vector in $V_c$ and assume $R_if_cu=0$. It implies $f_cu=f_{c'}\circ f^*_{c'}\circ f_c u$. Next, we show that the intersection of the image $({\mathbb{R}}_i f)_c(V_c)$ and $f_{c'}V_{c'}$ is $(0)$. Let $v$ be a vector in $V_c$ such that ${\mathbb{R}}_i f v\in V_{c'}$. It implies that $f_cv-f_{{{c'}}} f_{{{c'}}}^*f v \in V_{c'}$. Hence $f_c v\in V_{{c'}}$, and it implies $v=0$. The first and the second assertions show that ${\mathbb{R}}_i f$ defines an isomorphism of vector spaces. Finally, we show that $({\mathbb{R}}_i f, {\mathbb{R}}_i f^*)$ satisfies the conditions (1) and (2) in Definition \[stokes data\]. We only need to consider the case where $\tau({c})=i-1$ and $\tau({{c'}})=i$. We have $$\begin{aligned}
\label{A1}
              ({\mathbb{R}}_i f^*)_{c}\circ ({\mathbb{R}}_i f)_{c}&=(f^*_c-f^*_cT f_{c'} T^{-1}_{c'}f^*_{c'})(f_c-f_{c'}f_{c'}^*f_c)\\ 
               \label{A2}                       &=f^*_cf_c-f^*_cTf_{c'}T^{-1}_{c'}f^*_{c'}f_c
                                                     +f^*_cTf_{c'}T^{-1}_{c'}f^*_{c'}f_c\\
               \label{A3}                       &={\mathrm{id}}_{V_c}. 
              \end{aligned}$$ Here, we used $f_c^* f_{c'}=0$ from (\[A1\]) to (\[A2\]) and $f_c^* f_c={\mathrm{id}}\ (f_{c'}^*f_{c'}={\mathrm{id}})$ from (\[A2\]) to (\[A3\]). We also have $$\begin{aligned}
              ({\mathbb{R}}_i f^*)_{c'}\circ ({\mathbb{R}}_i f)_{c}&=f_{c'}^*\circ(f_c-f_{c'} f^*_{c'} f_c)\\
                                                    &=0.
              \end{aligned}$$ by using $f_{c'}^* f_{c'}={\mathrm{id}}$. They show the condition (1). The condition (2) is shown similarly, which proves the proposition (see Remark \[2.4\]).

We put $\sigma_i:{{\sf Std}}_{\bm k}(C,\tau)\to {{\sf Std}}_{\bm k}(C,(i\ i+1)\circ \tau)$ to be the functor $\big{(}(V,T),(V_{c},T_{c})_{{c}\in {C}}, f, f^*\big{)}\mapsto 
              \big{(}(V,T),(V_{c},T_{c})_{{c}\in {C}}, {\mathbb{R}}_{i+1}f, {\mathbb{R}}_{i+1}f^*\big{)}$ and put $\sigma_i^{-1}:{{\sf Std}}_{\bm k}(C,\tau)\to {{\sf Std}}_{\bm k}(C,(i\ i+1)\circ \tau)$ 