Closed image of locally compact space If $X$ is locally compact and $f : X \rightarrow Y$ is continuous closed and surjective, must $Y$ be locally compact? This seems like it should be relatively simply to answer, but I am unable to find either a proof or a counterexample. Any ideas?
 A: 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.  $\Box$
