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I want to prove that $\mathbb R^2$ is the universal covering space of the connected sum of $m$ torus $mT=T\sharp T\sharp ...\sharp T$.

I try to argue as follows: first, prove that $\mathbb R^2$ is the universal covering space of the double torus $2T=T\sharp T$; then prove that $mT$ is the covering space of $(m-1)T$. At last, by the universal property of universal covering space and induction, we have $\mathbb R^2$ is the universal covering space of the connected sum of $m$ torus $mT=T\sharp T\sharp ...\sharp T$.

But how to prove the above two statement (the inductive hypothesis)? It seems that we can't find an explicit expression of the covering projection.

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    $\begingroup$ The universal covering space for $T\#T$ is homeomorphic to $\mathbb{R}^2$ but its construction involves the Poincaré disk and hyperbolic geometry. This part of the argument is not trivial, see for example the last section of Lee - Introduction to Topological Manifolds Chapter 12. $\endgroup$
    – Tob Ernack
    Jun 9, 2022 at 15:25
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    $\begingroup$ Does this answer your question? math.stackexchange.com/questions/4296146/…' $\endgroup$
    – Lee Mosher
    Jun 9, 2022 at 17:02
  • $\begingroup$ In general, $mT$ doesn't cover $(m-1)T$. To see this, recall that if a space finitely $X$ covers a space $Y$ and they both have Euler characteristics, then $\chi(X)/\chi(Y)$ is a positive integer (the number of sheets). Because $mT$ and $(m-1)T$ are both compact, and possible covering would have finitely many sheets, so this applies. But $\chi(mT) = 2-2m$, and so$(1-m)/(2-m)$ must be a positive integer. This implies $m=3$. On the other hand, something similar to what you're trying will work: for every $m\geq 2$, $mT$ does cover $2T$. $\endgroup$ Jun 9, 2022 at 18:27
  • $\begingroup$ Unless I am mistaken, we can build on @JasonDeVito's argument to get an answer: the universal covering space of $mT$ must be a simply-connected surface, so it must be either $\Bbb{R}^2$ or $S^2$, but Jason's observations about Euler characteristics imply that it cannot be $S^2$, so it must be $\Bbb{R}^2$. $\endgroup$
    – Rob Arthan
    Jun 9, 2022 at 21:07
  • $\begingroup$ @Rob: I was suggesting a rather easier approach, since I find that fact that the only simply connected surfaces are $S^2$ and $\mathbb{R}^2$ to be a hard theorem. Here's the key: If $X$ covers $Y$, then $X$ and $Y$ have the same universal cover. Then, if the OP can prove that $\mathbb{R}^2$ is the universal cover of $2T$, it follows from my other comment that $\mathbb{R}^2$ is the universal cover of $mT$ for all $m\geq 2$. Lee Mosher gives an argument that $\mathbb{R}^2$ is the universal cover of $2T$ here:mathoverflow.net/questions/131863/…. $\endgroup$ Jun 9, 2022 at 21:41

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