I'm just wondering about my proof for the following fact. I get the feeling it is almost trivial but I am still getting a feel for geometry and so it doesn't seem 'obvious' to me just yet.
The projection $\pi$ of the fiber bundle $(E,\pi,M,F)$ is a submersion. Is the below argument a correct proof of this fact? Would it be considered 'the obvious' proof?
To show $\pi$ is a submersion, we must show that for each $p\in E$, the differential $d\pi_p:T_pE\rightarrow T_{\pi(p)}M$ is surjective.
Since, $(E,\pi,M,F)$ is a fiber bundle, there exists a neighbourhood $U\subseteq M$ containing $\pi(p)$ and diffeomorphism $\phi:\pi^{-1}(U)\rightarrow U\times F$ , such that $\pi|_{\pi^{-1}(U)}=\pi_1\circ \phi:\pi^{-1}(U)\rightarrow U$. Thus, showing $d\pi_p$ is surjective is equivalent to showing $d(\pi_1\circ \phi)_p={d\pi_1}_{\phi(p)}\circ d\phi_p$ is surjective. Since $\phi$ is a diffeomorphism, $d\phi_{p}$ is an isomorphism, so it is surjective. Furthermore, ${d\pi_1}_{\phi(p)}$ is surjective since projection onto the first factor is a submersion. Since the composition of two surjections is a surjection itself, it follows that ${d\pi_1}_{\phi(p)}\circ d\phi_p=d\pi_p$ is surjective. Thus, $\pi$ is a submersion.
