Quotient Map From Compactification to One-Point Compactification Let $X$ be a Hausdorff and locally compact space, and let $Y = X \cup \{\infty\}$ denote its one-point compactification. Let $Z$ be any Hausdorff compactification of $X$. I want to show the following:

The map $\pi: Z \to Y$ that acts as the identity on $X$ and maps any
other element to $\infty$, is a quotient map.

By Section 22 of Munkres, we know that a quotient map $\pi$ must be surjective, and satisfy $U$ is open in $Y$ if and only if $\pi^{-1}(U)$ is open in $Z$. The surjective property of $\pi$ can be verified immediately. It is known that open sets in $Y$ are either open sets of $X$ or of the form $\{\infty\} \cup (X-K)$ (with $K$ compact). For sets $U$ in $X$, we know that $U$ is open in $X$ if and only if $\pi^{-1}(U) = U$ is open in $X$.
My concern arises when dealing with sets of the form $\{\infty\} \cup (X-K)$ (with $K$ compact). I know that $$\pi^{-1}(\{\infty\} \cup (X-K)) = \pi^{-1}(\{\infty\}) \cup (\pi^{-1}(X)\cap\pi^{-1}(K^c)) = \pi^{-1}(\{\infty\}) \cup (X\cap\pi^{-1}(K^c)).$$
We know that $K^c$ is open as $K$ is closed in $X$. If we can show that $\pi^{-1}(\{\infty\})$ is open, then $\pi^{-1}(\{\infty\} \cup (X-K))$ will also be open. Is my thought process correct? If so, how can I verify that $\pi^{-1}(\{\infty\})$ is open?
 A: You certainly consider a non-compact locally compact Hausdorff space.
You cannot expect that $\pi^{-1}(\infty)$ is open. As a counterexample consider $X = \mathbb R$ and the extended real line $Z = \mathbb R \cup \{-\infty, +\infty\}$ which is a two-point compactification of $X$. Thus your approach does not work.
Here comes a proof. Let us first show the following
Lemma. If $X$ is a dense subset of a space $Z$, then for each open $U \subset Z$ we have $\overline U = \overline{U \cap X}$.
Proof. $\overline{U \cap X} \subset \overline{U}$ is trivial. To prove that $\overline U \subset \overline{U \cap X}$, consider $x \in \overline U$. This means that for each open neigborhood $V$ of $x$ in $Z$ we have $V \cap U \ne \emptyset$. Since $X$ is dense and $V \cap U$ is open, we see that $V \cap U \cap X \ne \emptyset$. Therefore $x \in \overline{U \cap X}$.
Let us next show that a locally compact Hausdorff $X$ is open in any Hausdorff compactification $Z$:
For each $x \in X$ there exists an open subset $U \subset X$ with $x \in U$ whose closure $C$ in $X$ is compact. Since $C$ is compact, it is a closed subspace of $Z$. Let $\overline{U}^Z$ be the closure in $Z$. Since $U \subset C$, we get $\overline{U}^Z \subset C \subset X$. Let $W \subset Z$ be open such that $W \cap X = U$. The Lemma shows that $x \in W \in  \overline{W}^Z = \overline{W \cap X}^Z = \overline{U}^Z \subset X$. In other words, each $x \in X$ has an open neighborhood in $Z$ which is contained in $X$. Thus $X$ is open in $Z$.
We now prove that $\pi$ is a quotient map.

*

*$\pi$ is continuous.
Let $U \subset X^+ = X \cup \{\infty\}$ be open. If $U \subset X$, then $\pi^{-1}(U) = U$ which is open in $Z$. If $U = X^+ \setminus K$, then $\pi^{-1}(U) = Z \setminus K$ which is also open in $Z$.


*$\pi$ is a quotient map.
Let $U \subset X^+$ be a set such that $\pi^{-1}(U)$ is open in $Z$. If $U \subset X$, then $\pi^{-1}(U) = U \subset X$. Thus $U$ is open in $X$ and therefore open in $X^+$. If $\infty \in U$, then $Z \setminus X \subset \pi^{-1}(U)$. Thus $K = Z \setminus \pi^{-1}(U)$ is contained in $X$ and closed in $Z$, hence compact. We conclude that $U = \pi(\pi^{-1}(U)) = \pi(Z \setminus K) = X^+ \setminus K$ is open in $X^+$.
