**Please excuse, very naive question:**

Suppose $g$ is a topological Lie algebra over **Q** and $G$ = $exp(g)$ the associated group

(take free group on formal symbols $exp(X)$, X $\in$ $G$, and impose all relations formally coming from the BCH-formula).

Suppose I have a short exact sequence

$0\longrightarrow b_{1}\cap b_{2}\longrightarrow b_{1}\oplus b_{2}\longrightarrow g\longrightarrow 0$

of g-modules, but special in the sense that

- $g$ is the full Lie algebra,
- $b_1$, $b_2$ (and then their intersection) are supposed to be proper Lie
*ideals*(and not just any kind of g-module) - and $b_1 + b_2 = g$ ; this can happen if $g$ is weird enough

It seems to me (in some cases)/(always)/(never? ;-) such a sequence should induce something like an exact sequence

1 -> A -> B -> G -> 1

(exact in the obvious classical sense, clearly groups are not an abelian category...)

where B could be something like the coproduct/free product of the normal subgroups associated to the Lie ideals b1, b2; and A the subgroup associated to the intersection of b1 and b2.

Is that true or is it complete nonsense? Is it trivially totally ridiculously false?

[please note, even though it may sound so, I do *not* want to go in
the direction of 'integrating' g-modules to G-modules, I would like to
transfer Lie algebra decompositions to 'nonlinear' group 'decompositions',
whatever that means....]