# Projection of fiber bundle is a submersion

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.

• You convinced me. – Amitai Yuval Feb 7 '15 at 17:52

Yes your proof is correct. You can formulate it a little bit shorter as follows: a map is a submersion if and only if there exists local charts such that the representation of the map in this charts is a projection (one direction is very easy, the converse follows from the inverse function theorem). Now for the projection of a fibre bundle, we just take a local trivialization $\pi^{-1}(U) \to U \times F$ and observe that the local representation of the bundle projection is the projection on the first factor. Essentially, this boils down to your argument.