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Let $X$ be a locally compact Hausdorff space.

A compactification $c(X)$ of $X$ is a compact Hausdorff space $c(X)$ paired with a map $i:X\to c(X)$ such that $i$ is an embedding and $i(X)\subseteq c(X)$ is dense.

A finite point compactification is a compactification $c(X)$ such that $c(X)-X$ is finite. In particular if $|c(X)-X|=n$, then $c(X)$ is called n-point compactification

A compactification map is a continuous quotient map $\pi:c_1(X)\to c_2(X)$ with $\pi|X=id_X$ (restriction to $X$) and $\pi(c(X)-X)=c_2(X)-X$.

I am trying to see two things.

  1. If $c(X)$ is compactification that is not finite point and if $\hat{X}$ is an n point compactification, then there is compactification map $\pi:c(X)\to \hat{X}$

  2. If $c_1(X)$ is an n-point compactification and $c_2(X)$ m-point compactification with $n>m$, then there is compactification map $\pi:c_1(X)\to c_2(X)$

In short, when X is locally compact, there is always a unique compactification to 1 point compactification from every (bigger) compactification, can this property extended to other finite point compactifications.

talekhine
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1 Answers1

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Both are false:

#1 is false:
$\beta \mathbb N \setminus \mathbb N$ contains a countable infinite, discrete subset $E$. Let $f: E \rightarrow [0,1] \cap \mathbb Q$ be a surjective map. By Gillman, Jerison, Rings of continuous functions, 6O, $E$ is $C^\star$-embedded in $\beta \mathbb N$. Hence, there is also a continuous extension $f^\prime: \beta \mathbb N \setminus \mathbb N \rightarrow [0,1]$ of $f$. Hence, $f^\prime$ is surjective. By Engelking, General topology, 3.5.13, $\mathbb N$ has a compactification $c(\mathbb N)$ with remainder homeomorphic to $[0, 1]$. Since this is connected, there is no compactification map to a 2-point compactification of $\mathbb N$.

#2 is false:
Let $X = (0, 2) \setminus \{1\}$. Then $[0,2]$ is a connected, 3-point compactification of $X$. Moreover, $X$ is homeomorphic to $\mathbb R \oplus \mathbb R$. Hence, $S^1 \oplus S^1$ is a 2-point compactification of $X$. Since the latter is not connected, there is no compactification map $[0,2] \rightarrow S^1 \oplus S^1$.

Addendum to #1
Here is an explicit construction of a compactification of $\mathbb N$ with remainder $[0,1]$:

Let $(F_n)_{n \in \mathbb N}$ be an increasing sequence of finite sets such that $ \bigcup_{n \in \mathbb N} F_n = [0,1] \cap \mathbb Q$.
$D = \bigcup_{n \in \mathbb N} (F_n \times \{\frac{1}{n}\}) \subset \mathbb R^2$ with the Euclidean topology is countable infinite and discrete, hence homeomorphic to $\mathbb N$. Moreover, $\overline{D} = D \cup ([0,1] \times \{0\})$ is closed, bounded, hence compact. Thus, $\overline{D}$ is a compactification of $D$ with remainder $[0,1] \times \{0\}$.

Ulli
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