Let $f : A \to B$ be a based fibration. Suppose $\Omega f$ has a section $\Gamma$. How do I show that $\mu \circ (\Gamma \times \Omega i) : \Omega B \times \Omega \mathsf{fib}(f) \to \Omega A$, where $i$ is the inclusion of the fiber and $\mu$ loop composition, is an equivalence?

I tried constructing an inverse $(f \times 0) \circ \Delta$ but to show that this is a section it seems that I need to also know that $\Gamma$ is a retraction. I don't think I have the right map but I can't think of any other one.