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Suppose $G$ is a Lie group and $\mathfrak{g}$ its Lie algebra. It is not so difficult to see that if $G$ is abelian and connected then $\exp:\mathfrak{g}\rightarrow G$ is a universal covering map. What if $G$ is non-abelian? Is there a characterization of when $\exp$ is a universal covering map?

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  • $\begingroup$ About surjectivity of $\exp$ see here. More can be found in Terry Tao's blog. $\endgroup$ – Dietrich Burde Apr 17 '19 at 8:05
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First of all, you should assume that $G$ is connected, since otherwise $\exp$ cannot be surjective. Then a complete characterization of Lie connected groups for which $\exp$ is a covering map is given by YCor in his answer here:

$\exp$ is a covering map if and only if ${\mathfrak g}$ is solvable and does not contain two particular Lie subalgebras ${\mathfrak e}$, $\tilde{\mathfrak e}$.

The reduction to Yves' answer is easy: $\exp: {\mathfrak g}\to G$ is a covering map if and only if $\exp: {\mathfrak g}\to \tilde{G}$ is a diffeomorphism, where $\tilde{G}$ is the universal covering group of $G$.

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