In my general topology course, we've briefly touched the subject of one-point compactifications (Alexandroff); we've stated that every topological space $X$ has a one-point compactification $(X^+,\varphi)$, where $X^+=X\cup\{\infty\}$ has the topology that consists of complements of closed compact sets in $X$ (in addition to open sets in $X$) and $\varphi$ is an embedding of $X$ into $X^+$. We've also stated that when $X$ is a locally compact Hausdorff space and $Y$ is a compact Hausdorff space, it's sufficient to check that $\exists b\in Y$ such that $Y\setminus\{b\}\cong X$, to see that $Y\cong X^+$.

The question may be naïve, but I'm interested in whether or not it holds that we can always embed any locally compact $A\subset\mathbb R^n$ into $\mathbb S^{n}$ in such a way that the north pole of $\mathbb S^{n}=\{(x,t)\in\mathbb R^n\times\mathbb R;\ \Vert x\rVert^2+t^2=1\}$, i.e. $N=(0, 1)$, takes the role of the compactification point $\infty$. I've considered some cases where it is possible to one-point compactify the space ($\mathbb Z$ with $\mathbb S^1$, $\mathbb Z\times(-1, 1)$, $\mathbb Z\times(-1, 1]$ with $\mathbb S^2$, tested how stereographic projections work with these, also proved any closed non-compact subset $Z\subset\mathbb R^n$ can be compactified in this way).

Then I began to consider spaces that aren't simply-connected, such as $\mathbb R^2\setminus\{(0,0)\}$, and certain non-connected spaces, such as $\mathbb R^2\setminus(\{0\}\times(-\infty,\infty))$ and quickly got stuck on how to compactify them in such a manner. I'm guessing it's because you can't, but missing what's the condition to be able to do so.


No. Your example of $\mathbb{R}^2 \backslash \{0\}$ is one of such cases. Its one-point compactification is homotopically equivalent to $S^2 \vee S^1$ (see here* before going on). So it suffices to argue why no space $A$ homotopically equivalent to $S^2 \vee S^1$ can be embedded in $S^2$.

One way to see this is via Alexander duality. We would need to have $$0 \simeq\widetilde{H}_{-1}(S^n-A;\mathbb{Z}) \simeq H^{2}(A;\mathbb{Z}) \simeq \mathbb{Z},$$ a contradiction (here we are using the fact that $S^n-A$ can't be empty. Note that $\widetilde{H}_{-1}(\emptyset) \simeq \mathbb{Z},$ allowing the only case $A=S^n$ of a CW-complex with non-trivial second cohomology to be embedded in $S^n$).

*One problem with the one-point compatification is that it is rarely a manifold ($S^2 \vee S^1$ for example is not a manifold). If you take an arbitrary open connected set of Euclidean space and one-point-compactify it, it will usually not be a manifold. If it were, we could try to attack the problem in a way simpler than what is done above in the following way: the only compact manifold of dimension $n$ which embeds in $S^n$ is $S^n$ itself; this is due to the invariance of domain theorem, which tells us that the embedding must be an open map (and the image is closed since it is the image of a compact set under a continuous map with Hausdorff codomain), hence the image must be the whole $S^n$. So it would suffice to pick an open subset of Euclidean space such that its one-point compactification is a manifold which is not $S^n$. As of now, I don't know if it is possible. I've asked this question here.

  • $\begingroup$ The one point compactification of $\mathbb R^2 -0$ is the sphere with two points identified. Is it the same as the wedge you mentioned? $\endgroup$ – user99914 Jan 21 '18 at 20:27
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    $\begingroup$ @JohnMa Homotopically equivalent, yes. For a reference, example $0.8$ in Hatcher deals with this. $\endgroup$ – Aloizio Macedo Jan 21 '18 at 20:31
  • $\begingroup$ Thank you for your answer. I've yet to have a course in geometric topology, so I can only understand your arguments intuitively and can't thouroughly question them. Will return to do so when I get there. $\endgroup$ – suhogrozdje Jan 21 '18 at 23:01

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