I am learning about quantum groups, and I have a question about the construction of certain antiautomorphism of an quantized universal enveloping algebra, $U_q(\mathfrak{g})$.
In the book $\ulcorner$Introduction to Quantum Groups and Crystal Bases$\lrcorner$ by Hong & Kang, in chapter 5, in prepation for the proof of exsistence of crystal bases, it constructs a symmetric bilinear form on $\mathcal{L}(\lambda)$, an irreducible representation of heighest weight $\lambda$. For the construction, it defines an antiautomorphism of $U_q(\mathfrak{g})$. It is written in p.94, and it is the map defined by
$q^h \longmapsto q^h$, $\quad$ $e_i \longmapsto q_i K^{-1}_i f_i$, $\quad$ $f_i \longmapsto q^{-1}_i K_i e_i$
where notations $K_i=q^{s_i h_i}$ and $q_i=q^{s_i}$ follow the book.
But I'm confused if it really defines an antihomomorphism. For example, if we denote the map by $\tau$,
$0=\tau(0)=\tau(e_i f_i - f_i e_i - \frac{K_i-K^{-1}_i }{q_i-q^{-1}_i}) \\=\tau(f_i) \tau(e_i) - \tau(e_i)\tau(f_i )-\tau(\frac{K_i-K^{-1}_i }{q_i-q^{-1}_i}) \\=q^{-1}_i K_i e_i q_i K^{-1}_i f_i-q_i K^{-1}_i f_i q^{-1}_i K_i e_i-\frac{K_i-K^{-1}_i }{q_i-q^{-1}_i} \\=K_i e_i K^{-1}_i f_i- K^{-1}_i f_i K_i e_i-\frac{K_i-K^{-1}_i }{q_i-q^{-1}_i} \\=q^{2}_i e_i f_i-q^{2}_i f_i e_i-\frac{K_i-K^{-1}_i }{q_i-q^{-1}_i} \\=(q^{2}_i-1)\frac{K_i-K^{-1}_i }{q_i-q^{-1}_i}=q_i(K_i-K^{-1}_i )$
, which is absurd. I think I made a mistake somewhere, but I cannot figure out. Any helps will be appreciated. Thank you.