Question regarding one-point compactification and homeomorphisms

Solution 1:

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 $\tau_0=\tau_1$.