# Are locally compact Hausdorff spaces with the homeomorphic one-point compactification necessarily homeomorphic themselves?

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.

• 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. Apr 11, 2013 at 1:34

• 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. Apr 11, 2013 at 2:15
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).