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It is well known that every topological space has a compactification. We therefore consider the one-point extension $(Y, T)$ of a topological space $(X, V)$ which is a compact space with a inclusion map $\iota$ as the embedding. Therefore $(\overline{\iota(X)}, T_{\iota(X)}, \iota)$ is the compactification of $(X, V)$.

We also know that the one-point extension is hausdorff if and only if $(X,V)$ is hausdorff and locally compact.

Is it true that for a compact hausdorff space $(Y, T)$ we can construct a topological space $(X, V)$ and an embedding $\iota$ such that $(Y, T, \iota)$ is a compactification of $(X,V)$ and $(Y,T)$ is a one-point extension of $(X,V)$ up to homoemorphism? So is it possible to somehow reverse the construction of a one-point extension of a topological space $(X, V)$?

I thought that we can reverse the construction of a one-point extension by taking a limit point $x \in Y$ of $Y$ and look at $X := Y \setminus \lbrace x \rbrace$ where $Y = \lbrace x \rbrace \cup Y \setminus \lbrace x \rbrace$ and show that \begin{align} T &= \lbrace U \subseteq Y : x \not\in U, \iota^{-1}(U) \mbox{ is open in } X \rbrace \\&\cup \lbrace U \subseteq Y : x \in U, X \setminus \iota^{-1}(U) \mbox{ is compact and closed in } X \rbrace \end{align} Since we look at one point extensions up to homeomorphism we can suppress the notation where $Y := Y \setminus \lbrace x \rbrace \sqcup \lbrace x \rbrace$ and $\iota: Y \setminus \lbrace x \rbrace \rightarrow Y \setminus \lbrace x \rbrace \sqcup \lbrace x \rbrace$ where $X \sqcup Y$ means the union of $\iota_X(X)$ and $\iota_Y(Y)$ where $\iota_X: x \mapsto (x,0)$ and $\iota_Y: y \mapsto (y,1)$. It is clear that since $(Y, T)$ is a one-point extension of $(X,V)$ that it is unique up to homeomorphism.

Are my thoughts so far true or am I misunderstanding something?

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  • $\begingroup$ The Aleksandrov extension is only a compactification (in the sense that $\iota[Y]$ is dense in $X$) when $Y$ is not compact to start with, of course. I take it in the most liberal sense, so I allow compact $Y$; i.e. the Aleksandrov extension always is defined and is compact and $\iota$ is an embedding. $\endgroup$ – Henno Brandsma Apr 9 at 15:14
  • $\begingroup$ If someone is interested in the proof of the construction I recently found an answer: math.stackexchange.com/a/427131/904739. $\endgroup$ – Orb Apr 9 at 15:50
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It's true: for a compact Hausdorff space $Y$ we can prove that $Y$ is the Aleksandrov extension (the common name for the one-pont extension) of $Y\setminus \{x\}$ for any $x \in Y$.

Of course $Y$ can be the Aleksandrov extension of several non-homomorphic spaces when $Y$ is non-homogeneous, so it's only a one-sided inverse..

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