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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.

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    $\begingroup$ Since the exponential map of the group is locally a diffeo, up to replacing $U$ by a smaller open set, always. $\endgroup$ – Mariano Suárez-Álvarez Dec 7 '15 at 17:11
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    $\begingroup$ It's not necessary, for $\exp$ to be surjective. For example, if $f$ maps everything to the identity in $G$, then $\phi$ mapping everything to $0$ works, regardless of whether or not $\exp$ is surjective. In addition, the domain of $f$ is an open subset of $\mathbb{R}^n$, so is a smooth manifold. So I don't understand your last sentence. $\endgroup$ – Jason DeVito Dec 7 '15 at 17:18

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