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

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.

   • 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.

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.

@Qmechanic: I belive there are problems with Baker's discussion of the surjectivityy in "Matrix groups". Here is a quote from Jean Gallier and Jocelyn Quaintance's lecture notes at U Penn: (http://www.seas.upenn.edu/~jean/diffgeom.pdf)

"We warn our readers about Chapter 6 of Baker’s book [16]. Indeed, this chapter is seriously flawed.The main two Theorems (Theorem 6.9 and Theorem 6.10) are false, and as consequence, the proof of Theorem 6.11 is wrong too. Theorem 6.11 states that the exponential map exp: so(n,1) → SO0(n,1) is surjective, which is correct, but known proofs are nontrivial and quite lengthy (see Section 6.2). The proof of Theorem 6.12 is also false, although the theorem itself is correct (this is our Theorem 6.17, see Section 6.2). The main problem with Theorem 6.9 (in Baker) is that the existence of the normal form for matrices in SO0(n,1) claimed by this theorem is unfortunately false on several accounts. Firstly, it would imply that every matrix in SO0(n, 1) can be diagonalized, but this is false for n ≥ 2. Secondly, even if a matrix A ∈ SO0(n,1) is diagonalizable as A = PDP−1, Theorem 6.9 (and Theorem 6.10) miss some possible eigenvalues and the matrix P is not necessarily in SO0(n,1) (as the case n = 1 already shows). For a thorough analysis of the eigenvalues of Lorentz isometries (and much more), one should consult Riesz [146] (Chapter III)".