The exponential map associated with the ()  connection on a Lie group is generally not surjective. This is because, for this connection, the oneparameter subgroups and geodesics coincide. If we drop the requirement that oneparameter subgroups are geodesics, is it possible to put a left invariant connection on a Lie group for which the geodesic exponential map is surjective?
1 Answer
Yes. Just take the LeviCivita connection of any leftinvariant Riemannian metric on the Lie group. The metric is complete, so any two points can be joined by a geodesic (HopfRinow). Thus, the geodesic exponential map of that connection starting from the identity is surjective.

$\begingroup$ Thanks for the answer, Robert. One implication of this is that the $(0)$connection on a noncompact Lie group will generally not be metric. $\endgroup$ Jul 29, 2018 at 23:15

3$\begingroup$ True, its holonomy is noncompact, so it is not the LeviCivita connection of any Riemannian metric. On the other hand if the group is semisimple, then the (0)connection does preserve the biinvariant pseudoRiemannian metric. $\endgroup$ Jul 29, 2018 at 23:18