When practicing old qualifying exam problems, I had trouble with this one. Thanks for any help! Is it true that if the $1$-point compactifications of two locally compact Hausdorff spaces $X$, $Y$ are homeomorphic, then $X$ and $Y$ are necessarily homeomorphic? Give a proof or counterexample, as appropriate.

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    $\begingroup$ Do you have a guess? There are some very elementary examples that will tell you the answer. Try to think of familiar compact spaces and remove a point. Do you always get the same space? You can then turn this into an answer. $\endgroup$ Commented Apr 11, 2013 at 1:34

2 Answers 2


Always think about the simplest cases first! What possible one point comp. could give rise to, say, [0,1]?

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    $\begingroup$ Just because I'll be studying for Quals myself soon enough, could one take as examples $X=[0,\frac{1}{2})\cup(\frac{1}{2},1]$ and $Y=[0,1)$? It seems the one-point compactifications will both give rise to something homeomorphic to $[0,1]$, while $X$ and $Y$ are not homeomorphic as they don't even have the same number of components. $\endgroup$
    – anon271828
    Commented Apr 11, 2013 at 2:15
  • $\begingroup$ Yes, that's the example I was thinking of! The key idea in constructing this example is that whilst restrictions of homeomorphisms are homeomorphisms, one needs "the one point that is added to be the same in each case". Trying to violate this gives rise to lots of similar examples. $\endgroup$ Commented Apr 11, 2013 at 2:19

Sharkos's answer tells the story simply and eloquently. I think it's worth noting that the converse does hold. That is, if $X$ and $Y$ are homeomorphic, then their respective one-point compactifications $\alpha X$ and $\alpha Y$ will be homeomorphic as well. Indeed, let $f:X\to Y$ be a homeomorphism and let $g:Y\hookrightarrow\alpha Y$ the canonical inclusion. Show that $g\circ f:X\to\alpha Y$ is continuous and injective. Since $X$ is dense (when canonically included) in $\alpha X,$ then there is a unique continuous function $h:\alpha X\to\alpha Y$ whose restriction to $X$ is $g\circ f$. Show that $h$ is a homeomorphism. You may need to consider separately the case where $X$ and $Y$ are compact Hausdorff and the case where they are not (one is trivial).

  • $\begingroup$ It seems to me that this is just true because the Alexandroff construction is indeed topological: i.e., makes use of only homeomorphism-invariant properties of the topological space. $\endgroup$ Commented Apr 11, 2013 at 5:58
  • $\begingroup$ True enough, Pete. The proof I outlined is very brief, for just that reason. $\endgroup$ Commented Apr 11, 2013 at 12:44
  • $\begingroup$ I just remembered that I wrote the wikipedia article en.wikipedia.org/wiki/Alexandroff_extension. Take a look at the last sentence! $\endgroup$ Commented Apr 11, 2013 at 14:01
  • $\begingroup$ Interesting! I've not had much experience with category theory or functors, but that seems an apt perspective. $\endgroup$ Commented Apr 11, 2013 at 15:07

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