lies in $G_{s}$ so does $a$. Thus $ab\in G\setminus G_{s}$ and $ a\o e(s)\cdot b\o e(s)\in N(s)$.

3)-4) are clear.

Let $A$ be an algebra (eg., associative algebra, Lie algebra or vertex operator algebra). We denote the module category of $A$ by $A$-Mod.

For convenience we set $d_{g,s}=g\otimes e(s)$ for $g\in G$ and $s\in S.$

\[t3.3\] The functors $$\begin{aligned}
 M&\stackrel{f}{\mapsto}D(s)&\bigotimes_{S(s)}M\\
N&\stackrel{g}{\mapsto}&d_{1,s}N\\\end{aligned}$$ ($M\in \C^{\a_s}[G_s]$-[*Mod*]{}, $N\in D({\O}_s)$-[*Mod*]{}) define an equivalence between categories $\C^{\a_s}[G_s]$-[*Mod*]{} and $D({\O}_s)$-[*Mod*]{}. In particular, the simple $\C^{\a_s}[G_s]$-modules are mapped to simple $D({\O}_s)$-modules and conversely.

Recall that $G=\bigcup_{i=1}^k G_s g_i$ be a coset decomposition with $g_1=1.$ Then $G=\bigcup_{i=1}^k g_i^{-1}G_s.$ Note that $g_i^{-1}a\otimes e(s)$ can be rewritten as $\alpha_s(g_i^{-1},a)^{-1}g_i^{-1}\otimes e(s)\cdot a\otimes e(s)$ for any $i$ and $a\in G_s.$ Let $M\in \C^{\a_s}[G_s]$-Mod. Then an arbitrary vector in $D(s)\bigotimes_{S(s)}M$ has expression $\sum_{i=1}^k d_{g_i^{-1},s}\o_{S(s)} m_i$ for some $m_i\in M.$ From $$d_{1,s}(\sum_{i=1}^kd_{g_i^{-1},s}\o_{S(s)} m_i)=d_{1,s}\o_{S(s)} m_1$$ we see that $d_{1,s}D(s)\bigotimes_{S(s)}M$ is a subset of $d_{1,s}\otimes_{S(s)}M.$ On the other hand, $d_{1,s}\o_{S(s)} m= d_{1,s}(d_{1,s}\o_{S(s)} m)$ belongs to $d_{1,s}(D(s)\bigotimes_{S(s)}M)$. Thus $d_{1,s}\otimes_{S(s)}M$ and $d_{1,s}(D(s)\bigotimes_{S(s)}M)$ are equal. Since $d_{1,s}\bigotimes_{S(s)}M$ is isomorphic to $M$ as $\C^{\a_s}[G_s]$-modules . This implies that $g\circ f(M)$ and $M$ are isomorphic as $\C^{\a_s}[G_s]$-modules.

Now take $N\in D({\O}_s)$-Mod. We are going to show that $D(s)\otimes_{S(s)}d_{1,s}N$ and $N$ are isomorphic 