{\rangle}_A )=\Psi(E_1^{\sigma, v}(\Psi^{-1}(z))) \\
& =E_1^{\sigma, v}(\Psi^{-1}(z))=E_1^{\rho, u}(z)=\theta_{1,1}^C (z)=(1\rtimes_{\widehat{\rho}}\tau)(z) .\end{aligned}$$ Thus, $\widehat{\Psi}(1\rtimes_{\widehat{\sigma}}\tau)=1\rtimes_{\widehat{\rho}}\tau$. Furthermore, for any $x, y\in D$, $$\begin{aligned}
(E_2^{\rho, u}\circ\widehat{\Psi})((x\rtimes_{\widehat{\sigma}}1^0 )
(1\rtimes_{\widehat{\sigma}}\tau)(y\rtimes_{\widehat{\sigma}}1^0 ))
& =E_2^{\rho, u}((\Psi(x)\rtimes_{\widehat{\rho}}1^0 )
(1\rtimes_{\widehat{\rho}}\tau)(\Psi(y)\rtimes_{\widehat{\rho}}1^0 )) \\
& =\Psi(x)\tau(e)\Psi(y)=\Psi(xy)\tau(e) .\end{aligned}$$ On the other hand $$(\Psi\circ E_2^{\sigma, v})((x\rtimes_{\widehat{\sigma}}1^0 )(1\rtimes_{\widehat{\sigma}}\tau)
(y\rtimes_{\widehat{\sigma}}1^0 ))=\Psi(x\tau(e)y)
=\Psi(xy)\tau(e) .$$ Hence $E_2^{\rho, u}\circ\widehat{\Psi}=\Psi\circ E_2^{\sigma, v}$. Therefore, we obtain the conclusion.

\[lem:equation3\]With the above notations and assumptions, there is an isomorphism $\widehat{\widehat{\Psi}}$ of $D_2$ onto $C_2$ satisfying that $$\widehat{\widehat{\Psi}}|_{D_1} =\widehat{\Psi}, \quad
\widehat{\widehat{\Psi}}(1_{D_1}\rtimes_{\widehat{\widehat{\sigma}}}e)
=1_{C_1}\rtimes_{\widehat{\widehat{\rho}}}e , \quad
E_3^{\rho, u}\circ\widehat{\widehat{\Psi}}=\widehat{\Psi}\circ E_3^{\sigma, v} .$$

We can prove this lemma in the same way as in the proof of Lemma \[lem:equation2\]

By Lemmas \[lem:equation2\] and \[lem:equation3\], $\widehat{\Psi}|_{A' \cap D_1}$ and $\widehat{\widehat{\Psi}}|_{D' \cap D_2}$ are isomorphisms of $A' \cap D_1$ and $D' \cap D_2$ onto $A' \cap C_1$ and $C' \cap C_2$, respectively.

\[lem:Ciso\]With the above notations and assumptions, $A' \cap D_1 \cong H^0$ as $C^*$-algebras.

Since $H^0 \cong 1\rtimes_{\widehat{\sigma}}H^0$ as $C^*$-algebras, it suffices to show that $A' \cap D_1 =1\rtimes_{\widehat{\sigma}}H^0$, where we identify $A$ with $A\rtimes_{\sigma, v}1\rtimes_{\widehat{\sigma}}1^0$. Let $x\in A' \cap D_1$. Then we can write that $x=\sum_i x_i \rtimes_{\widehat{\sigma}}\phi_i$, where $x_i \in D$ and $\{\phi_i \}_i$ is a basis of $H^0$. For any $a\in A$, $$\begin{aligned}
(a\rtimes_{\sigma, v}1\rtimes_{\widehat{\sigma}}1^0 )x 