$A$-action and the right $A$-valued inner product as follows: For any $a\in A$, $x, y\in X_{\pi}$, $$x\cdot a=x\pi^{-1}(a) , \quad
{\langle}x, y {\rangle}_A =\pi(x^* y) .$$ Let $\nu$ be the linear map from $X_{\pi}$ to $X_{\pi}\otimes H^0$ defined by $\nu(x)=\rho(x)$ for any $x\in X_{\pi}$. Then $\nu$ is a twisted coaction of $H^0$ on $X_{\pi}$ with respect to $(B, A, \rho, u, \rho_{\pi}, u_{\pi})$. Indeed, for any $a\in A$, $b\in B$, $x, y\in X_{\pi}$, $$\begin{aligned}
& \nu(b\cdot x) =\rho(b)\nu(x)=\rho(b)\cdot \nu(x) , \\
& \nu(x\cdot a) =\nu(x) ((\pi^{-1}\otimes{{\rm{id}}})\circ(\pi\otimes{{\rm{id}}})\circ\rho\circ\pi^{-1})(a)
=\nu(x)\cdot\rho_{\pi}(a) , \\
& {}_{B\otimes H^0} {\langle}\nu(x), \nu(y) {\rangle}=\rho(xy^* )=\rho({}_B {\langle}x, y {\rangle}) , \\
& {\langle}\nu (x), \nu(y) {\rangle}_{A\otimes H^0 } =(\pi\otimes{{\rm{id}}})(\rho(x^* y))=\rho_{\pi}({\langle}x, y {\rangle}_A ), \\
& ({{\rm{id}}}\otimes\epsilon^0 )(\nu(x))= ({{\rm{id}}}\otimes\epsilon^0 )(\rho(x))=x , \\
& ((\nu\otimes{{\rm{id}}})\circ\nu)(x)=u({{\rm{id}}}\otimes\Delta^0 )(\rho(x))u^*
=u\cdot ({{\rm{id}}}\otimes\Delta^0 )(\rho(x))\cdot u_{\pi}^* .\end{aligned}$$ Thus $\nu$ is a twisted coaction of $H^0$ on $X_{\pi}$ with respect to $(B, A, \rho, u, \rho_{\pi}, u_{\pi})$. Therefore we obtain the conclusion.

\[remark:review\]Let $(\rho, u)$ be a twisted coaction of $H^0$ on $A$. By Lemma \[lem:iso\], [@Kodaka:equivariance Lemma 4.10] and [@KT2:coaction Theorem 3.3], we can see that $(\rho, u)$ is strongly Morita equivalent to $\widehat{\widehat{\rho}}$, the second dual coaction of $(\rho, u)$. This is obtained in [@KT3:equivalence the proof 