\otimes\Delta_H^0 )((\pi\otimes\lambda^0 )(w^* )) .$$ Since $w\in B\otimes H^0$, we can write that $w=\sum_i b_i\otimes\phi_i $, where $b_i \in B$, $\phi_i \in H^0$. Then $$(\pi\otimes\lambda^0 \otimes\lambda^0 )((\rho\otimes{{\rm{id}}}_{K^0})(w))
=\sum_i (\pi\otimes\lambda^0 )(\rho(b_i ))\otimes\lambda^0 (\phi_i ) .$$ On the other hand $$\begin{aligned}
(\rho_{\pi, \lambda^0}\otimes{{\rm{id}}}_{H^0})((\pi\otimes\lambda^0 )(w)) & =
(((\pi\otimes\lambda^0 )\circ\rho\circ\pi^{-1})\otimes{{\rm{id}}}_{H^0})((\pi\otimes\lambda^0 )(w)) \\
& =(((\pi\otimes\lambda^0 )\circ\rho\circ\pi^{-1})\otimes{{\rm{id}}}_{H^0})(\sum_i \pi(b_i )\otimes\lambda^0 (\phi_i )) \\
& =\sum_i ((\pi\otimes\lambda^0 )\circ\rho)(b_i )\otimes\lambda^0 (\phi_i ) \\
& =\sum_i (\pi\otimes\lambda^0 )(\rho(b_i ))\otimes\lambda^0 (\phi_i ) .\end{aligned}$$ Thus $$(\pi\otimes\lambda^0 \otimes\lambda^0 )((\rho\otimes{{\rm{id}}}_{K^0})(w))
=(\rho_{\pi, \lambda^0 }\otimes{{\rm{id}}}_{H^0}((\pi\otimes\lambda^0 )(w)) .$$ Hence $$\begin{aligned}
v_{\pi, \lambda^0 } &= ((\pi\otimes\lambda^0 )(w)\otimes 1^0 )(\rho_{\pi, \lambda^0 }\otimes{{\rm{id}}}_{H^0 })
((\pi\otimes\lambda^0 )(w))u_{\pi, \lambda^0} \\
& \times ({{\rm{id}}}_A \otimes\Delta_H^0 )((\pi\otimes\lambda^0 )(w^* )) .\end{aligned}$$Therefore, we obtain the conclusion.

Let $(\rho, u)$ be a twisted coaction of $H^0$ on $B$ and let $\pi$ be an isomorphism of $B$ onto $A$. Let $$\rho_{\pi}=(\pi\otimes{{\rm{id}}})\circ\rho\circ\pi^{-1} , \quad
u_{\pi}=(\pi\otimes{{\rm{id}}}_{H^0}\otimes{{\rm{id}}}_{H^0})(u) .$$ By Lemma \[lem:induced\], $(\rho_{\pi}, u_{\pi})$ is a twisted coaction of $H^0$ on $A$.

\[lem:iso\]With the above notations, $(\rho_{\pi}, u_{\pi})$ is strongly Morita equivalent to $(\rho, u)$.

Let $X_{\pi}$ be the $B-A$-equivalence bimodule induced by $\pi$, that is, $X_{\pi}=B$ as vector spaces and the left $B$-action on $X_{\pi}$ and the left $B$-valued inner product are defined in the evident way. We define the right 