Is it true that every locally trivial fibration(fiber bundle) in smooth category admits an Ehresmann connection? The converse is obviously true, and it's one of the conclusion of the Ehresmann's theorem.
Suppose $(U_\alpha \subset M, \Phi_\alpha : \pi^{-1}(U_\alpha) \to U_\alpha \times F)$ be a trivialization of the fiber bundle $(E, \pi, M)$. Then, for each $U_\alpha$, we could give a natural connection(splitting) of $\mathrm{T} E|U_\alpha$ by $$ \mathrm{H} E|U_\alpha:= \{ \partial_t|_{t=0}\Phi^{-1}_\alpha(\gamma(t),x) \mid x \in M \} $$ However, there always be a gluing problem. Since transition map doesn't preserve the horizontal splitting, we couldn't construct the Ehresmann connection in this way.
So, here is the question. Are there a canonical way to construct an Ehresmann connection from smooth fibration? If we are dealing with vector bundle, then we could canonically gives a metric, and we could choose a compatible connection structure. So, it's done. However, if the fiber is an arbitrary manifold $F$, then what would be happen? Are there exists an obstruction?
Given a surjective submersion $\pi\colon E\to B$, you have the vertical subbundle $VE:=\ker T\pi\subset TE$. Now, if you pick a Riemannian metric $g$ on $E$, then $H:=(VE)^\perp$, the orthogonal complement of $VE$ with respect to $g$, will give you an Ehresmann connection on $E$