# Submersion admitting a complete Ehresmann connection is a fiber bundle

In the third paragraph of this paper it's claimed that a submersion admitting a complete Ehresmann connection is a fiber bundle. The proof sketch says to reduce to the base space an open ball in Euclidean space, and then perform parallel transport along radial segments.

Since the connection is complete, any radial segment lifts to a curve with horizontal velocity field. Flowing along the velocity field gives parallel transport along a single radial segment. To construct a trivialization however, I need a global horizontal field (flowing along it is the trivialization). I don't see how to combine the individual horizontal fields over radial segments into a horizontal vector field over the entire ball.

I'm sure the underlying principle here is smooth dependence of ODE on parameters, but I'm struggling to write things down. The only ODE that comes to mind is that of parallel transport given by the covariant derivative induced by the Ehresmann connection, but I just can't compile things.

So we have a submersion $$\pi:M\to B$$, where $$B$$ is an open ball in $$\mathbb{R}^n$$. Write $$F:=\pi^{-1}(0)$$. Then $$F$$ is a smooth submanifold of $$M$$. For $$x\in B$$, let $$P_x:F\to\pi^{-1}(x)\subset M$$ denote parallel transport along the radial line segment connecting the origin with $$x$$. Define $$\Psi:F\times B\to M,\qquad(q,x)\mapsto P_x(q).$$ Then $$\Psi$$ is the desired trivialization, and the proof of this is contained in your post.
• I don't see how to actually apply smooth dependence on parameters to get smoothness of $\Psi$.. – Arrow May 1 at 21:13
• @Arrow Smoothness of $\Psi$ follows from smooth dependence on parameters once you write all the data in coordinates. In short, the line segment connecting the origin with $x$ is generated by the constant vector field $x$. The horizontal lift of this vector field depends smoothly on $x$, and hence so does the associated flow. – Amitai Yuval May 1 at 21:17