It is known that the exponential map is surjective for compact, connected Lie groups. This is generally proved through the introduction of some bi-invariant metric etc. etc. My question is related to the following: Can this be shown using only Lie theory and not deferring to other concepts like Riemannian geometry, etc.? In particular, can I use some properties of the Lie algebra of compact Lie groups? Details:
Let $G$ be a compact, connected Lie group. We know $\exp:\mathfrak{g}\to G$ is a local homeomorphism. So I can find a subset of the Lie algebra, $S$, containing $0$ such that $\exp{S}\subset G$ is open in $G$. Now, consider an open subset of $\exp{S}$, $U$, with the property that $U^{-1}:=\{g^{-1}:g\in U\}=U$ and $e\in U$. Then, with a bit of work, you can show $\exists n\in\mathbb{N}$ such that $G=U^n$ (length $n$ words in the elements of $U$). This fact is equivalent to compactness (finite subcover).
So, given $g\in G, \exists g_1,...,g_n\in U$ such that $g=g_1g_2...g_n$. By local homeomorphism property of $\exp$, $\exists x_1,...,x_n \in \exp^{-1}{(U)}\subset S\subset \mathfrak{g}$ such that $e^{x_i}=g_i \forall i$. Therefore, $g=e^{x_1}...e^{x_n}=e^{\Omega_n(x_1,...,x_n)}$ where $\Omega_n$ denotes the Baker-Campbell-Hausdorff formula of $n$ elements. In particular, $$\Omega_2(A,B):=A+B+\frac{1}{2}[A,B]+\frac{1}{12}[A,[A,B]]+...$$
However, if $\exp$ is to be surjective, $\Omega_n$ must converge for all $x_1,...,x_n$ in $\exp^{-1}{(U)}$. But in order to prove surjectivity, is there any way I can use the properties of Lie algebras of compact Lie groups to show convergence of the BCH formula in this set?
Please let me know if this question is ill-posed/ill-formed. Thanks!