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Suppose $X$ is a space with a one point compactification $f: X\to Y$, where $x_0$ is the point that gets added to $X$, and suppose we have another one point compactification $f': X\to Y'$ where $x_1$ is the point that gets added to $X$.

I think we can define a homeomorphism $g: Y\to Y'$ in the following way: $g(x_0)=x_1, g(x)=f'f^{-1}(x)$ for all other $x$. Intuitively I'm thinking this should be a homeomorphism (if it's not correct me), but I'm not seeing how to prove it. I think it's fairly clear that it should be bijective, but how to show continuity (of both it and its inverse) isn't coming to me. If anyone could give a hint that would be swell.

Michael Hardy
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Alex
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  • The maps $f,f'$ must have some additional properties like being open or something.. – Berci Dec 04 '20 at 01:36
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    Do you know the definition of the one point compactification topology? I.e. What are the open subsets of $Y$ and $Y'$? That would be a good start... – Lee Mosher Dec 04 '20 at 01:37
  • I do not know the definition of the one point compactification topology. I've pretty much just seen the definition that Y is a compactification of X is there is an embedding f taking X into Y such that cl(f(X))=Y – Alex Dec 04 '20 at 02:41
  • A categorical approach would be to find a defining universal property of the on-point compactification. Then maybe from that, in a nonconstructive way, you can find the homeomorphisms. – Behnam Esmayli Dec 04 '20 at 03:24
  • Does "compact" include Hausdorff? If not, then it is wrong (with your definition). – Paul Frost Dec 04 '20 at 11:06
  • I wonder if it isn't possible to prove that two one-point compactifications of the same space are homeomorphic, even without knowing a specific construction. I think there might be a proof in Munkres. – MJD Dec 04 '20 at 14:57
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    @MJD For Hausdorff compactifications yes. In the non-Hausdorff case the universal property is not sufficient to prove uniqueness. – Paul Frost Dec 05 '20 at 00:28

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The homeomorphisms $f$ and $f'$ are continuous and their inverses are continuous; therefore $f'f^{-1}$ is continuous and its inverse $ff'^{-1}$ is continuous.

That is enough to show that $f'f^{-1}$ and $ff'^{-1}$ are continuous at every point except $x_0$ and $x_1.$

Continuity of a function $f$ at a point $x$ in its domain can be characterized by saying that for every open neighborhood $V$ of $f(x)$ there is some open neighborhood $U$ of $x$ such that the image of $U$ under $f$ is a subset of $V.$

To show that $f'f^{-1}$ is continuous at $x_0,$ suppose $V$ is some open neighborhood of $f(x_0) = x_1.$ An open neighborhood of the point $x_1$ that gets added is the complement (within $X\cup\{x_1\}$) of a closed set in $X.$ So $X\smallsetminus V$ is closed in $X;$ hence the complement of $X\smallsetminus V$ within $X\cup\{x_0\}$ is an open neighborhood of $x_0.$ It can readily be shown that the image of that open neighborhood of $x_0$ is that open neighborhood of $x_1.$

Continuity of the inverse $ff'^{-}$ is shown in the same way.

Michael Hardy
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