Factoring a the exponential form of a group element of a Lie group, using subgroups

I am working on non linear realization of Goldstone bosons, as is done by Weinberg in section 19.6 of Quantum theory of fields, volume II.

We have a real, compact and connected Lie group $$G$$ with as subgroup $$H$$. Let $$t_a$$ be the generators of $$H$$, and $$x_i$$ be the generators left over (broken generators) such that they together span the Lie group of $$G$$, $$\mathfrak g$$. A general element of $$G$$ can then, to my understanding, be written $$g = \exp\{i(\xi_i x_i + \theta_a t_a)\}.$$ However, Weinberg claims we can write (eq 19.6.12) $$g = \exp\{i \xi_i' x_i \} \exp\{i\theta_a' t_a\}.$$ This is also used in other sources that tackles the same material. Why is this true? It seems plausible to me, by looking at $$SO(3)$$ as an example, however I have not seen a proof nor any real justification for this form.

Edit: So I investigated the origin of the claim, which as Cosmas states in the comments is this paper by CWZ. The statement here is qualified as "in some neighbourhood of $$G$$, any element $$g \in G$$ can be uniquely decomposed as $$g = \exp(i \xi_i xi)\exp(i \theta_a t_a)$$. This is a weaker statement, and seems straight forward from the inverse function theorem.

• The statement entered the mainstream of physics in CWZ. Commented Nov 15, 2021 at 23:19
• You are asking for a proof the the CBH map $(\xi',\theta ')\mapsto (\xi,\theta)$ is invertible, right? Commented Nov 15, 2021 at 23:39
• Or that $(\xi', \theta') \rightarrow G$ is surjective, I think. Commented Nov 16, 2021 at 6:38
• Right. But this is a strictly math question, best covered in the MSE. If you've got the stomach for it, you ought to go through Fulton & Harris. Commented Nov 16, 2021 at 14:21

• Is this enough to conclude that any element of $G$ can be written in this way? Commented Nov 15, 2021 at 21:39
• Well, that is not strong enough. Weinberg claims "Because $t_a$ and $x_i$ span the Lie algebra of $G$, any finite element of $G$ may be expressed in the form $g = \exp(i \xi_i x_i)\exp(i \theta_a t_a)$". Are you saying this is not true? As far as i can tell, this is necessary for the construction of the realization of the Goldstone bosons. Commented Nov 15, 2021 at 22:01