Closed image of locally compact space

Solution 1:

Here is a counterexample.

First note that $\mathbb{R}$ is locally compact.

Consider the quotient space $Y = \mathbb{R} / \mathbb{N}$ (i.e., identifying all natural numbers to a point $*$). Note that the quotient mapping $f : \mathbb{R} \to Y$ is closed (and continuous). (This essentially follows because we are identifying a discrete subset of $\mathbb{R}$.)

Claim: $Y$ is not locally compact.

proof: If $U$ is a neighbourhood of $*$, we may without loss of generality assume that it is of the form $$U = \bigcup_{n\in \mathbb{N}} ( (n-\varepsilon_n , n + \varepsilon_n ) \setminus \{ n \} ) \cup \{ * \},$$ where $\varepsilon_n < \frac{1}{4}$ for all $n$. It follows that $$\overline{U} = \bigcup_{n\in \mathbb{N}} ( [n-\varepsilon_n , n + \varepsilon_n ] \setminus \{ n \} ) \cup \{ * \}.$$ For each $m \in \mathbb{N}$ define the open set $V_m$ to be $$\left( \bigcup_{n < m} ( ( n - \varepsilon_n - \frac{1}{4} , n + \varepsilon_n + \frac{1}{4} ) \setminus \{ n \} ) \right) \cup \left( \bigcup_{n > M} ( (n-\varepsilon_n , n + \varepsilon_n ) \setminus \{ n \} ) \right) \cup \{ * \}.$$ It is clear that $\{ V_m : m \in \mathbb{N} \}$ is an open cover of $\overline{U}$, however it has no finite subcover, and so $\overline{U}$ is not compact. $\Box$