Proof. Let $k\geq 1$ be an integer. We claim that $\prod_{\to}^{(k)}$ is a bijection of sets (this implies the proposition by taking the projective limit as $k\to+\infty$). We will prove the bijection by induction in $k$ . Case $k=1$ is obvious because all the groups under considerations are trivial.

We would like to prove that $\prod_{\to}^{(k+1)}$ is a bijection assuming that $\prod_{\to}^{(k)}$ is a bijection. Let $h$ be an element of $G/G^{\geq k+1}$ and $\overline{h}$ its image in $G/G^{\geq k}$. By the induction assumption there exist unique $\overline{h}_{i}\in G_{\lambda_{i}}/G_{\lambda_{i}}^{\geq k+1},1\leq i\leq N_{k+1}$ such that $\overline{h}_{1}\dots\overline{h}_{N_{k+1}}=\overline{h}$. Let $h_{i},1\leq i\leq N_{k+1}$ be any liftings of $\overline{h}_{i}$ to $G_{i}/G_{i}^{\geq k}$. Then $h_{1}\dots h_{N_{k+1}}=h\pmod{G^{\geq k}}$, hence $c:=h_{1}\dots h_{N_{k+1}}h^{-1}$ belongs to $G^{\geq k}/G^{\geq k+1}\subset Center(G/G^{\geq k+1})$. The last inclusion holds because $[{\bf g},{\bf g}^{\geq k}]=[{\bf g}^{\geq 1},{\bf g}^{\geq k}]\subset{\bf g}^{\geq k+1}$.

Next we observe that the isomorphism of abelian Lie algebras

implies an isomorphism of the corresponding abelian groups

Hence we can write uniquely $c=c_{1}\dots c_{N_{k+1}}$, where $c_{i}\in G^{\geq k}/G^{\geq k+1}\subset Center(G/G^{\geq k+1})$. It follows that $\prod_{\to}^{(k+1)}\left((h_{i}c_{i}^{-1})\right)=h$. Also it is now clear that this decomposition of $h$ is unique. This concludes the proof. $\blacksquare$