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I'm trying to prove the next result:

Let $X$ be a connected covering space of the torus $T:=\mathbb{R}^2/\mathbb{Z}^2$. If $X$ is compact then It is homeomorphic to a product of two circles.

Probably the most direct way of proving this is to show that there are only $3$ possible connected covering spaces of $T$, to know: $\mathbb{R}\times\mathbb{R}$, $\mathbb{R}\times S^1$ and $S^1\times S^1$ (the last one is the only one which is compact, so the result follows). As far as I know this can be deduced from the classification theorem of covering spaces (see this answer), however I need to prove first:

  1. $\mathbb{R}^2$ is the universal covering space of $T=\mathbb{R}^2/\mathbb{Z}^2$
  2. The fibers of this covering by the subgroups of $\pi_1(T)$ are precisely those $3$ spaces (up to conjugacy).

I'd appreciate any hints to prove these two facts, or any ideas to prove the main result in a simpler way. Thanks in advance.

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We want to show that the only covering spaces of the $2$-dimensional torus are tori, cylinders and the plane. We know that $\pi_1(S^1\times S^1)=\mathbb{Z}\times \mathbb{Z}$, which has the following subgroups: 1. the trivial subgroup, 2. free abelian groups with one generator $(p, q)$, 3. free abelian groups with two generators $(p, q)$ and $(r, s)$ such that $ps − qr\neq 0$. For each subgroup one constructs now a covering space. If the subgroup is trivial we obtain the universal covering space $\mathbb{R}^2$ with covering map $p : \mathbb{R}^2\rightarrow S_1 × S_1$, $$p(x, y) = (e^{2πix}, e^{2πiy}).$$ For the second case we obtain $S^1\times \mathbb{R}$, and for the third case $S^1\times S^1$. As you said, only the third one is compact.

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    $\begingroup$ I'll throw in the reason these calculations answer the question, namely, the covering space is determined by the subgroup. If $f_i : Y_i \to T$ are two covering maps $(i=1,2)$, if $Y_1,Y_2$ are connected, and if $$\text{image}\bigl((f_1)_* : \pi_1(Y_1) \to \pi_1(T) \bigr) = \text{image}\bigl((f_2)_*:\pi_1(Y_2)\to\pi_1(T)\bigr)$$ then $Y_1,Y_2$ are homeomorphic. $\endgroup$ – Lee Mosher Nov 20 '17 at 0:29
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A slightly different way of doing this is to note that, for a covering space to be compact here, it has to be finite sheeted (since the fiber over a point is a discrete set). And of course, the only finite-index subgroups of $\mathbb{Z}\times\mathbb{Z}$ are isomorphic to $\mathbb{Z}\times\mathbb{Z}$. We can go even further and note that we can always choose a basis $(x,y)$ of $\mathbb{Z}\times\mathbb{Z}$, so that for integers $n$ and $m$, $nx$ and $my$ are a basis for the subgroup. Then the covering is just the usual $n$-fold map ($z\mapsto z^n$) on the first $S^1$ factor, and $m$-fold on the second $S^1$ factor. [I should mention that changing the basis changes the $S^1$ factors that made up the original torus; in other words, we might have to parameterize the torus slightly differently to get the covering map to be so nice.]

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