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):
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.
The case $d=1$ is trivial.
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.
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.
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.
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