The title is not complete, since it would be too long. Consider the following statement:

Let $U \subset \mathbb{R}^n$ be open, connected and such that its one-point compactification is a manifold. Then, this compactification must be (homeomorphic to) the sphere $S^n$.

Is the statement above true? If so, why?

  • $\begingroup$ In order to apply the result you are quoting, you need to know that $U$ itself is homeomorphic to $R^n$. I am not sure that the statement you are making is true. $\endgroup$ Jan 22, 2018 at 0:34
  • $\begingroup$ @MoisheCohen Yes, you are correct, my blunder. This may be worse, then... $\endgroup$
    – Aloizio Macedo
    Jan 22, 2018 at 0:39
  • $\begingroup$ After thinking a bit, the statement, seems to be true and is a corollary of the topological Schoenflies theorem, which is a very hard result, still open in the smooth category in dimension 4. $\endgroup$ Jan 22, 2018 at 0:42

2 Answers 2


I do not expect any simple proofs of this result. A good exercise would be to prove this for domains in $R^2$ without using anything about the classification of surfaces or the Schoenflies theorem in $R^2$.

Here is a proof of your statement (in the topological category). I will assume that $n\ge 2$ since there is nothing to prove in dimension 1.

Let $U$ be any open (connected, which is not really necessary) domain in $R^n$ whose 1-point compactification is an $n$-manifold $M$. First of all, it is easy to see that the complement of $U$ has to be connected (since a point cannot locally separate a manifold of dimension $\ge 2$).

Let $p\in M$ be such that $M-p\cong U$. Let $B$ be a metric ball centered at $p$ (with respect to a Euclidean metric on a neighborhood of $p$ in $M$). Let $Y=\partial B$. Then $Y$ as an $n-1$-dimensional tame sphere is contained in $U$ and separating $R^n$ in two components, a bounded component $C$ and an unbounded one. The bounded component is necessarily contained in $U$ (since the complement of $U$ is connected). Then, by topological Schoenflies theorem, $C$ is homeomorphic to $B^n$. Thus, $M$ is obtained by gluing two balls ($B$ and $C$) along their common boundary sphere $Y$ and, hence, homeomorphic to $S^n$.

  • $\begingroup$ Thanks for the answer! Just to clarify (and to see if I understood correctly), when you say "in the topological category" you intend to mean only that we can get a homeomorphism with $S^n$, and not that we can suppose that the manifold (the compactification) is a topological manifold correct? Because you need that the manifold is smooth to arrange that tame sphere. $\endgroup$
    – Aloizio Macedo
    Jan 25, 2018 at 19:24
  • 1
    $\begingroup$ @AloizioMacedo: Topological category is the one of topological manifolds (objects are topological manifolds, morphisms are continuous maps, embeddings are tame embeddings, etc.), as opposed to the one of smooth (or PL) manifolds; submanifolds in this category are (by definition) assumed to be tame. To get a tame sphere I do not need smoothness (there are other ways to accomplish tameness!), $Y$ is contained in a single chart of $M$, so it is easy to arrange for $Y$ to be tame by taking $Y$ to be a round sphere with respect to the chart. $\endgroup$ Jan 25, 2018 at 20:04

I can imagine an ellementary approach only for the special case of $\mathbb{R^2} $ and $\mathbb{R}$.

For $\mathbb{R^2}$:We know that all the compact surfaces arise from adding to the sphere a finite amount of handles or Mobius-strips. In any case, if you remove a point from a compact surface which is something of the above but not a sphere then you wouldn't get something homemoprhic to $\mathbb{R^2}$. So the only compactification could be to a sphere.

Similar arguments goes for $\mathbb{R}$ since the only compact $1$ manifolds are a closed line segment and the circle.


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