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I know that any topological manifold has a universal cover (a link: Existence of a universal cover of a manifold.). I was wondering whether $\mathbb{R}^n$ is a universal cover for any $n$-dimensional manifold? I was trying to form a counter example by showing that $\mathbb{R}^2$ doesn't cover $S^2$ but didn't make it.

user5721565
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    Every simply connected manifold is its own universal cover, so already $S^2$ is a counterexample. $\mathbb{R}^n$ is the universal cover of any complete connected flat manifold (https://en.wikipedia.org/wiki/Flat_manifold) or complete connected hyperbolic manifold (https://en.wikipedia.org/wiki/Hyperbolic_manifold), which implies that among closed connected surfaces $S^2$ is actually the only counterexample. – Qiaochu Yuan Jul 02 '18 at 06:24
  • Right! just to make sure I understand the contradiction, any two universal covers for space are hoeomorphic? (and $S^2$ is not homeomorphic to $\mathbb{R^2}$) – user5721565 Jul 02 '18 at 06:28
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    Yes, that's right. – Qiaochu Yuan Jul 02 '18 at 06:33
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    @dan "any two universal covers for space are hoeomorphic" That's what the word "universal" means. They are necessarily homeomorphic, and there is (in some sense) only one homeomorphism that plays well with the covering projetions. – Arthur Jul 02 '18 at 06:33
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    $\mathbb{R}^n$ is the universal cover of $\mathbb{R}^n$, $\mathbb{T}^n (=(S^1)^n)$, and more generally for any discrete group $G$ acting properly and freely on $\mathbb{R}^n$, you get an $n$-dimensional manifold $M=\mathbb{R}^n/G$ whose universal cover is $\mathbb{R}^n$ – Maxime Ramzi Jul 02 '18 at 09:22

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