The theorem I am referring to here says that if we start with a Lie bialgebra $\mathfrak g$ determined by some skew-symmetric element $r \in \mathfrak g \otimes \mathfrak g$ satisfying classical Yang-Baxter equation then there exists a Hopf algebra deformation $U_h (\mathfrak g)$ of the universal enveloping algebra $U(\mathfrak g)$ of $\mathfrak g$ such that $\mathfrak g$ is the classical limit of $U_h (\mathfrak g)$ (in other words $U_h (\mathfrak g)$ is a quantization of $\mathfrak g$) and $U_h (\mathfrak g)$ is a triangular Hopf algebra which is isomorphic to $U(\mathfrak g)[[h]]$ as an algebra over $\mathbb R [[h]].$
The proof goes as follows $:$
Start with the trivial deformation $U(\mathfrak g)[[h]]$ of $U(\mathfrak g).$ Then Lemma $6.2.10$ guarantees that there exists an invertible element $\mathcal F_h \in U(\mathfrak g) [[h]] \otimes U(\mathfrak g) [[h]]$ satisfying $$(\mathcal F_h)_{12} (\Delta_h \otimes \text {id}) (\mathcal F_h) = (\mathcal F_h)_{23} (\text {id} \otimes \Delta_h) (\mathcal F_h)$$ and,
$$(\varepsilon_h \otimes \text {id}) (\mathcal F_h) = 1 = (\text {id} \otimes \varepsilon_h) (\mathcal F_h)$$ Hence by Theorem $4.2.14$ it follows that the twisted Hopf algebra $U(\mathfrak g)[[h]]^{\mathcal F_h}$ is triangular. The remaining part of the proof goes on by showing that $\mathfrak g$ is the classical limit of $U(\mathfrak g)[[h]]^{\mathcal F_h}.$ For that the authors claimed that, by virtue of Theorem $6.2.7$, it is enough to show that $$\mathcal F_h \equiv 1 \otimes 1\ (\text {mod}\ h)\ \ \text {and}\ \ \frac {\mathcal F_h - 1 \otimes 1} {h} \equiv - \frac {1} {2} r\ (\text {mod}\ h)$$ This is the part which I didn't follow properly. Could anyone please shed some light on it?
Thanks for your time.
