If a principal bundle has a global section, then it is the trivial bundle

Question

Let be $(P, M, \pi, G)$ a principal G-bundle. Suppose that there is a section $\sigma: M \to P$. I want to prove that $P \cong M \times G$. I built $f: M\times G \to P$ such that $f(p, g) = \sigma(p) \cdot g$ where $\cdot$ is the right action on the bundle. I can prove that this is smooth and bijective, but I cannot prove that this is a diffeomorphism. My idea is to use the global rank theorem, but I can prove only that the map has constant rank on each fiber, where the action is transitive. How can I prove that the rank is the same everywhere?

Answer 1

You have to use the fact that $P$ is locally trivial. To check that $f$ is a diffeomorphism, we can work locally over $M$, and thus assume that $P$ is actually the trivial bundle $M\times G$. Our smooth section $\sigma$ then has the form $\sigma(p)=(p,\tau(p))$ for some smooth function $\tau:M\to G$, so then our putative diffeomorphism is the function $f:M\times G\to M\times G$ given by $f(p,g)=(p,\tau(p)\cdot g)$. You can then think about the rank of this map as you suggest, but the easiest way to see it is a diffeomorphism is to just write down its inverse: $f^{-1}(p,g)=(p,\tau(p)^{-1}\cdot g)$. This inverse function is smooth since $\tau$ and the inverse map on $G$ are both smooth.