The following result seems to be well-known.
If $\mathfrak{g}$ is a real finite-dimensional solvable Lie algebra and $G$ is the corresponding connected simply connected Lie group, then $G$ is diffeomorphic to the Euclidean space via the canonical coordinates of the second kind.
I think that I am misunderstanding this result.
Consider the solvable three-dimensional Lie algebra (Bianchi $VII_\beta$ with $\beta = 0$, reference) whose commutators satisfy $$ [e_1,e_3] = - e_2, \quad [e_2,e_3] = e_1, \quad [e_1, e_2] = 0. $$ We can represent this algebra as the matrix algebra spanned by the matrices $$ e_1 = \begin{pmatrix} 0 & 0& 1\\ 0 & 0 & 0\\ 0 & 0 &0 \end{pmatrix}, \quad e_2 = \begin{pmatrix} 0 & 0& 0\\ 0 & 0 & 1\\ 0 & 0 &0 \end{pmatrix}, \quad e_3 = \begin{pmatrix} 0 & -1& 0\\ 1 & 0 & 0\\ 0 & 0 &0 \end{pmatrix}. $$ However, the canonical coordinates of the second kind $$ \mathbb{R}^3 \ni x \mapsto \text{Exp}(x_1e_1) \, \text{Exp}(x_2e_2) \,\text{Exp}(x_3e_3) = \begin{pmatrix} \cos x_3 & -\sin x_3 & x_1\\ \sin x_3 & \cos x_3 & x_2\\ 0 & 0 & 1 \end{pmatrix} $$ do not form a diffeomorphism, as the map is not even injective.
Question: Where does this example break down?