\end{aligned}$$

Recall that for $\sigma\in \mathrm{Br}_m$, we have $\mathbb{M}_\sigma:{\sf Std}_{\bm k}({C},\tau){\xrightarrow{\sim}}{\sf Std}_{\bm k}({C},\bar{\sigma}\circ\tau)$ (Definition \[braid action\]). Define $\sigma f$ and $\sigma f^*$ by $\mathbb{M}_\sigma\big{(}(V,T),(V_{c},T_{c})_{{c}\in {C}}, f, f^*\big{)}=
           \big{(}(V,T),(V_{c},T_{c})_{{c}\in {C}}, \sigma f, \sigma f^*\big{)}$.

\[braidact\] For $s\in S_m$, we have the following formulas: $$\begin{aligned}
\label{fc}
           \big{(}(s)_R{f}\big{)}_c&=\left( \prod_{i\in I_{\tau(c)}(s)}^{\leftarrow}R_i\right)\circ f_c,\\
           \label{fcc}
           \big{(}(s)_R{f^*}\big{)}_c&=f^*_c\circ\left( \prod_{i\in I_{\tau(c)}(s)}^{\rightarrow}R_i^* \right).
           \end{aligned}$$

For $s\in S_m$, let $\ell(s)$ be the length of $s$. We shall show the lemma by induction on $\ell(s)$. If $\ell(s)=0$, $s$ is identity and the lemma is obvious. If $\ell(s)>0$, then there exists some $s_j$ such that $s=s'\cdot s_j$ with $\ell(s)=\ell(s')+1$. Set $$\begin{aligned}
           &f'_c:=\begin{cases}
                      f_c & (\tau(c)\neq j)\\
                      R_{j+1}\circ f_c& (\tau(c)=j),
                   \end{cases} \\
           &R_i':=\begin{cases}
                    R_i& (i\neq j, j+1)\\
                    R_{j+1}& (i=j)\\
                    {\mathrm{id}}-R_{j+1}\circ f_{c_j}\circ f_{c_j}^*\circ R_{j+1}^* &(i=j+1)
                    \end{cases}           
           \end{aligned}$$ where $c_j=\tau^{-1}(j)$. By induction hypothesis, we have $$\begin{aligned}
           \big{(}(s)_R{f}\big{)}_c=\left(\prod_{i\in I_{s_j\circ\tau(c)}(s')}^{\leftarrow}R'_i\right)\circ f'_c.
           \end{aligned}$$ Since $\ell(s)=\ell(s')+1$, we have $j+1\notin I_j(s')$, and $$\begin{aligned}
            I_k(s)=\begin{cases}
                       s_j(I_{s_j(k)}(s')) & k\neq j\\
                       s_j(I_{s_j(k)}(s'))\cup\{j+1\} & k=j.
                       \end{cases}
           \end{aligned}$$ If $j+1\notin I_{s_j\circ \tau(c)}(s')$, then $$\begin{aligned}
           \prod_{i\in I_{s_j\circ \tau(c)}(s')}^{\leftarrow}R_i'&=
           \prod^{\leftarrow}_{i\in s_j(I_{s_j\circ\tau(c)}(s'))}R_i,          
           \end{aligned}$$ hence we have (\[fc\]). If $j+1\in I_{s_j\circ\tau(c)}(s')$, then $j\in I_{s_j\circ\tau(c)}(s')$ since $s'(j)<s'(j+1)$ by $j+1\notin I_j(s')$. By the proof of Proposition \[braid relation\], we have $R'_{j+1}\circ R'_j=R_{j+1}\circ R_j$. This implies (\[fc\]). The equation (\[fcc\]) is shown similarly.

