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?
Best Answer
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..