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Say $X^{*}$ represents the one-point compactification of a space $X$ that is locally compact, non-compact, and Hausdorff. Also suppose that $Y$ is compact Hausdorff and $Y-\{p\}$, for a single point $p$, is homeomorphic to $X$. So then would $Y$ be homeomorphic to $X^{*}$? I think it is, but I'm not sure how it can be shown.

My attempt so far: Correct me if I'm wrong, but I started with inclusions $i: X \rightarrow X^{*}$ and $j: Y-\{p\} \rightarrow Y$, along with the homeomorphism $h: X \rightarrow Y-\{p\}$. So we need to find a homeomorphism $g: X^{*} \rightarrow Y$. So I tried using a commutative diagram and so far, I've concluded that $g$ (if it is to exist) would have to be continuous since the continuous image of a compact space is compact (I think this follows from $X^{*}$ being compact), but then one of my inclusions, i.e. $i$, can't possibly be continuous, can it (since $X$ is not compact)? How can this be fixed?

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$g$ must extend $h$ and assign $p$ to the 'one point' of $X^*$. The only thing to show is that $g$ and $g^{-1}$ are continious in that point. – Berci Dec 3 '12 at 0:12
up vote 4 down vote accepted

You have the homeomorphism $h:X\to Y\setminus\{p\}$. Let $q$ be the point at infinity in $X^*$, and define

$$\hat h:X^*\to Y:x\mapsto\begin{cases}h(x),&\text{if }x\in X\\p,&\text{if }x=q\;.\end{cases}$$

Let $U$ be an open nbhd of $p$ in $Y$. Then $Y\setminus U$ is compact, so $\hat h^{-1}[Y\setminus U]=h^{-1}[Y\setminus U]$ is compact, and therefore $\hat h^{-1}[U]=X\setminus\hat h^{-1}[Y\setminus U]$ is an open nbhd of $q$. It follows that $\hat h$ is continuous, since it is clearly continuous at each point of $X$.

Now let $U$ be an open nbhd of $q$ in $X^*$. Then $X\setminus U$ is compact, so $\hat h[X\setminus U]=h[X\setminus U]$ is compact in $Y$, and $\hat h[U]=Y\setminus\hat h[X\setminus U]$ is an open nbhd of $p$ in $Y$. If $V\subseteq X$ is open in $X^*$, then $V$ is open in $X$, so $\hat h[V]=h[V]$ is open in $Y\setminus\{p\}$. But $Y\setminus\{p\}$ is open in $Y$, so $\hat h[V]$ is open in $Y$. Thus, $h$ is also open. Clearly $h$ is a bijection, so $h$ is a homeomorphism.

Another approach is to identify $X$ and $Y\setminus\{p\}$ and simply imagine that you have two compact Hausdorff topologies, $\tau_0$ and $\tau_1$, on $X^*=X\cup\{p\}$, where $p$ is some point not in $X$, such that $\langle X^*,\tau_0\rangle$ is the one-point compactification, and $\{U\cap X:U\in\tau_0\}=\{U\cap X:U\in\tau_1\}$. Compactness of $\langle X^*,\tau_1\rangle$ then implies that $\tau_0\subseteq\tau_1$. But it’s well-known that a compact Hausdorff topology is maximal compact: no strictly larger topology is compact. Since $\tau_1$ is a compact topology on $X^*$, it cannot be strictly larger than $\tau_0$, and therefore $\tau_0=\tau_1$.

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Ahhh… The point is to construct a certain continuous map that extends $h$. Can't you simply say that $\hat h$ is continuous from the fact that the continuous image of a compact space is compact? – Libertron Dec 3 '12 at 0:32
@Sachin: No, because the non-continuous image of a compact space can also be compact. – Brian M. Scott Dec 3 '12 at 0:33

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