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I am trying to prove that if $N$ is a compact manifold that covers the torus $T$, then $N$ must be homeomorphic to $T$. I pretty much have the proof (used euler characteristic property and classification of 2-manifolds). However, I am left with one difficulty. I need to show that $S^2$ cannot cover $T$. Is there an easy way to show this? Thank you in advance.

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Observe that the universal cover of $T$ is $\mathbb R^2$, via the map $\mathbb R^2\to \mathbb R^2/\mathbb Z^2\cong T$. If $S^2$ also covered $T$, then since it is simply connected it would also be a universal cover of $T$. Since universal covers are unique, this would imply $S^2\cong \mathbb R^2$, which is clearly false as $S^2$ is compact and $\mathbb R^2$ is not.

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    $\begingroup$ That's an amusing, very elementary argument Alex: bravo! $\endgroup$ – Georges Elencwajg Apr 14 '13 at 20:46
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First observe that if $\pi: Y \rightarrow X$ is a covering map and $Y$ is compact and connected, then the degree must be finite.

Second: yes, use Euler characteristics! Remember that if $\pi: Y \rightarrow X$ is a finite degree covering map, then $\chi(Y) = (\operatorname{\deg} \pi) \chi(X)$. I think you'll find that you're done.

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  • $\begingroup$ This is exactly what I did. However, $\chi(T) = 0$. We also have that $\chi(Klein) = \chi(S^2) = 0$. $\endgroup$ – gary Apr 14 '13 at 20:30
  • $\begingroup$ Does this look right to you? $\endgroup$ – gary Apr 14 '13 at 20:33
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    $\begingroup$ Dear gary, $\chi (S^2)=2$ $\endgroup$ – Georges Elencwajg Apr 14 '13 at 20:43
  • $\begingroup$ @Gary: Ah! Georges has identified your mistake. $\endgroup$ – Pete L. Clark Apr 14 '13 at 21:02
  • $\begingroup$ oh yes sorry my mistake. 1 0-cell and 1 2-cell $\endgroup$ – gary Apr 15 '13 at 0:26
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The other answers are great and so I thought I'd just add a 'high powered'* method for getting the same result. As Alex said in his answer, if $S^2$ covered $T$ then $S^2$ would be the universal cover of $T$. Note though, that if $p\colon X\rightarrow Y$ is a covering map, then the induced map $p_*\colon\pi_n(X)\rightarrow\pi_n(Y)$ on homotopy groups is an isomorphism for $n\geq 2$.

The second homotopy group of $S^2$ is cyclic abelian on one generator but the second homotopy group of $T$ is trivial as $T$ is covered by the plane $\mathbb{R}^2$ which has trivial homotopy groups at all levels. That is, $\pi_2(T)\cong\pi_2(\mathbb{R}^2)=0$ and $\pi_2(S^2)\cong\mathbb{Z}$ and so no isomorphism exists between homotopy groups. It follows that no such covering map can exist.

*high powered only in the sense that it uses the higher homotopy groups of a space and results about induced maps on them from covering space theory.

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Ok! I think there is an elementary proof of the original fact you are proving. First of all, note that, if $ p: E\to T $ is a covering, with $E$ compact, then the index of $ \pi _1(E,e) $ in $ \pi _1 (T,t) $ is finite.

Since you know $\pi _1 (T,t) $, you know all subgroups of finite index... Then, to prove what you want, you only have to realize all the coverings inducing such subgroups (all these coverings will be different maps with the same domain : torus).

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Another proof:

Endow both $T, S^{2}$ with compatible complex structures. Then since $S^{2}$ is simply connected, the map lifts to the universal cover of $T$, which is $\mathbb{C}$. Moreover, the lift is holomorphic, and so we obtain a holomorphic map $S^2 \cong \mathbb{P}^1 \to \mathbb{C}$. However, by the maximum modulus principle, the only such maps are constant. QED.

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