Let $P$ be any smooth manifold (if it helps, I'm happy to set $P = S^1$ but I'm sure it doesn't matter). I'm working with a continuous map $F:[0,1] \times P \to [0,1] \times P$ which takes $(t, b)$ to $(f(t), b)$ ($f$ is some continuous function into the interval [0,1]. I know a little bit about this map: I know that it is invertible (with inverse $F^{-1}: [0,1] \times P \to [0,1] \times P$) and I know in particular that when you fix $p \in P$ then the map $F^{-1}(\cdot, p): [0,1] \to [0,1]$ is continuous.
I just want to show that the inverse $F^{-1}: [0,1] \times P \to [0,1] \times P$ is continuous! I do strongly believe it is true but it is a bit of a struggle to show it explicitly. If anyone has any counterexamples I'm happy to consider it!
If $P=S^1$ or any other compact space then the domain $[0,1]\times P$ is compact. Therefore $F$ is closed and so the inverse is continuous regardless of the definition of $F$. In other words: continuous bijections from compact spaces are homeomorphisms.
Another approach is this: your function $F=f\times\text{id}$ where $f:[0,1]\to[0,1]$ and $\text{id}:P\to P$ is the identity. Obviously $\text{id}$ is a homeomorphism and so is $f$ because it has to be invertible and the domain $[0,1]$ is compact. And the product of homeomorphisms is a homeomorphism.