TITLE: Exponential of Lie Groups.
QUESTION [6 upvotes]: When the exponential map  defined  a bijection between the group G and their Lie algebra? The only example  I know is the Heisenberg group.

REPLY [5 votes]: There is a really cool example that I love, given in a paper by Arsigny, et al. which shows that the symmetric positive definite matrices can be given a Lie group structure, and this is one such example where the exponential map is a bijection from the Lie group to its Lie algebra.  Let $SPD(n)$ be the manifold of symmetric positive definite matrices.  This can be realized as a manifold both as an open subset of the symmetric matrices $Symm(n)$, and also as a homogeneous space $SPD(n) \approx GL_n(\mathbb{R})/O(n)$ under the action $GL_n(\mathbb{R}) \times SPD(n) \to SPD(n)$ given by $(A, P) \to APA^T$.  Using the spectral theorem we can diagonalize any $P \in SPD(n)$ as $P = QDQ^T$ with $D$ diagonal with positive elements and $Q \in O(n)$.  Furthermore, the tangent space $T_PSPD(n) \cong Symm(n)$ at each point, hence the tangent vectors are symmetric matrices.  The matrix exponential map $\exp:Symm(n) \to SPD(n)$ given by the Taylor series has the convenient computational property that 
$$
\exp(P) \;\; =\;\; Q\exp(D) Q^T
$$
and similarly the logarithm map $\log:SPD(n) \to Symm(n)$ defined by the usual Taylor expansion is defined for all matrices in $SPD(n)$ and serves as an inverse for $\exp$.
Now, the part that Arsigny and his collaborators introduced in their paper is how to turn $SPD(n)$ into a Lie group.  This can be done by introducing the binary operation $\odot: SPD(n) \times SPD(n) \to SPD(n)$ given by 
$$
A\odot B \;\; =\;\; \exp(\log(A) + \log(B)).
$$
One can easily show that this operation is smooth and satisfies all of the conditions to turn $(SPD(n), \odot)$ into a Lie group.  What's also exciting too is that in the paper, the authors introduce a scalar multiplication that turns $SPD(n)$ into a vector space.
Update 9/15/2019
I wanted to update this example with a couple more features that will make this analysis a little more complete.  

Notice that the Lie group product turns $SPD(n)$ into a commutative Lie group.  This implies that the Lie algebra $Symm(n)$ is commutative, and rightfully so the bracket $[\cdot, \cdot]: Symm(n) \times Symm(n) \to Symm(n)$ is trivial.  This fact is correctly reflected in the Campbell-Baker-Hausdorff formula:

$$
\log\left (\exp(A) \odot \exp(B)\right ) \;\; =\;\; A+B.
$$

Observe now that the Lie group structure $(SPD(n), \odot)$ is really just the pushforward of vector addition on $Symm(n)$ via $\exp$, and this is afforded to us by the fact that $\exp$ is a global diffeomorphism.  We can extend this to a vector space structure on $SPD(n)$ by also pushing forward the scalar multiplication.  Arsigny et al. defined the operation $\circledast:\mathbb{R} \times SPD(n) \to SPD(n)$ given by 
$$
\lambda\circledast P \;\; =\;\; \exp \left (\lambda \log(P)\right ).
$$

One can show that $\circledast$ satisfies the distributive properties of scalar multiplication and in fact serves as a scalar multiplication on $SPD(n)$.