Let $G$ be a connected Lie group and suppose that its Lie algebra ${\frak g}$ splits into a direct sum of ideals $${\frak g} = {\frak a}\oplus{\frak b}.$$ Let $A$ be the connected Lie subgroup of $G$ with Lie algebra ${\frak a}$ and $B$ the one with Lie algebra ${\frak b}$. Let $\tilde{A}$ and $\tilde{B}$ be the universal covering groups of $A$ and $B$, respectively.
Then, $\tilde{A}\times\tilde{B}$ is a simply connected Lie group with Lie algebra ${\frak g}$, so it is isomorphic to the universal covering group $\tilde{G}$ of $G$. Hence, we have a covering homomorphism $$\varphi:\tilde{A}\times\tilde{B}\longrightarrow G.$$
Question: Do we have that $\varphi(\tilde{A}\times\{1\})=A$ and $\varphi(\{1\}\times B)=B$?
This would imply that $G=AB$ since $\varphi$ is a surjective group homomorphism. Hence, if this is true we would have that:
Proposition. If ${\frak g}={\frak a}\oplus{\frak b}$ then $G=AB$.
Is that true?
What about $G=(S^1)^2$, $S^1$ the Lie group of complex number of modulus 1. Let $A=(u,1)$ with $u\in S^1$, $B= (v^n,v^m)$ where $(n,m)=1$, $v\in S^1$. The kernel of the map $A\times B$ to $G$ is the finite group $\bf Z/mZ$ $(r^{-n}, (r^n,1))$, $r$ a $m-$th root of $1$.