sorry for my bad english.
I am in a proof and I get stuck in the following step.
Let $f : X \to Y$ be a continuous function between two topological spaces, $G$ a locally compact topological group, and $\phi : G \times X \to X$ a continuous action on $X$. If $g \in G$, $K \subset X$ is a compact set and $U \subset Y$ is an open set such that $f(gK) \subset U$ then I need to prove that exists an open set $Q$ such that $g \in Q \subset \overline{Q}$, $\overline{Q}$ is a compact set and $f(Q K) \subset U$.
What I have tried is to use the fact that $G$ is locally compact, then for every neighborhood $V$ with $g \in V$ exists an open set $O$ with $g \in O \subset \overline{O} \subset V$ but I don't know how to use it in my problem.
Do you believe $f(gK)$ is compact? Can you cover that image with a finite set of open sets? Can you pull those back through $f \circ \phi$ to $G$? Does the union of those pullbacks produce an open set containing $g$? Can you finish from there?