same. – Arthur Jun 28 '16 at 22:08