Suppose $A$ is a $G$-space and we let $\pi:A \rightarrow A/G$ be a principle $G$ bundle with cross section that is continuous $s:A/G \rightarrow A$. That means we have $\pi \circ s = id_{A/G}$. I need to show that $\pi:A \rightarrow A/G$ is a trivial bundle.
Do I have to find a homeomorphism $h$ such that $h:A \rightarrow A/G \times F$ where $F$ is the fibre? How can I construct the map $h$? Or I have to show that $A=A/G \times F$?