to be the functor $\big{(}(V,T),(V_{c},T_{c})_{{c}\in {C}}, f, f^*\big{)}\mapsto
               \big{(}(V,T),(V_{c},T_{c})_{{c}\in {C}}, \mathbb{L}_i f,\mathbb{L}_i f^*\big{)}$. These functors acts trivially on the morphisms. It can easily be checked that $\sigma_i, \sigma_i^{-1}$ actually define functors.

\[braid relation\] The functors $\{\sigma_i,\sigma_i^{-1}\}_{i=1}^{m-1}$ satisfy the following braid relations.

1.  \[braid-1\] $\sigma_i\circ\sigma_i^{-1}=\mathrm{id},$ $\sigma_i\circ\sigma_i^{-1}=\mathrm{id}$,

2.  \[braid-2\] $\sigma_i\circ\sigma_{i+1}\circ\sigma_i=\sigma_{i+1}\circ\sigma_i\circ\sigma_{i+1} $, and

3.  \[braid-3\] $\sigma_i\circ\sigma_j=\sigma_j\circ\sigma_i\ \ \  (\ |i-j|\geq 2\ )$.

In particular, by the relation $(\ref{braid-1})$, $\sigma_i$ and $\sigma_i^{-1}$ are equivalences of categories.

The component $(\mathbb{L}_{i}\mathbb{R}_{i+1}f)_c$ is the composition $L_{i+1}R_{i+1}f_c$ if $\tau(c)=i$, otherwise $f_c$. Easy calculation shows that $L_{i+1} R_{i+1} f_c=f_c$. Similarly, we have $\mathbb{L}_{i}\mathbb{R}_{i+1}f^*=f^*$. This implies $\sigma_i^{-1}\circ\sigma_i={\mathrm{id}}$. The relation $\sigma_i\circ\sigma_i^{-1}={\mathrm{id}}$ is shown similarly. Hence we get $(\ref{braid-1})$.

We shall prove (\[braid-2\]). Let $\big{(}(V,T),(V_{c},T_{c})_{{c}\in {C}}, f, f^*\big{)}$ be an object of ${{\sf Std}}_{\bm k}(C,\tau)$. Then we have the following: $$\begin{aligned}
               ({\mathbb{R}}_{i+1} {\mathbb{R}}_{i+2} {\mathbb{R}}_{i+1} f)_c=
                 \begin{cases}
                 R_{i+2} R_{i+1} f_c & \tau(c)=i      \\   
                 R_{i+1} f_c & \tau(c)=i+1 \\     
                  f_c& \text{otherwise},              
                 \end{cases}\\
                                 ({\mathbb{R}}_{i+2} {\mathbb{R}}_{i+1} {\mathbb{R}}_{i+2} f)_c=
                 \begin{cases}
                 R'_{i+2} R_{i+2} f_c & \tau(c)=i      \\   
                 R_{i+1} f_c & \tau(c)=i+1 \\     
                  f_c& \text{otherwise},
                 \end{cases}                                                
              \end{aligned}$$ where $R'_{i+2}:={\mathrm{id}}_V- R_{i+2}f_{\tau^{-1}(i+1)}f_{\tau^{-1}(i+1)}^*R_{i+2}^*$. Put $R'_{i+1}:=R_{i+2}$. Then we have $R^{\prime}_{i+2}\circ R^{\prime}_{i+1}=R_{i+2}\circ R_{i+1}$, which implies ${\mathbb{R}}_{i+1} {\mathbb{R}}_{i+2} {\mathbb{R}}_{i+1} f={\mathbb{R}}_{i+2} {\mathbb{R}}_{i+1} {\mathbb{R}}_{i+2} f$. Indeed, we have $$\begin{aligned}
              R^{\prime}_{i+2}\circ R^{\prime}_{i+1}
              &=({\mathrm{id}}-R_{i+2}\circ({\mathrm{id}}-R_{i+1})\circ R^*_{i+2})\circ R_{i+2}\\
              &=R_{i+2}-R_{i+2}\circ({\mathrm{id}}-R_{i+1})\circ R_{i+2}\\
              &=R_{i+2}\circ 