H$, $\psi\in H^0$, $y\in Y$, $$\psi\triangleright(a\rtimes_{\rho, u}h)\cdot y=[\psi_{(1)}\cdot_{\widehat{\rho}}(a\rtimes_{\rho, u}h)]\cdot
[\psi_{(2)}\triangleright y] .$$

For any $a\in A$, $h\in H$, $\psi\in H^0$ , $y\in Y$, $$\begin{aligned}
& \psi\triangleright(a\rtimes_{\rho, u}h)\cdot y =
NE^Y ((1\rtimes_{\rho, u}1\rtimes_{\widehat{\rho}}\psi)\cdot
\phi((a\rtimes_{\rho, u}h)\cdot y)\cdot(1\rtimes_{\sigma, v}1\rtimes_{\widehat{\sigma}}\tau)) \\
& =NE^Y ((1\rtimes_{\rho, u}1\rtimes_{\widehat{\rho}}\psi)(a\rtimes_{\rho, u}h\rtimes_{\widehat{\rho}}1^0 )
\cdot \phi(y)\cdot(1\rtimes_{\sigma, v}1\rtimes_{\widehat{\sigma}}\tau)) \\
& =NE^Y (([\psi_{(1)}\cdot_{\widehat{\rho}}(a\rtimes_{\rho, u}h)]\rtimes_{\widehat{\rho}}\psi_{(2)})
\cdot\phi(y)\cdot(1\rtimes_{\sigma, v}1\rtimes_{\widehat{\sigma}}\tau)) \\
& =[\psi_{(1)}\cdot_{\widehat{\rho}}(a\rtimes_{\rho, u}h)]
\cdot NE^Y ((1\rtimes_{\rho, u}
1\rtimes_{\widehat{\rho}}\psi_{(2)})\cdot\phi(y)\cdot (1\rtimes_{\sigma, v}1\rtimes_{\widehat{\sigma}}\tau)) \\
& =[\psi_{(1)}\cdot_{\widehat{\rho}}(a\rtimes_{\rho, u}h)]\cdot[\psi_{(2)}\triangleright y]\end{aligned}$$ by [@KT4:morita Lemma 6.1].

\[lem:definition2\]With the above notations, for any $\psi, \chi\in H^0$, $y\in Y$, $$\psi\triangleright[\chi\triangleright y]=\psi\chi\triangleright y .$$

By Remark \[remark:calculus\], for any $\psi, \chi\in H^0$, $y\in Y$, $$\psi\triangleright[\chi\triangleright y]=\psi\triangleright\sum_{I\in \Lambda}[\chi\cdot_{\widehat{\rho}}W_I^{\rho*}]
\cdot E^X (W_I^{\rho}\cdot y) .$$ Since $\chi\cdot_{\widehat{\rho}}W_I^{\rho*}\in C$, by Lemma \[lem:definition1\], $$\psi\triangleright[\chi\triangleright y]=\sum_{I\in\Lambda}[\psi_{(1)}\cdot_{\widehat{\rho}}[\chi\cdot_{\widehat{\rho}}
W_I^{\rho*}]]\cdot[\psi_{(2)}\triangleright E^X (W_I^{\rho}\cdot y)] .$$ Since $E^X (W_I^{\rho} \cdot y)\in X$, by Lemma \[lem:fix\] and Remark \[remark:calculus\] $$\begin{aligned}
\psi\triangleright[\chi\triangleright y] & =\sum_{I\in\Lambda}[\psi_{(1)}\chi\cdot_{\widehat{\rho}}W_I^{\rho*}]
\cdot\epsilon^0 (\psi_{(2)})E^X (W_I^{\rho}\cdot y) \\
& =\sum_{I\in \Lambda}[\psi\chi\cdot_{\widehat{\rho}}W_I^{\rho*}]\cdot E^X (W_I^{\rho}\cdot y)
=\psi\chi\triangleright y .\end{aligned}$$

\[lem:definition3\]With the above notations, for any $\psi\in H^0$, $y, z\in Y$, $$\psi\cdot_{\widehat{\rho}}{}_C {\langle}y, z {\rangle}={}_C {\langle}\psi_{(1)}\triangleright y \, , \, S^0 (\psi_{(2)}^* )\triangleright z {\rangle}.$$

By Remark \[remark:calculus\] and [@KT4:morita Lemma 5.4], for any $\psi\in H^0$, $y, z\in Y$, $$\begin{aligned}
& {}_C {\langle}\psi_{(1)}\triangleright y \, , \, S^0 (\psi_{(2)}^* )\triangleright 