special relativity - Is there a general theorem stating why the restricted Lorentz group's exponential map is surjective?

The exponential map for the restricted Lorentz group is surjective. An outline of why is shown on the wiki page Representation Theory of the Lorentz Group.

Is there a more general theorem that states that for some class of Lie groups or Riemannian manifolds (which includes the restricted Lorentz group), the exponential map is surjective?

There is a theorem stating that compact, connected Lie groups have surjective exponential maps. But as the restricted Lorentz group is not compact, this isn't applicable.

Answer

Comments to the question (v2):

1. 
   

   The consensus in the literature seems to be that the surjectivity of the exponential map $$\tag{1}\exp: so(1,d;\mathbb{R}) \to SO^+(1,d;\mathbb{R})$$ for the restricted Lorentz group for general spacetime dimensions $D=d+1$ does not have a short proof.

   
   
   
   
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     The case $d=1$ is trivial.

     
     
   
   
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     The case $d=2$ can be proved via the isomorphism $SO^+(1,2;\mathbb{R})\cong SL(2,\mathbb{R})/\mathbb{Z}_2$, cf. e.g. this Phys.SE post.

     
     
     
   
   
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     The case $d=3$ can be proved via the isomorphism $SO^+(1,3;\mathbb{R})\cong SL(2,\mathbb{C})/\mathbb{Z}_2$, cf. e.g. Wikipedia and this Phys.SE post.

     
     
   
   
   
   


2. 
   

   Already the exponential map $\exp: sl(2,\mathbb{R}) \to SL(2,\mathbb{R})$ is not surjective, cf. e.g. this MO.SE answer and this Phys.SE post. Note that the Lie algebras $$\tag{2}so(1,2;\mathbb{R}) ~\cong~ sl(2,\mathbb{R})$$ are isomorphic, but only the Lie group $SO^+(1,3;\mathbb{R})$ for the lefthand-side of the isomorphism (2) has a surjective exponential map; not the Lie group $SL(2,\mathbb{R})$ for the righthand-side. A counterexample such as (2) undoubtedly makes it delicate to try to formulate a generalization of (1) beyond the restricted Lorentz groups $SO^+(1,d;\mathbb{R})$ and case-by-case-proofs. See also this Math.SE post.