The exponential map on a Lie algebra which is a direct sum

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I am asking the following question as I am trying to better understand Lie theory and the exponential map

Assume $\mathfrak{g}$ is a (finite-dimensional) Lie algebra of the (matrix) Lie group $G\subset GL_{n}(\mathbb{R})$. Say we write $\mathfrak{g}=\bigoplus_{i=1}^{k}\mathfrak{g_i}$ as a direct sum.

Let $U\subset\mathfrak{g}$ be a small neighborhood of $0$ so that $\exp:\mathfrak{g}\to G$ is a diffeomorphism when restricted to $U$.

My question is can every element $g\in\exp(U)$ be written uniquely as a product $g=g_1\cdots g_k$ for $g_i\in\exp(\mathfrak{g}_i\cap U)$?

A similar constraint on $g_i$ is fine as well. Or perhaps choosing $U$ to be a ball around $0$ rather than a general neighborhood.

I guess this should be true. The main (but not only) problem I am having with writing a proof is the fact that in general $\exp(\sum \alpha_i)\not=\prod\exp(\alpha_i)$. But maybe it is true if the sum is direct?

P.S - my full attempt - please verify:

Let $r$ be such that $\exp$ is a diffeomorphism from $B(kr,0)$ the ball of radius $kr$ around $0$ in $\mathfrak{g}$ (with some arbitrary metric compatible with the topology). Let $U=B(r,0)$.

Let $g\in\exp(U)$. Then there are $\alpha_i\in\mathfrak{g}_i$ such that $\exp(\sum\alpha_i)=g$. Since $\alpha_i$ commute, we get $g=\prod \exp(\alpha_i)$.

Assume $g=\prod\exp\beta_i$ for $\beta_i\in\mathfrak{g}_i\cap U$. Then $g=\exp(\sum\beta_i)$ and $\sum\beta_i\in B(kr,0)$ by the triangle inequality. From injectivity on $B(kr,0)$ we get $\sum\beta_i=\sum\alpha_i$ so $\beta_i=\alpha_i$ for all $i$, so $\exp(\beta_i)=\exp(\alpha_i)$, so we get what we wanted.