If $f\in C_0$ and $g$ is continuous, can we show that $f\circ g\in C_0$?

Question

Let $E$ be a locally compact Hausdorff space and $$C_0(E):=\left\{f\in C(E):\left\{|f|\ge\varepsilon\right\}\text{ is compact for all }\varepsilon>0\right\}.$$

Let $f\in C_0(E)$, $E'$ be another locally compact Hausdorff space and $g\in C(E',E)$. Are we able to show $h:=f\circ g\in C_0(E')$?

By continuity, $h\in C(E')$. Now let $\varepsilon>0$. $\left\{|g|\ge\varepsilon\right\}=E'\setminus\left\{|g|<\varepsilon\right\}$ is clearly closed (as the complement of an open set), but is it compact?

Answer 1

The answer is (trivially) yes if $f(y) = 0$ for all $y \in E$, or if $E'$ is compact (since a closed subset of a compact set is compact).

In all other cases the answer is no: Choose any $y_0 \in E$ with $f(y_0) \ne 0$ and define $g: E' \to E$ as the constant function $g(x) = y_0$. Then with $\varepsilon = \frac 12 |f(y_0)|$, $$ \{x \in E' : |f(g(x))| \ge \varepsilon \} = E' $$ is not compact.

So the desired conclusion does not hold in general because for continuous functions, the preimage of a compact set is not necessarily compact.