Let $U$ be an open subset of $\mathbb{R}^n$, $G$ a Lie group and $f:U\rightarrow G$ a smooth surjective map.
Under which conditions there is a smooth function $\phi:U\rightarrow\mathfrak{g}$ such that $f=\exp(\phi)$?
A necessary condition is for the exponential map to be surjective, which is the case if $G$ is compact and connected. But is surjectivity of the exponential map also a sufficient condition? Is it possible to say something generally if the domain of $f$ is some smooth manifold?
EDIT: As Jason pointed out in the comments, for the problem as originally stated surjectivity of exp is not needed, so I added the requirement, which I had in mind, of $f$ being surjective. To address his second point, I was wondering wether the domain of $f$ having some non trivial topological structure (e.g. a sphere or projective plane) makes any difference in the answer.