There are many closed manifolds with universal cover homotopy equivalent to $\mathbb{R}^n$, they are precisely the closed aspherical manifolds. There are also many closed smooth manifolds with universal cover diffeomorphic to $\mathbb{R}^n$, e.g. those which admit a metric of non-positive curvature. If one weakens diffeomorphic to homeomorphic, then the only additional examples one could possibly obtain would be four-dimensional, but no such examples are known to exist, see this question.

If one considers $\mathbb{R}^n\setminus\{x\}$ with $n > 2$ instead as a universal cover, there are plenty of examples. Any quotient of $S^{n-1}$ will have universal cover $S^{n-1}$ which is homotopy equivalent to $\mathbb{R}^n\setminus\{x\}$. If one upgrades to diffeomorphism, then the product of a smooth quotient of $S^{n-1}$ with $S^1$ yields a suitable manifold. Unlike the case of $\mathbb{R}^n$, one can construct smooth manifolds with universal cover homeomorphic but not diffeomorphic to $\mathbb{R}^n\setminus\{x\}$. For example, for any exotic $(n-1)$-sphere $\Sigma$, the universal cover of $\Sigma\times S^1$ is diffeomorphic to $\Sigma\times\mathbb{R}$ which is homeomorphic to $S^{n-1}\times\mathbb{R}$, and hence $\mathbb{R}^n\setminus\{x\}$, but is not diffeomorphic to it, see these comments by Igor Belegradek.

What if we remove more than one point from $\mathbb{R}^n$?

Is there a closed manifold whose universal cover is homotopy equivalent/homeomorphic/diffeomorphic to $\mathbb{R}^n\setminus\{x_1, \dots, x_k\}$ for some $k > 1$?

There are no such manifolds in dimension one or two, but I don't even know if such examples can arise in dimension three.

The space $\mathbb{R}^n\setminus\{x_1,\dots, x_k\}$ is homotopy equivalent to $\bigvee_{i=1}^kS^{n-1}$. If $M$ is a closed manifold with the given universal cover, one might hope that an analysis of the natural $\pi_1(M)$-action on $\pi_{n-1}(M) \cong \mathbb{Z}^k$ could provide some insight.

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    $\begingroup$ For homeomorphic/diffeomorphic the answer should be no, by the Stallings theorem on ends of groups. The fundamental group is finitely generated; the assumption implies it has $k+1>1$ ends; Stallings' theorem tells you it has either $1$ or infinitely many. (I am sorry for answering in a comment. It is late, and I don't have time to check that I am not making a stupid error or to write anything in more detail.) $\endgroup$ Mar 30, 2022 at 0:18
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    $\begingroup$ The theorem that the number of ends is $0,1,2$ or $\infty$ is due to Freudenthal and Hopf, decades before Stallings. By the way, this "ends" result says that the space of ends, if infinite, is a Cantor. Hence, for every $n\ge 3$ and every totally disconnected compact space $K$ with at least 3 elements and at least one isolated point, $S^n-K$ is not homeomorphic to the universal cover of any compact space. $\endgroup$
    – YCor
    Mar 30, 2022 at 8:21
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    $\begingroup$ @YCor Thanks! I did forget the case of 2 ends last night. The correct form is just now typed in an answer... But I did not remember it is due to Freudenthal. $\endgroup$ Mar 30, 2022 at 8:22
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    $\begingroup$ 2 ends can be achieved (universal cover of $\mathbf{S}^{n-1}\times \mathbf{S}^1$). References for the Freudenthal-Hopf theorem can be found on Wikipedia: en.wikipedia.org/wiki/Stallings_theorem_about_ends_of_groups And of course 1 end (universal covering $\mathbf{R}^n$) and 0 end (universal covering $\mathbf{S}^n$) can be achieved too. I'm not sure about $\mathbf{S}^n$ minus Cantor (by the way this is not uniquely defined up to homeomorphism — let's say, minus a Cantor that fits inside a line). $\endgroup$
    – YCor
    Mar 30, 2022 at 8:24
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    $\begingroup$ @DavidESpeyer Actually there exists a Cantor subset in $\mathbf{S}^3$ whose complement is not simply connected (Antoine necklace). I don't know what can be done if the complement is required to be simply connected. (At the positive side all Cantor subsets in the line or plane are topologically equivalent.) $\endgroup$
    – YCor
    Mar 30, 2022 at 13:38

1 Answer 1


If we demand that the universal cover is homeomorphic / diffeomorphic to $\mathbb{R}^n \setminus \{x_1,\ldots,x_k\}$ with $k>1$ the answer is no, there are no such closed manifolds. Each missing point (together with the "infinity" of the one-point compactification of $\mathbb{R}^n$) is an end of the covering space, and these are all the ends. Therefore the universal cover has $k+1 \ge 3$ ends.

Freudenthal and Hopf proved that a finitely generated group has $0$, $1$, $2$, or infinitely many ends. The ends of the fundamental group biject with the ends of the universal cover, so the theorem contradicts our assumption.

I suspect (but am really not sure) that it may be possible to extend the argument to closed $n$-manifolds with universal cover only homotopy equivalent to $\mathbb{R}^n \setminus \{x_1,\ldots,x_k\}$ with $k>1$. Perhaps it can be shown that the universal cover must have $k+1$ ends by thinking of the separation properties of representatives of $H_{n-1}(\bigvee_i S^{n-1})$.

  • $\begingroup$ Thanks. I really should have thought of this myself (I was looking at Stallings' Theorem yesterday). The homotopy equivalence version of the question is still open, but that's probably the least interesting of the three. $\endgroup$ Mar 30, 2022 at 8:34

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