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

Question

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.

Answer 1

Both are true actually. In a neighborhood of the identity both maps define local diffeomorphisms from the tangent space at the identity element to the Lie group. Notice that the maps are different: the same element of the group is determined by two different sets of values. One speaks of coordinates of first and second type. The proof relies on the inverse function theorem: in both cases the differential of the map at the origin of the tangent space is non singular and then the map is a local diffeomorphism.