---
abstract: 'Let $A$ and $B$ be unital $C^*$-algebras and let $H$ be a finite dimensional $C^*$-Hopf algebra. Let $H^0$ be its dual $C^*$-Hopf algebra. Let $(\rho, u)$ and $(\sigma, v)$ be twisted coactions of $H^0$ on $A$ and $B$, respectively. In this paper, we shall show the following theorem: We suppose that the unital inclusions $A\subset A\rtimes_{\rho, u}H$ and $B\subset B\rtimes_{\sigma, v}H$ are strongly Morita equivalent. If $A''\cap (A\rtimes_{\rho, u}H)={\mathbf C}1$, then there is a $C^*$-Hopf algebra automorphism $\lambda^0$ of $H^0$ such that the twisted coaction $(\rho, u)$ is strongly Morita equivalent to the twisted coaction $(({{\rm{id}}}_B \otimes\lambda^0 )\circ\sigma \, , \, ({{\rm{id}}}_B \otimes\lambda^0 \otimes\lambda^0 )(v))$ induced by $(\sigma, v)$ and $\lambda^0$.'
address:
- 'Department of Mathematical Sciences, Faculty of Science, Ryukyu University, Nishihara-cho, Okinawa, 903-0213, Japan'
- 'Faculty of Education, Gunma University, 4-2 Aramaki-machi, Maebashi City, Gunma, 371-8510, Japan'
- '*[E-mail address]{}: [kodaka@math.u-ryukyu.ac.jp]{}*'
- '*[E-mail address]{}: [teruya@gunma-u.ac.jp]{}*'
author:
- Kazunori Kodaka and Tamotsu Teruya
title: 'Coactions of a finite dimensional $C^*$-Hopf algebra on unital $C^*$-algebras, unital inclusions of unital $C^*$-algebras and the strong Morita equivalence'
---

Introduction {#sec:intro}
============

In the previous papers [@KT3:equivalence] and [@KT4:morita], we discussed the strong Morita equivalences for twisted coactions of a finite dimensional $C^*$-Hopf algebra on unital $C^*$-algebras and unital inclusions of unital $C^*$-algebras. In this paper, we shall discuss the relation between the strong Morita equivalence for twisted coactions of a finite dimensional $C^*$-Hopf algebra on unital $C^*$-algebras and the strong Morita equivalence for the unital inclusions of the unital $C^*$-algebras induced by the twisted coactions of the finite dimensional $C^*$-Hopf algebra on the unital $C^*$-algebras.

Let us explain the problem in detail. Let $A$ and $B$ be unital $C^*$-algebras. Let $H$ be a finite dimensional $C^*$-Hopf algebra and $H^0$ its dual $C^*$-Hopf algebra. Let $(\rho, u)$ and $(\sigma, v)$ be twisted coactions of $H^0$ on $A$ and $B$, respectively. Then we can obtain the unital inclusions of unital $C^*$-algebras $A\subset A\rtimes_{\rho, u}H$ and $B\subset B\rtimes_{\sigma, v}H$. In the same way as in [@KT4:morita Example], we can see that if $(\rho, u)$ and $(\sigma ,v)$ are strongly Morita equivalent, then the unital nclusions $A\subset A\rtimes_{\rho, u}H$ and $B\subset B\rtimes_{\sigma, v}H$ are strongly Morita equivalent. In this paper, we shall discuss the inverse implication. Our main theorem is as follows: We suppose that the unital inclusions $A\subset A\rtimes_{\rho, u}H$ and $B\subset B\rtimes_{\sigma, v}H$ are strongly Morita equivalent in the sense of [@KT4:morita Definition 2.1]. If $A'\cap (A\rtimes_{\rho, u}H)={\mathbf C}1$, then there is a $C^*$-Hopf algebra automorphism $\lambda^0$ of $H^0$ such that the twisted coaction $(\rho, u)$ is strongly Morita equivalent to the twisted coaction $(({{\rm{id}}}_B \otimes\lambda^0 )\circ\sigma \, , \, ({{\rm{id}}}_B \otimes\lambda^0 \otimes\lambda^0 )(v))$ induced by $(\sigma, v)$ and $\lambda^0$.

For a unital $C^*$-algebra $A$, let $M_n (A)$ be the $n\times n$-matrix algebra over $A$ and $I_n$ denotes the unit element in $M_n (A)$. We identify $M_n (A)$ with $A\otimes M_n ({\mathbf C})$.

Let $A$ and $B$ be $C^*$-algebras and $X$ an $A-B$-bimodule. We denote its left $A$-action and right $B$-action on $X$ by $a\cdot x$ and $x\cdot b$ for any $a\in A$, $b\in B$, $x\in X$, respectively. Also, we denote by $\widetilde{X}$ the dual $B-A$-bimodule of $X$ and we denote by $\widetilde{x}$ the element in $\widetilde{X}$ induced by $x\in X$.

Preliminaries {#sec:pre}
=============

Let $H$ be a finite dimensional $C^*$-Hopf algebra. We denote its comultipication, counit and antipode by $\Delta$, $\epsilon$ and $S$, respectively. We shall use Sweedler’s notation, $\Delta(h)=h_{(1)}\otimes h_{(2)}$ for any $h\in H$ which suppresses a possible summation when we write comultiplications. We denote by $N$ the dimension of $H$. Let $H^0$ be the dual $C^*$-Hopf algebra of $H$. We denote its comultiplication, counit and antipode by $\Delta^0$, $\epsilon^0$ and $S^0$, respectively. There is the distinguished projection $e$ in $H$. We note that $e$ is the Haar trace on $H^0$. Also, there is the distinguished projection $\tau$ in $H^0$ which is the Haar trace on $H$. Since $H$ is finite dimensional, $H\cong\oplus_{k=1}^L M_{f_k}({\mathbf C})$ and $H^0 \cong \oplus_{k=1}^K M_{d_k }({\mathbf C})$ as $C^*$-algebras. Let $\{v_{ij}^k \, | \, k=1,2, \dots L , \, i, j=1,2,\dots, f_k \}$ be a system of matrix units of $H$. Let $\{w_{ij}^k \, | \, k=1,2,\dots,K ,\, i, j=1,2, \dots , d_k \}$ be a basis of $H$ satisfying Szymański and Peligrad’s [@SP:saturated Theorem 2.2,2], which is called a system of *comatrix units of $H$, that is, the dual basis of the system of matrix units of $H^0$. Also, let $\{\phi_{ij}^k \, | \, k=1,2,\dots,K, \, i,j=1,2,\dots,d_k \}$ and $\{\omega_{ij}^k \, | \, k=1,2,\dots, L, \, i, j=1,2,\dots,f_k \}$ be systems of matrix units and comatrix units of $H^0$, respectively. Let $A$ be a unital $C^*$-algebra. We recall the definition of a twisted coaction $(\rho, u)$ of $H^0$ on $A$ (See [@KT1:inclusion], [@KT2:coaction]). Let $\rho$ be a weak coaction of $H^0$ on $A$ and $u$ a unitary element in $A\otimes H^0 \otimes H^0$. Then we say that $(\rho, u)$ is a *twisted coaction of $H^0$ on $A$ if (1) $(\rho\otimes{{\rm{id}}})\circ\rho={{\rm{Ad}}}(u)\circ({{\rm{id}}}\otimes\Delta^0 )\circ\rho$, (2) $(u\otimes 1^0 )({{\rm{id}}}\otimes\Delta^0 \otimes{{\rm{id}}})(u)=(\rho\otimes{{\rm{id}}}\otimes{{\rm{id}}})(u)
({{\rm{id}}}\otimes{{\rm{id}}}\otimes\Delta^0 )(u)$, (3) $({{\rm{id}}}\otimes h\otimes\epsilon^0 )(u)=({{\rm{id}}}\otimes\epsilon^0 \otimes h)(u)=\epsilon^0 (h)1$ for any $h\in H$.**

Let ${{\rm{Hom}}}(H, A)$ be the linear space of all linear maps from $H$ to $A$. Then by Sweedler [@Sweedler:Hopf pp67-70], it becomes a unital convolution $*$-algebra. Since $H$ is finite dimensional, ${{\rm{Hom}}}(H, A)$ is isomorphic to $A\otimes H^0$. For any element $x\in A\otimes H^0$, we denote by $\widehat{x}$ the element in ${{\rm{Hom}}}(H, A)$ induced by $x$. Similarly, we define ${{\rm{Hom}}}(H\times H, A)$. We identify $A\otimes H^0 \otimes H^0$ with ${{\rm{Hom}}}(H\times H, A)$. For any element $y\in A\otimes H^0 \otimes H^0$, we denote by $\widehat{y}$ the element in ${{\rm{Hom}}}(H\times H, A)$ induced by $y$. Furthermore, for a Hilbert $C^*$-bimodule $X$, let ${{\rm{Hom}}}(H, X)$ be the linear space of all linear maps from $H$ to $X$. We also identify ${{\rm{Hom}}}(H, X)$ with $X\otimes H^0$. For any element $x\in X\otimes H^0$, we denote by $\widehat{x}$ the element in ${{\rm{Hom}}}(H, X)$ induced by $x$.

For a twisted coaction $(\rho, u)$, we can consider the twisted action of $H$ on $A$ and its unitary element $\widehat{u}$ defined by $$h\cdot_{\rho, u}x=\widehat{\rho(x)}(h)=({{\rm{id}}}\otimes h)(\rho(x))$$ for any $x\in A$, $h\in H$. We call it the twisted action induced by $(\rho, u)$. Let $A\rtimes_{\rho, u}H$ be the twisted crossed product by the twisted action of $H$ on $A$ induced by $(\rho, u)$. Let $x\rtimes_{\rho, u}h$ be the element in $A\rtimes_{\rho, u}H$ induced by elements $x\in A$, $h\in H$. Let $\widehat{\rho}$ be the dual coaction of $H$ on $A\rtimes_{\rho, u}H$ defined by $$\widehat{\rho}(x\rtimes_{\rho, u}h)=(x\rtimes_{\rho, u}h_{(1)})\otimes h_{(2)}$$ for any $x\in A$, $h\in H$. Let $E_1^{\rho, u}$ be the canonical conditional expectation from $A\rtimes_{\rho, u}H$ onto $A$ defined by $$E_1^{\rho, u}(x\rtimes_{\rho, u}h)=\tau(h)x$$ for any $x\in A$, $h\in H$. Let $\Lambda$ be the set of all triplets $(i, j, k)$, where $i, j=1,2,\dots,d_k$ and $k=1,2,\dots,K$ with $\sum_{k=1}^K d_k^2 =N$. Let $W_I=\sqrt{d_k}\rtimes_{\rho, u}w_{ij}^k$ for any $I=(i, j, k)\in\Lambda$. By [@KT1:inclusion Proposition 3.18], $\{(W_I^{\rho *}, W_I^{\rho})\}_{I\in \Lambda}$ is a quasi-basis for $E_1^{\rho, u}$.

\[lem:pre\]With the above notations, for any $\psi\in H^0$, $$1\rtimes_{\rho, u}1\rtimes_{\widehat{\rho}}\psi
=\sum_{I\in\Lambda}([\psi\cdot_{\widehat{\rho}}W_I^{\rho*}]\rtimes_{\widehat{\rho}}1^0 )
(1\rtimes_{\rho, u}1\rtimes_{\widehat{\rho}}\tau)(W_I^{\rho}\rtimes_{\widehat{\rho}}1^0 ) .$$

Since $\{(W_I^{\rho*}, W_I^{\rho})\}_{I\in\Lambda}$ is a quasi-basis for $E_1^{\rho, u}$ by [@KT1:inclusion Proposition 3.18], $$\sum_{I\in\Lambda}(W_I^{\rho*}\rtimes_{\widehat{\rho}}1^0 )(1\rtimes_{\rho, u}1\rtimes_{\widehat{\rho}}\tau)
(W_I^{\rho}\rtimes_{\widehat{\rho}}1^0 )=1\rtimes_{\rho, u}1\rtimes_{\widehat{\rho}}1^0 .$$ Hence for any $\psi\in H^0$, $$\begin{aligned}
1\rtimes_{\rho, u}1\rtimes_{\widehat{\rho}}\psi &
= \sum_{I\in \Lambda}(1\rtimes_{\rho, u}1\rtimes_{\widehat{\rho}}\psi)(W_I^{\rho*}\rtimes_{\widehat{\rho}}1^0 )
(1\rtimes_{\rho, u}1\rtimes_{\widehat{\rho}}\tau)(W_I^{\rho}\rtimes_{\widehat{\rho}}1^0 ) \\
& =\sum_{I\in\Lambda}([\psi_{(1)}\cdot_{\widehat{\rho}}W_I^{\rho*}]\rtimes_{\widehat{\rho}}\psi_{(2)})
(1\rtimes_{\rho, u}1\rtimes_{\widehat{\rho}}\tau)(W_I^{\rho}\rtimes_{\widehat{\rho}}1^0 ) \\
& =\sum_{I\in\Lambda}([\psi\cdot_{\widehat{\rho}}W_I^{\rho*}]\rtimes_{\widehat{\rho}}1^0 )
(1\rtimes_{\rho, u}1\rtimes_{\widehat{\rho}}\tau)(W_I^{\rho}\rtimes_{\widehat{\rho}}1^0 ) .\end{aligned}$$ Therefore, we obtain the conclusion.

A left coaction of a finite dimensional $C^*$-Hopf algebra on an equivalence bimodule {#sec:left}
=====================================================================================

Let $A$ and $B$ be unital $C^*$-algebras and let $(\rho, u)$ and $(\sigma, v)$ be twisted coactions of $H^0$ on $A$ and $B$, respectively. Let $A\rtimes_{\rho, u}H$ and $B\rtimes_{\sigma, v}H$ be the twisted crossed products of $A$ and $B$ by $(\rho, u)$ and $(\sigma, v)$, respectively. We denote them by $C$ and $D$, respectively. Then we obtain unital inclusions $A\subset C$ and $B\subset D$ of unital $C^*$-algebras. We suppose that $A\subset C$ and $B\subset D$ are strongly Morita equivalent with respect to a $C-D$-equivalence bimodule $Y$ and its closed subspace $X$ in the sense of [@KT4:morita Definition 2.1]. Also, by [@KT4:morita Section 2], there is a conditional expectation $E^X$ from $Y$ onto $X$ with respect to $E_1^{\rho, u}$ and $E_1^{\sigma, v}$ satisfying Conditions (1)-(6) in [@KT4:morita Definition 2.2]. Furthermore, by [@KT4:morita Section 6], we can see that the unital inclusions $C\subset C_1$ and $D\subset D_1$ are strongly Morita equivalent with respect to the $C_1 -D_1$-equivalence bimodule $Y_1$ and its closed subspace $Y$, where $C_1 =C\rtimes_{\widehat{\rho}}H^0$ and $D_1 =D\rtimes_{\widehat{\sigma}}H^0$. As defined in [@KT4:morita], we define $Y_1$ as follows: We regard $C$ and $D$ as a $C_1 -A$-equivalence bimodule and a $D_1 -B$-equivalence bimodule in the usual way as in [@KT4:morita Section 4], respectively. Let $Y_1 =C\otimes _A X\otimes_B \widetilde{D}$. Let $E^Y$ be the conditional expectation from $Y_1$ onto $Y$ with respect to $E_2^{\rho, u}$ and $E_2^{\sigma, v}$ defined by $$E^Y (c\otimes x\otimes\widetilde{d})=\frac{1}{N}c\cdot x\cdot d^*$$ for any $c\in C$, $d\in D$, $x\in X$, where $E_2^{\rho, u}$ and $E_2^{\sigma, v}$ are the canonical conditional expectations from $C_1$ and $D_1$ onto $C$ and $D$ defined by $$E_2^{\rho, u}(c\rtimes_{\widehat{\rho}}\psi) =c\psi(e)  , \quad
E_2^{\sigma, v}(d\rtimes_{\widehat{\sigma}}\psi)=d\psi(e) .$$ for any $c\in C$, $d\in D$, $\psi\in H^0$, respectively. We regard $Y$ as a closed subspace of $Y_1$ by the injective linear map $\phi$ from $Y$ to $Y_1$ defined by $$\phi(y)=\sum_{I, J\in \Lambda}W_I^{\rho*}\otimes E^X (W_I^{\rho}\cdot y\cdot W_J^{\sigma*})
\otimes\widetilde{W_J^{\sigma*}}$$ for any $y\in Y$. In this section, we construct a left coaction of $H$ on $Y$ with respect to $(C, \widehat{\rho})$. First, we define the bilinear map $`` \triangleright "$ from $H^0 \times Y$ to $Y$ as follows: For any $\psi\in H^0$, $y\in Y$, $$\psi\triangleright y =NE^Y ((1\rtimes_{\rho, u}1\rtimes_{\widehat{\rho}}\psi)\cdot
\phi(y)\cdot (1\rtimes_{\sigma, v}1\rtimes_{\widehat{\sigma}}\tau)) .$$

\[remark:calculus\]For any $\psi\in H^0$, $y\in Y$, $$\psi\triangleright y=\sum_{I\in \Lambda}[\psi\cdot_{\widehat{\rho}}W_I^{\rho*}]
\cdot E^X (W_I^{\rho}\cdot y) .$$ Indeed, since $1\rtimes_{\rho, u}1\rtimes_{\widehat{\rho}} \tau$ is the Jones projection for the conditional expectation $E_1^{\rho, u}$, for any $c\in C$ $(1\rtimes_{\rho, u}1\rtimes_{\widehat{\rho}}\tau)\cdot c=E_1^{\rho, u}(c)$. Also, $(1\rtimes_{\sigma, v}\rtimes_{\widehat{\sigma}}\tau)\cdot d =E_1^{\sigma, v}(d)$ for any $d\in D$. Hence by Lemma \[lem:pre\], for any $\psi\in H^0$, $y\in Y$, $$\begin{aligned}
&\psi\triangleright y \\
&=\sum_{I, J\in \Lambda}NE^Y ((1\rtimes_{\rho, u}1\rtimes_{\widehat{\rho}}\psi)\cdot W_I^{\rho*}
\otimes E^X (W_I^{\rho}\cdot y\cdot W_J^{\sigma*} )\otimes\widetilde{W_J^{\sigma*}}\cdot
(1\rtimes_{\sigma, v}1\rtimes_{\widehat{\sigma}}\tau)) \\
& =\sum_{I, I_1 , J \in\Lambda}NE^Y ([\psi\cdot_{\widehat{\rho}}W_{I_1}^{\rho*}]
E_1^{\rho, u}(W_{I_1}^{\rho}W_I^{\rho*})\otimes E^X (W_I^{\rho}\cdot y\cdot W_J^{\sigma*})
\otimes E_1^{\sigma, v}(W_J^{\sigma*})^{\widetilde{}}) \\
& =\sum_{I, I_1 , J\in \Lambda}NE^Y ([\psi\cdot_{\widehat{\rho}}W_{I_1}^{\rho*}]
\otimes E_1^{\rho, u}(W_{I_1}^{\rho}W_I^{\rho*})\cdot E^X (W_I^{\rho}\cdot y\cdot W_{J}^{\sigma*})\cdot
E_1^{\sigma, v}(W_J^{\sigma})\otimes \widetilde{1}) \\
& =\sum_{I, I_1 , J\in\Lambda}[\psi\cdot_{\widehat{\rho}}W_{I_1}^{\rho*}]\cdot E^X (E_1^{\rho, u}(W_{I_1}^{\rho}
W_I^{\rho*})W_I^{\rho}\cdot y\cdot W_J^{\sigma*}E_1^{\sigma, v}(W_J^{\sigma})) \\
& =\sum_{I_1 \in \Lambda}[\psi\cdot_{\widehat{\rho}}W_{I_1}^{\rho*}]\cdot E^X (W_{I_1}^{\rho}\cdot y).\end{aligned}$$

\[lem:fix\]With the above notations, for any $x\in X$, $\psi\in H^0$, $\psi\triangleright x=\epsilon^0 (\psi)x$.

By Remark \[remark:calculus\], [@SP:saturated Theorem 2.2] and [@KT4:morita Lemma 6.9] $$\begin{aligned}
\psi\triangleright x & =NE^Y ((1\rtimes_{\rho, u}1\rtimes_{\widehat{\rho}}\psi)\cdot\phi(x)\cdot
(1\rtimes_{\sigma, v}1\rtimes_{\widehat{\sigma}}\tau)) \\
& =NE^Y ((1\rtimes_{\rho, u}1\rtimes_{\widehat{\rho}}\psi\tau)\cdot\phi(x)) \\
& =N\epsilon^0 (\psi)E^Y ((1\rtimes_{\rho, u}1\rtimes_{\widehat{\rho}}\tau)\cdot\phi(x))
=\epsilon^0 (\psi)\sum_{I\in\Lambda}[\tau\cdot_{\widehat{\rho}}W_I^{\rho*}]\cdot E^X (W_I^{\rho}\cdot x) \\
& =\epsilon^0 (\psi) ( \sum_{I\in\Lambda}E_1^{\rho, u}(W_I^{\rho*}E_1^{\rho, u}(W_I^{\rho})))\cdot x
=\epsilon^0 (\psi)x .\end{aligned}$$

\[lem:definition1\]With the above notations, for any $a\in A$, $h\in H$, $\psi\in H^0$, $y\in Y$, $$\psi\triangleright(a\rtimes_{\rho, u}h)\cdot y=[\psi_{(1)}\cdot_{\widehat{\rho}}(a\rtimes_{\rho, u}h)]\cdot
[\psi_{(2)}\triangleright y] .$$

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 $\psi\in H^0$, $y, z\in Y$, $$\begin{aligned}
& {}_C {\langle}\psi_{(1)}\triangleright y \, , \, S^0 (\psi_{(2)}^* )\triangleright z {\rangle}\\
& =\sum_{I, J\in\Lambda}{}_C {\langle}[\psi_{(1)}\cdot_{\widehat{\rho}}W_I^{\rho*}]\cdot E^X (W_I^{\rho}\cdot y) \, , \,
[S^0 (\psi_{(2)}^* )\cdot_{\widehat{\rho}}W_J^{\rho*}]\cdot E^X (W_J^{\rho}\cdot z) {\rangle}\\
& =\sum_{I, J\in\Lambda}[\psi_{(1)}\cdot_{\widehat{\rho}}W_I^{\rho*}]
{}_A {\langle}E^X (W_I^{\rho}\cdot y) \, , \, E^X (W_J^{\rho}\cdot z) {\rangle}[\psi_{(2)}\cdot_{\widehat{\rho}}W_J^{\rho}] \\
& =\sum_{I, J\in\Lambda}[\psi\cdot_{\widehat{\rho}}W_I^{\rho*} \, {}_A
{\langle}E^X (W_I^{\rho}\cdot y) \, , \, E^X (W_J^{\rho}\cdot z ) {\rangle}W_J^{\rho}] \\
& =\sum_{I, J\in\lambda}[\psi\cdot_{\widehat{\rho}} \, {}_C {\langle}W_J^{\rho*}\cdot E^X (W_I^{\rho}\cdot y) \, ,\, 
W_J^{\rho*}\cdot E^X (W_J^{\rho}\cdot z ) {\rangle}] \\
& =\psi\cdot_{\widehat{\rho}}{}_C {\langle}y, \, z {\rangle}.\end{aligned}$$

\[prop:leftcoaction\]With the above notations, the linear map from $Y$ to $Y\otimes H$ induced by the bilinear map $ ``\triangleright"$ from $H^0 \times Y$ to $Y$ defined by $$\psi\triangleright y =NE^Y ((1\rtimes_{\rho, u}1\rtimes_{\widehat{\rho}}\psi)\cdot \phi(y)\cdot
(1\rtimes_{\sigma, v}1\rtimes_{\widehat{\sigma}}\tau))$$ for any $y\in Y$, $\psi\in H^0$ is a left coaction of $H$ on $Y$ with respect to $(C, \widehat{\rho})$ satisfying that $$\psi\triangleright x = \epsilon^0 (\psi)x$$ for any $x\in X$, $\psi\in H^0$.

By Lemmas \[lem:fix\], \[lem:definition1\], \[lem:definition2\] and \[lem:definition3\], it suffices to show that $1^0 \triangleright y=y$ for any $y\in Y$. Indeed, by Remark \[remark:calculus\] and [@KT4:morita Lemma 5.4] $$1^0 \triangleright y=\sum_{I\in \Lambda}W_I^{\rho*}\cdot E^X (W_I^{\rho}\cdot y)=y$$ for any $y\in Y$.

We denote by $\mu$ the above left coaction induced by $``\triangleright"$.

A coaction of a finite dimensional $C^*$-Hopf algebra on a unital $C^*$-algebra induced by a left coaction {#sec:coaction}
==========================================================================================================

We use the same notations as in the previous section and also we suppose that the same assumptions as in the previous section. We note that $C=A\rtimes_{\rho, u}H$ and $D=B\rtimes_{\sigma, v}H$. We also note that $D$ is anti-isomorphic to ${}_C {\mathbf B}(Y)$, the $C^*$-algebra of all adjointable left $C$-modules maps on $Y$. We identify $D$ with ${}_C {\mathbf B}(Y)$. Hence for any $T, R\in {}_C {\mathbf B}(Y)$ we can write their product as follows: For any $y\in Y$, $$(TR)(y)=R(T(y)) .$$ For any $y, z\in Y$, let $\Theta_{y, z}$ be the rank-one left $C$-module map on $Y$ defined by $$\Theta_{y, z}(x)={}_C {\langle}x, z {\rangle}\cdot y$$ for any $x\in Y$. Then since $Y$ is of finite type in the sense of Kajiwara and Watatani [@KW1:bimodule], ${}_C {\mathbf B}(Y)$ is the linear span of the above rank-one left $C$-module maps on $Y$ by [@KW1:bimodule]. We define a bilinear map $``\rightharpoonup"$ from $H^0 \times D$ to $D$ so that $$\psi\rightharpoonup{\langle}y, z {\rangle}_D ={\langle}S^0 (\psi_{(1)}^* )\triangleright y \, , \, \psi_{(2)}\triangleright z {\rangle}_D$$ for any $\psi \in H^0$ and $y, z\in Y$. Hence since ${\langle}y, z {\rangle}_D =\Theta_{z, y}$ for any $y, z\in Y$, we define $``\rightharpoonup"$ as follows: For any $\psi\in H^0$, $y, z\in Y$, $$\psi\rightharpoonup \Theta_{y, z}=\Theta_{[\psi_{(2)}\triangleright y]\,, \,[S^0 (\psi_{(1)}^* )\triangleright z]} .$$

\[lem:identity\]With the above notations, for any $\psi\in H^0$ and $T\in {}_C {\mathbf B}(Y)$, $$[\psi\rightharpoonup T](x)=\psi_{(2)}\triangleright T(S^0 (\psi_{(1)})\triangleright x)$$ for any $x\in Y$.

Since ${}_C {\mathbf B}(Y)$ is the linear span of the rank-one left $C$-module maps on $Y$, it suffices to show that $$[\psi\rightharpoonup \Theta_{y, z}](x)=\psi_{(2)}\triangleright \Theta_{y, z}(S^0 (\psi_{(1)})\triangleright x)$$ for any $\psi\in H^0$ and $x, y, z\in Y$. For any $\psi\in H^0$ and $x, y, z\in Y$, $$\begin{aligned}
[\psi\rightharpoonup\Theta_{y, z}](x) & =\Theta_{[\psi_{(2)}\triangleright y] , [S^0 (\psi_{(1)}^* )\triangleright z]}(x)
={}_C {\langle}x \, , \, S^0 (\psi_{(1)}^* )\triangleright z {\rangle}\cdot [\psi_{(2)}\triangleright y] \\
& ={}_C {\langle}\psi_{(2)}S^0 (\psi_{(1)})\triangleright x \, , \, S^0 (\psi_{(3)}^* )\triangleright z {\rangle}\cdot [\psi_{(4)}\triangleright y] \\
& =[\psi_{(2)}\cdot_{\widehat{\rho}}{}_C {\langle}S^0 (\psi_{(1)})\triangleright x \, , \, z {\rangle}]\cdot [\psi_{(3)}\triangleright y] \\
& =\psi_{(2)}\triangleright ({}_C {\langle}S^0 (\psi_{(1)})\triangleright x \, ,\, z {\rangle}\cdot y) \\
& =\psi_{(2)}\triangleright \Theta_{y, z}(S^0 (\psi_{(1)})\triangleright x) .\end{aligned}$$ Hence we obtain the conclusion.

\[lem:action\]With the above notations, the bilinear map $`` \rightharpoonup "$ from $H^0 \times D$ to $D$ is an action of $H^0$ on $D$.

Let $T, R\in {}_C {\mathbf B}(Y)$ and $\psi, \chi \in H^0$. For any $y\in Y$, $$\begin{aligned}
([\psi_{(1)}\rightharpoonup T][\psi_{(2)}\rightharpoonup R])(y))
& =[\psi_{(2)}\rightharpoonup R]([\psi_{(1)}\rightharpoonup T](y)) \\
& =\psi_{(3)}\triangleright R(S^0 (\psi_{(2)})\triangleright[\psi_{(1)}\rightharpoonup T](y)) \\
& =\psi_{(4)}\triangleright R(S^0 (\psi_{(3)})\psi_{(2)}\triangleright T(S^0 (\psi_{(1)})\triangleright y)) \\
& =\psi_{(2)}\triangleright R(T(S^0 (\psi_{(1)})\triangleright y)) \\
& =\psi_{(2)}\triangleright(TR)(S^0 (\psi_{(1)})\triangleright y) \\
& =[\psi\rightharpoonup TR](y) .\end{aligned}$$ Hence we obtain that $\psi\rightharpoonup TR=[\psi_{(1)}\rightharpoonup T][\psi_{(2)}\rightharpoonup R]$. For any $y\in Y$, $$\begin{aligned}
[\psi\rightharpoonup[\chi\rightharpoonup T]](y) & =
\psi_{(2)}\triangleright[\chi\rightharpoonup T](S^0 (\psi_{(1)})\triangleright y) \\
& =\psi_{(2)}\triangleright\chi_{(2)}\triangleright T(S^0 (\chi_{(1)})\triangleright S^0 (\psi_{(1)})\triangleright y) \\
& =\psi_{(2)}\chi_{(2)}\triangleright T(S^0 (\psi_{(1)}\chi_{(1)})\triangleright y) \\
& =[\psi\chi\rightharpoonup T](y) .\end{aligned}$$ Hence we obtain that $\psi\rightharpoonup[\chi\rightharpoonup T]=\psi\chi\rightharpoonup T$. Let $I_Y$ be the identity map on $Y$. Then for any $y\in Y$ $$[\psi\rightharpoonup I_Y ](y)=\psi_{(2)}\triangleright I_Y (S^0 (\psi_{(1)})\triangleright y)
=\epsilon^0 (\psi)y .$$ Hence $\psi\rightharpoonup I_Y =\epsilon^0 (\psi)I_Y$. Also, for any $y\in Y$, $[1^0 \rightharpoonup T](y)=T(y)$. Thus $1^0 \rightharpoonup T=T$. Furthermore, for any $y, z\in Y$, $$\begin{aligned}
{}_C {\langle}[\psi\rightharpoonup T]^* (y) \, , \, z {\rangle}& ={}_C {\langle}y \, , \, [\psi\rightharpoonup T](z) {\rangle}\\
& ={}_C {\langle}y \, , \, \psi_{(2)}\triangleright T(S^0 (\psi_{(1)})\triangleright z) {\rangle}\\
& ={}_C {\langle}S^0 (\psi_{(3)}^* )\psi_{(4)}^* \triangleright y \, , \, \psi_{(2)}
\triangleright T(S^0 (\psi_{(1)})\triangleright z) {\rangle}\\
& = S^0 (\psi_{(2)}^* )\cdot_{\widehat{\rho}}\,{}_C {\langle}\psi_{(3)}^* \triangleright y \, , \, 
T(S^0 (\psi_{(1)})\triangleright z) {\rangle}\\
& = S^0 (\psi_{(2)}^* )\cdot_{\widehat{\rho}}\,{}_C {\langle}T^* (\psi_{(3)}^* \triangleright y) \, , \, 
S^0 (\psi_{(1)})\triangleright z {\rangle}\\
& ={}_C {\langle}S^0 (\psi_{(3)}^* )\triangleright T^* (\psi_{(4)}^* \triangleright y) \, , \, \psi_{(2)}S^0 (\psi_{(1)})
\triangleright z {\rangle}\\
& ={}_C {\langle}S^0 (\psi_{(1)}^* )\triangleright T^* (\psi_{(2)}^* \triangleright y) \, , \, z {\rangle}\\
& ={}_C {\langle}[S^0 (\psi^* )\rightharpoonup T^* ](y) \, ,\, z {\rangle}.\end{aligned}$$ Hence we can see that $[\psi\rightharpoonup T]^* =S^0 (\psi^* )\rightharpoonup T^* $. Therefore we obtain the conclusion.

By Lemma \[lem:action\], the map $``\rightharpoonup"$ is an action of $H^0$ on $D$. We denote by $\beta$ the coaction of $H$ on $D$ induced by the action $``\rightharpoonup"$ of $H^0$ on $D$. By the definition of the action $``\rightharpoonup"$, the left coaction $\mu$ of $H$ on $Y$ induced by the action $``\triangleright"$ is also a right coaction of $H$ on $Y$ with respect to $(D, \beta)$. Thus $\mu$ is a coaction of $H$ on $Y$ with respect to $(C, D, \widehat{\rho}, \beta)$. Hence we can see that $\widehat{\rho}$ is strongly Morita equivalent to $\beta$. By [@KT4:morita Section 4], the unital inclusions $C\subset C_1 (=C\rtimes_{\widehat{\rho}}H^0)$ and $D\subset D\rtimes_{\beta}H^0$ are strongly Morita equivalent with respect to the $C_1 -D\rtimes_{\beta}H^0$-equivalence bimodule $Y\rtimes_{\mu}H^0$ and its closed subspace $Y$. Since the unital inclusions $C\subset C_1$ and $D\subset D_1$ are also strongly Morita equivalent with respect to the $C_1 -D_1$-equivalence bimodule $Y_1$ and its closed subspace $Y$ by [@KT4:morita Corollary 6.3], we can see that the unital inclusions $D\subset D_1$ and $D\subset D\rtimes_{\beta}H^0 $ are strongly Morita equivalent with respect to the $D_1 -D\rtimes_{\beta}H^0$-equivalence bimodule $\widetilde{Y_1}\otimes_{C_1} (Y\rtimes_{\mu}H^0 )$ and its closed subspace $\widetilde{Y}\otimes_C Y\cong D$ by [@KT4:morita Proposition 2.2].

The exterior equivalence and the strong Morita equivalence {#sec:exterior}
==========================================================

First, we shall present some basic properties on coactions of a finite dimensional $C^*$-Hopf algebra on a unital $C^*$-algebra and their exterior equivalence and strong Morita equivalence.

Let $H$ and $K$ be finite dimensional $C^*$-Hopf algebras and let $H^0$ and $K^0$ be their dual $C^*$-Hopf algebras, respectively. Let $A$ and $B$ unital $C^*$-algebras and let $\pi$ be an isomorphism of $B$ onto $A$. Let $\lambda^0$ be a $C^*$-Hopf algebra isomorphism of $K^0$ onto $H^0$. Let $(\rho, u)$ be a coaction of $K^0$ on $B$ and let $\rho_{\pi, \lambda^0}$ be the homomorphism of $A$ to $A\otimes H^0$ defined by $$\rho_{\pi, \lambda^0}=(\pi\otimes \lambda^0 )\circ\rho\circ\pi^{-1} .$$ Let $u_{\pi, \lambda^0}$ be the unitary element in $A\otimes H^0 \otimes H^0$ defined by $$u_{\pi, \lambda^0}=(\pi\otimes\lambda^0 \otimes\lambda^0 )(u) .$$

\[lem:induced\]With the above notations, $(\rho_{\pi, \lambda^0} ,\, u_{\pi, \lambda^0})$ is a twisted coaction of $H^0$ on $A$.

This is immediate by routine computations.

We call the above pair $(\rho_{\pi, \lambda^0}, u_{\pi, \lambda^0})$ the twisted coaction of $H^0$ on $A$ induced by $(\rho, u)$ and $\pi$, $\lambda^0$.

Let $(\sigma, v)$ be a coaction of $K^0$ on $B$ and $(\sigma_{\pi, \lambda^0}, v_{\pi, \lambda^0})$ the twisted coaction of $H^0$ on $A$ induced by $(\sigma, v)$ and $\pi$, $\lambda^0$.

\[lem:exterior\]With the above notations, if $(\rho, u)$ is exterior equivalent $(\sigma, v)$, then $(\rho_{\pi, \lambda^0} ,\, u_{\pi, \lambda^0})$ is exterior equivalent to $(\sigma_{\pi, \lambda^0}, v_{\pi, \lambda^0})$.

Let $\Delta_H^0$ and $\Delta_K^0$ be the comultiplications of $H^0$ and $K^0$, respectively. Since $(\rho, u)$ and $(\sigma, v)$ are exterior equivalent, there is a unitary element $w\in B\otimes K^0$ such that $$\sigma ={{\rm{Ad}}}(w)\circ \rho, \quad
v=(w\otimes 1^0 )(\rho\otimes {{\rm{id}}}_{K^0})(w)u({{\rm{id}}}_B \otimes\Delta_K^0 )(w)^* .$$ Since $\rho=(\pi\otimes\lambda^0 )^{-1}\circ\rho_{\pi, \lambda^0}\circ\pi$ and $\sigma=(\pi\otimes\lambda^0 )^{-1}\circ\sigma_{\pi. \lambda^0 }\circ\pi$, $$(\pi\otimes\lambda^0 )^{-1}\circ\sigma_{\pi, \lambda^0}\circ\pi
={{\rm{Ad}}}(w)\circ(\pi\otimes\lambda^0 )^{-1}\circ\rho_{\pi, \lambda^0}\circ\pi .$$ Hence $$\sigma_{\pi, \lambda^0 }={{\rm{Ad}}}((\pi\otimes\lambda^0 )(w))\circ\rho_{\pi, \lambda^0 } .$$ Furthermore, $$\begin{aligned}
v_{\pi, \lambda^0}
=(\pi\otimes\lambda^0 \otimes\lambda^0 )(v) 
&=((\pi\otimes\lambda^0 )(w)\otimes 1^0 )(\pi\otimes\lambda^0 \otimes\lambda^0 )((\rho\otimes{{\rm{id}}}_{K^0})(w)) \\
& \times u_{\pi, \lambda^0 }(\pi\otimes\lambda^0 \otimes\lambda^0 )(({{\rm{id}}}_B \otimes\Delta_K^0 )(w^* )) .\end{aligned}$$ Since $(\lambda^0 \otimes\lambda^0 )\circ\Delta_K^0 =\Delta_H^0 \circ\lambda^0 $, $$(\pi\otimes\lambda^0 \otimes\lambda^0 )(({{\rm{id}}}_B \otimes\Delta_K^0 )(w^* ))
=({{\rm{id}}}_A \otimes\Delta_H^0 )((\pi\otimes\lambda^0 )(w^* )) .$$ 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 $A$-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 of Corollary 4.8].

Let $A$ and $H$, $H^0$ be as before. Let $(\rho, u)$ and $(\sigma, v)$ be twisted coactions of $H^0$ on $A$. Let $C=A\rtimes_{\rho, u}H$ and $D=A\rtimes_{\sigma, v}H$. We also regard $A$ as an $A-A$-equivalence bimodule in the usual way. We suppose that the unital inclusions $A\subset C$ and $A\subset D$ are strongly Morita equivalent with respect to a $C-D$-equivalence bimodule $Y$ and its closed subspace $A$, that is, we assume that the $A-A$-equivalence bimodule $A$ is included in $Y$ as a closed subspace. Furthermore, we suppose that $A'\cap C={\mathbf C}1$. Then by [@KT4:morita Lemma 10.3], $A' \cap D={\mathbf C}1$. Let $E_1^{\rho, u}$ and $E_1^{\sigma, v}$ be the canonical conditional expectations from $C$ and $D$ onto $A$ defined by $$E_1^{\rho, u}(a\rtimes_{\rho, u}h)=\tau(h)a , \quad
E_1^{\sigma, v}(a\rtimes_{\sigma, v}h)=\tau(h)a$$ for any $a\in A$, $h\in H$. By [@KT4:morita Theorem 2.7], there are a conditional expectation $F^A$ of Watatani index-finite type from $D$ onto $A$ and a conditional expectation $G^A$ from $Y$ onto $A$ with respect to $E_1^{\rho, u}$ and $F^A$. Since $A' \cap D={\mathbf C}1$, by Watatani [@Watatani:index Proposition 4.1], $F^A =E_1^{\sigma, v}$. Hence $G^A$ is a conditional expectation from $Y$ onto $A$ with respect to $E_1^{\rho, u}$ and $E_1^{\sigma, v}$. By the discussions in [@KT4:morita Section 2] and the proof of Rieffel [@Rieffel:rotation Proposition 2.1], there is an isomorphism $\Psi$ of $D$ onto $C$ defined by $$\Psi(d)={}_C {\langle}1_A \cdot d \, , \, 1_A {\rangle}$$ for any $d\in D$, where $A$ is a closed subspace of $Y$ and the unit element in $A$ is regarded as an element in $Y$. Then for any $a\in A$ $$\Psi(a)={}_A {\langle}1_A \cdot a \, , \, 1_A {\rangle}={}_A {\langle}a , \, 1_A {\rangle}=a .$$ Also, for any $d\in D$ $$\begin{aligned}
(E_1^{\rho, u}\circ\Psi)(d) & =E_1^{\rho, u}({}_C {\langle}1_A \cdot d \, , \, 1_A {\rangle})
={}_A {\langle}G^A (1_A \cdot d) \, , \, 1_A {\rangle}={}_A {\langle}1_A \cdot E_1^{\sigma, v}(d) \, , \, 1_A {\rangle}\\
& =E_1^{\sigma, v}(d) .\end{aligned}$$ Thus, we obtain the following lemma:

\[lem:equation\]With the above notations, we suppose that $A' \cap C={\mathbf C}1$. Then $\Psi$ is an isomorphism of $D$ onto $C$ satisfying that $$\Psi|_A ={{\rm{id}}}_A, \quad E_1^{\rho, u}\circ\Psi=E_1^{\sigma, v}$$

Let $\widehat{\rho}$ and $\widehat{\sigma}$ be the dual coactions of $H$ on $C$ and $D$ induced by the twisted coactions $(\rho, u)$ and $(\sigma, v)$, respectively and let $C_1 =C\rtimes_{\widehat{\rho}}H^0$ and $D_1 =D\rtimes_{\widehat{\sigma}}H^0$ the crossed products of $C$ and $D$ by the actions of $H^0$ on $C$ and $D$ induced by $\widehat{\rho}$ and $\widehat{\sigma}$, respectively. Similarly, we define $\widehat{\widehat{\rho}}$ and $\widehat{\widehat{\sigma}}$, $C_2 =C_1\rtimes_{\widehat{\widehat{\rho}}}H$ and $D_2 =D_1 \rtimes_{\widehat{\widehat{\sigma}}}H$, respectively. Let $E_2^{\rho, u}$ and $E_2^{\sigma, v}$ be the canonical conditional expectations from $C_1$ and $D_1$ onto $C$ and $D$ defined in the same way as above. Also, let $E_3^{\rho, u}$ and $E_3^{\sigma, v}$ be the canonical conditional expectations from $C_2$ and $D_2$ onto $C_1$ and $D_1$ defined in the same way as above, respectively. Using $E_1^{\rho, u}$ and $E_1^{\sigma, v}$, we regard $C$ and $D$ as right Hilbert $A$-modules, respectively. Let ${\mathbf B}_A (C)$ and ${\mathbf B}_A (D)$ be the $C^*$-algebras of all right $A$-module maps on $C$ and $D$, respectively. We note that any right module map in ${\mathbf B}_A (C)$ or ${\mathbf B}_A (D)$ is adjointable by Kajiwara and Watatani [@KW1:bimodule Lemma 1.10] since $C$ and $D$ are of finite index in the sense of [@KW1:bimodule]. For any $x, y\in C$, let $\theta_{x, y}^C$ be the rank-one $A$-module map on $C$ defined by $\theta_{x, y}^C (z)=x\cdot {\langle}y, z {\rangle}_A$ for any $z\in C$. Similarly, we define $\theta_{x,y}^D $ the rank-one $A$-module map on $D$ for any $x, y\in D$. Then ${\mathbf B}_A (C)$ and ${\mathbf B}_A (D)$ are the linear spans of the above rank-one $A$-module maps on $C$ and $D$, respectively. Furthermore, we define ${\mathbf B}_C (C_1 )$, ${\mathbf B}_D (D_1 )$ and $\theta_{x, y}^{C_1}$, $\theta_{x',y'}^{D_1}$ for $x, y\in C_1$, $x', y' \in D_1$.

\[lem:equation2\]With the above notations and assumptions, there is an isomorphism $\widehat{\Psi}$ of $D_1$ onto $C_1$ satisfying that $$\widehat{\Psi}|_D =\Psi, \quad \Psi\circ E_2^{\sigma, v}=E_2^{\rho, u}\circ\widehat{\Psi}, \quad
\widehat{\Psi}(1\rtimes_{\widehat{\sigma}}\tau)=1\rtimes_{\widehat{\rho}}\tau .$$

We note that $C_1 \cong{\mathbf B}_A (C)$ and $D_1 \cong {\mathbf B}_A (D)$, respectively. Then $\Psi$ can be regarded as a Hilbert $A-A$-bimodule isomorphism of $D$ onto $C$ since $E_1^{\rho, u}\circ\Psi=E_1^{\sigma, v}$. Let $\widehat{\Psi}$ be the map from ${\mathbf B}_A (D)$ to ${\mathbf B}_A (C)$ defined by $\widehat{\Psi}(T)=\Psi\circ T\circ\Psi^{-1}$ for any $T\in {\mathbf B}_A (D)$. By routine computations, $\widehat{\Psi}$ is an isomorphism of $D_1$ onto $C_1$. For any $x\in D$, let $T_x$ be the element in ${\mathbf B}_A (D)$ defined by $T_x (y)=xy$ for any $y\in D$. Then for any $x\in D$ and $z\in C$, $$\widehat{\Psi}(T_x )(z) =(\Psi\circ T_x \circ\Psi^{-1})(z) =
\Psi(x\Psi^{-1}(z))=\Psi(x)z 
=T_{\Psi(x)}(z) .$$ Hence $\widehat{\Psi}|_D =\Psi$. Also, since $1\rtimes_{\widehat{\rho}}\tau$ and $1\rtimes_{\widehat{\sigma}}\tau$ are identified with $\theta_{1,1}^C$ and $\theta_{1,1}^D$, respectively. For any $z\in C$, $$\begin{aligned}
\widehat{\Psi}(1\rtimes_{\widehat{\sigma}}\tau)(z) & =(\Psi\circ\theta_{1,1}^D \circ\Psi^{-1})(z)
=\Psi (1\cdot {\langle}1, \, \Psi^{-1}(z) {\rangle}_A )=\Psi(E_1^{\sigma, v}(\Psi^{-1}(z))) \\
& =E_1^{\sigma, v}(\Psi^{-1}(z))=E_1^{\rho, u}(z)=\theta_{1,1}^C (z)=(1\rtimes_{\widehat{\rho}}\tau)(z) .\end{aligned}$$ 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 $a\in A$, $$\begin{aligned}
(a\rtimes_{\sigma, v}1\rtimes_{\widehat{\sigma}}1^0 )x & =\sum_i (a\rtimes_{\sigma, v}1)x_i
\rtimes_{\widehat{\sigma}}\phi_i , \\
x(a\rtimes_{\sigma, v}1\rtimes_{\widehat{\sigma}}1^0 ) & =\sum_i (x_i \rtimes_{\widehat{\sigma}}\phi_i )
(a\rtimes_{\sigma, v}1\rtimes_{\widehat{\sigma}}1^0 ) \\
& =\sum_i x_i [\phi_{i (1)}\cdot_{\widehat{\sigma}}a\rtimes_{\sigma, v}1]\rtimes_{\widehat{\sigma}}\phi_{i(2)} \\
& =\sum_i x_i (a\rtimes_{\sigma, v}1)\rtimes_{\widehat{\sigma}}\phi_i .\end{aligned}$$ Since $(a\rtimes_{\sigma, v}1\rtimes_{\widehat{\sigma}}1^0 )x
=x(a\rtimes_{\sigma, v}1\rtimes_{\widehat{\sigma}}1^0 )$, $(a\rtimes_{\sigma, v}1)x_i =x_i (a\rtimes_{\sigma, v}1)$ for any $a\in A$ and $i$. Since $A' \cap D={\mathbf C}1$, $x_i \in {\mathbf C}1$ for any $i$. Therefore, $A' \cap D_1 \subset 1\rtimes_{\widehat{\sigma}}H^0$. The inverse inclusion is clear. Hence we obtain the conclusion.

By Lemma \[lem:Ciso\], $\widehat{\Psi}|_{A' \cap D_1}$ can be regarded as a $C^*$-algebra automorphism of $H^0$. We denote it by $\lambda^0$.

\[lem:Hopfiso\]With the above notations, $\lambda^0$ is a $C^*$-Hopf algebra automorphism of $H^0$.

It suffices to show that for any $\phi\in H^0$, $$S^0 (\lambda^0 (\phi))=\lambda^0 (S^0 (\phi)), \quad
\Delta^0 (\lambda^0 (\phi))=(\lambda^0 \otimes\lambda^0 )(\Delta^0 (\phi)), \quad
\epsilon^0 (\lambda^0 (\phi))=\epsilon^0 (\phi) .$$ We note that for any $h\in H$, $\phi\in H^0$, $$\phi(h)=N^2 (E_2^{\rho, u}\circ E_3^{\rho, u})((1\rtimes_{\widehat{\rho}}\phi\rtimes_{\widehat{\widehat{\rho}}}1)
(1\rtimes_{\widehat{\rho}}1^0 \rtimes_{\widehat{\widehat{\rho}}}e)
(1\rtimes_{\widehat{\rho}}\tau\rtimes_{\widehat{\widehat{\rho}}}1)
(1\rtimes_{\widehat{\rho}}1^0\rtimes_{\widehat{\widehat{\rho}}}h))$$ by easy computations. Indeed, $$\begin{aligned}
& N^2 (E_2^{\rho, u}\circ E_3^{\rho, u})((1\rtimes_{\widehat{\rho}}\phi\rtimes_{\widehat{\widehat{\rho}}}1)
(1\rtimes_{\widehat{\rho}}1^0 \rtimes_{\widehat{\widehat{\rho}}}e)
(1\rtimes_{\widehat{\rho}}\tau\rtimes_{\widehat{\widehat{\rho}}}1)
(1\rtimes_{\widehat{\rho}}1^0\rtimes_{\widehat{\widehat{\rho}}}h)) \\
& =N^2 (E_2^{\rho, u}\circ E_3^{\rho, u})((1\rtimes_{\widehat{\rho}}\phi\rtimes_{\widehat{\widehat{\rho}}}e)
(1\rtimes_{\widehat{\rho}}\tau\rtimes_{\widehat{\widehat{\rho}}}h)) \\
& =N^2 (E_2^{\rho, u}\circ E_3^{\rho, u})((1\rtimes_{\widehat{\rho}}\phi)[e_{(1)}\cdot_{\widehat{\widehat{\rho}}}
(1\rtimes_{\widehat{\rho}}\tau)]\rtimes_{\widehat{\widehat{\rho}}}e_{(2)}h) \\
& =N^2 E_2^{\rho, u}((1\rtimes_{\widehat{\rho}}\phi)[e_{(1)}\cdot_{\widehat{\widehat{\rho}}}
(1\rtimes_{\widehat{\rho}}\tau)])
\tau' (e_{(2)}h) \\
& =N^2 E_2^{\rho, u}((1\rtimes_{\widehat{\rho}}\phi)(1\rtimes_{\widehat{\rho}}\tau_{(1)})\tau_{(2)}(e_{(1)}))
\tau' (e_{(2)}h) \\
& =N^2 E_2^{\rho, u}(1\rtimes_{\widehat{\rho}}\phi\tau_{(1)})\tau_{(2)}(e_{(1)})\tau' (e_{(2)}h) \\
& =N^2 (\phi\tau_{(1)})(e' )\tau_{(2)}(e_{(1)})\tau' (e_{(2)}h) \\
& =N^2 (\phi\tau_{(1)})(e' )\tau_{(2)}(e_{(1)}h_{(2)}S(h_{(1)}))\tau' (e_{(2)}h_{(3)}) \\
& =N^2 (\phi\tau_{(1)})(e' )\tau_{(2)}(e_{(1)}h_{(2)})\tau_{(3)}(S(h_{(1)}))\tau' (e_{(2)}h_{(3)}) \\
& =N^2 (\phi\tau_{(1)})(e' )(\tau_{(2)}\tau' )(eh_{(2)})\tau_{(3)}(S(h_{(1)})) \\
& =N(\phi\tau_{(1)})(e' )\tau_{(2)}(S(h)) \\
& =N(\phi_{(1)}\tau_{(1)})(e' )(S^0 (\phi_{(3)})\phi_{(2)}\tau_{(2)})(S(h)) \\
& =N(\phi_{(1)}\tau_{(1)})(e' )S^0 (\phi_{(3)})(S(h_{(2)}))(\phi_{(2)}\tau_{(2)})(S(h_{(1)})) \\
& =N(\phi_{(1)}\tau)(e' S(h_{(1)}))S^0 (\phi_{(2)})(S(h_{(2)})) \\
& =S^0 (\phi)(S(h)) \\
& =\phi(h) ,\end{aligned}$$ where $e' =e$ and $\tau' =\tau$. We also note that $$\phi(h)=N^2 (E_2^{\sigma, v}\circ E_3^{\sigma, v})
((1\rtimes_{\widehat{\sigma}}\phi\rtimes_{\widehat{\widehat{\sigma}}}1)
(1\rtimes_{\widehat{\sigma}}1^0 \rtimes_{\widehat{\widehat{\sigma}}}e)
(1\rtimes_{\widehat{\sigma}}\tau\rtimes_{\widehat{\widehat{\sigma}}}1^0 )
(1\rtimes_{\widehat{\sigma}}1^0 \rtimes_{\widehat{\widehat{\sigma}}}h))$$ for any $h\in H$, $\phi\in H^0$. Hence for any $h\in H$, $\phi\in H^0$, $$\begin{aligned}
& \lambda^0 (\phi)(h) \\
&  =N^2 (E_2^{\rho, u}\circ E_3^{\rho, u})(\widehat{\widehat{\Psi}}
(1\rtimes_{\widehat{\sigma}}\phi\rtimes_{\widehat{\widehat{\sigma}}}1)
(1\rtimes_{\widehat{\rho}}1^0 \rtimes_{\widehat{\widehat{\rho}}}e)
(1\rtimes_{\widehat{\rho}}\tau\rtimes_{\widehat{\widehat{\rho}}}1)
(1\rtimes_{\widehat{\rho}}1^0 \rtimes_{\widehat{\widehat{\rho}}}h)) \\
& =N^2 (E_2^{\rho, u}\circ E_3^{\rho, u}\circ\widehat{\widehat{\Psi}})
((1\rtimes_{\widehat{\sigma}}\phi\rtimes_{\widehat{\widehat{\sigma}}}1)
(1\rtimes_{\widehat{\sigma}}1^0 \rtimes_{\widehat{\widehat{\sigma}}}e)
(1\rtimes_{\widehat{\sigma}}\tau\rtimes_{\widehat{\widehat{\sigma}}}1) \\
& \times\widehat{\widehat{\Psi}}^{-1}(1\rtimes_{\widehat{\rho}}1^0 \rtimes_{\widehat{\widehat{\rho}}}h)) \\
& =N^2 (\Psi\circ E_2^{\sigma, v}\circ E_3^{\sigma, v})
((1\rtimes_{\widehat{\sigma}}\phi\rtimes_{\widehat{\widehat{\sigma}}}1)
(1\rtimes_{\widehat{\sigma}}1^0 \rtimes_{\widehat{\widehat{\sigma}}}e)
(1\rtimes_{\widehat{\sigma}}\tau\rtimes_{\widehat{\widehat{\sigma}}}1) \\
& \times\widehat{\widehat{\Psi}}^{-1}(1\rtimes_{\widehat{\rho}}1^0 \rtimes_{\widehat{\widehat{\rho}}}h)) \\
& =\Psi(\phi(\widehat{\widehat{\Psi}}^{-1}(h)))=\phi(\widehat{\widehat{\Psi}}^{-1}(h))\end{aligned}$$ by Lemmas \[lem:equation2\] and \[lem:equation3\]. Hence by the above equation, for any $h\in H$, $\phi\in H^0$, $$\begin{aligned}
S^0 (\lambda^0 (\phi))(h) & =\lambda^0 (\phi)(S(h))=\overline{\lambda^0 (\phi)^* (h^* )}
=\overline{\widehat{\Psi}(\phi^* )(h^* )}=\overline{\lambda^0 (\phi^* )(h^* )} \\
& =\overline{\phi^*(\widehat{\widehat{\Psi}}^{-1}(h^* ))}=\phi(S(\widehat{\widehat{\Psi}}^{-1}(h)))
=S^0 (\phi)(\widehat{\widehat{\Psi}}^{-1}(h)) =\lambda^0 (S^0 (\phi))(h) .\end{aligned}$$ Thus $S^0 (\lambda^0 (\phi))=\lambda^0 (S^0 (\phi))$ for any $\phi\in H^0 $. Also, for any $h, l\in H$, $\phi\in H^0$, $$\begin{aligned}
\Delta^0 (\lambda^0 (\phi))(h\otimes l) & = \lambda^0 (\phi)(hl)=\phi(\widehat{\widehat{\Psi}}^{-1}(hl))
=\Delta^0 (\phi)(\widehat{\widehat{\Psi}}^{-1}(h)\otimes\widehat{\widehat{\Psi}}^{-1}(l)) \\
& =\phi_{(1)}(\widehat{\widehat{\Psi}}^{-1}(h))\phi_{(2)}(\widehat{\widehat{\Psi}}^{-1}(l))
=\lambda^0 (\phi_{(1)})(h)\lambda^0 (\phi_{(2)})(l) \\
& =(\lambda^0 (\phi_{(1)})\otimes\lambda^0 (\phi_{(2)}))(h\otimes l) \\
& =(\lambda^0 \otimes\lambda^0 )(\Delta^0 (\phi))(h\otimes l) .\end{aligned}$$ Hence $\Delta^0 (\lambda^0 (\phi))=(\lambda^0 \otimes\lambda^0 )(\Delta^0 (\phi))$ for any $\phi\in H^0$. Furthermore, for any $\phi\in H^0 $, $$\epsilon^0 (\lambda^0 (\phi))=\lambda^0 (\phi)(1)=\phi(\widehat{\widehat{\Psi}}^{-1}(1))=\phi(1)=\epsilon(1) .$$ Therefore, we obtain the conclusion.

\[lem:conjugate\]With the above notations, $$\widehat{\widehat{\rho}}\circ\widehat{\Psi}=(\widehat{\Psi}\otimes\lambda^0 )\circ\widehat{\widehat{\sigma}}$$ on $D_1$.

For any $x\in D$, $$\begin{aligned}
(\widehat{\widehat{\rho}}\circ\widehat{\Psi})(x\rtimes_{\widehat{\sigma}}1^0 ) &=\widehat{\widehat{\rho}}
(\Psi(x)\rtimes_{\widehat{\rho}}1^0 )= (\Psi(x)\rtimes_{\widehat{\sigma}}1^ 0)\otimes 1^0
=\widehat{\Psi}(x\rtimes_{\widehat{\sigma}}1^0 )\otimes 1^0 \\
& =(\widehat{\Psi}\otimes\lambda^0 )((x\rtimes_{\widehat{\sigma}}1^0 )\otimes 1^0 )
=((\widehat{\Psi}\otimes\lambda^0 )\circ\widehat{\widehat{\sigma}})(x\rtimes_{\widehat{\sigma}}1^0 )\end{aligned}$$ by Lemma \[lem:equation2\]. Also, for any $\phi\in H^0$, $$\begin{aligned}
(\widehat{\widehat{\rho}}\circ\widehat{\Psi})(1\rtimes_{\widehat{\sigma}}\phi) & =
\widehat{\widehat{\rho}}(1\rtimes_{\widehat{\rho}}\lambda^0 (\phi))=
(1\rtimes_{\widehat{\rho}}\lambda^0 (\phi)_{(1)})\otimes\lambda(\phi)_{(2)} \\
& =(1\rtimes_{\widehat{\rho}}\lambda^0 (\phi_{(1)}))\otimes\lambda^0 (\phi_{(2)}) 
=\widehat{\Psi}(1\rtimes_{\widehat{\sigma}}\phi_{(1)})\otimes\lambda^0 (\phi_{(2)}) \\
& =(\widehat{\Psi}\otimes\lambda^0 )((1\rtimes_{\widehat{\sigma}}\phi_{(1)})\otimes\phi_{(2)})
=((\widehat{\Psi}\otimes\lambda^0 )\circ\widehat{\widehat{\sigma}})(1\rtimes_{\widehat{\sigma}}\phi)\end{aligned}$$ by Lemma \[lem:Hopfiso\]. Therefore, we obtain that $\widehat{\widehat{\rho}}\circ\widehat{\Psi}=(\widehat{\Psi}\otimes\lambda^0 )\circ\widehat{\widehat{\sigma}}$.

\[prop:key2\]Let $(\rho, u)$ and $(\sigma, v)$ be twisted coactions of $H^0$ on $A$. Let $C=A\rtimes_{\rho, u}H$ and $D=A\rtimes_{\sigma, v}H$. We suppose that there is an isomorphism $\Psi$ of $C$ onto $D$ satisfying that $$\Psi|_A ={{\rm{id}}}_A , \quad E_1^{\rho, u}\circ\Psi=E_1^{\sigma, v} ,$$ where $E_1^{\rho, u}$ and $E_1^{\sigma, v}$ are the canonical conditional expectations from $C$ and $D$ onto $A$ defined by $$E_1^{\rho, u}(a\rtimes_{\rho, u}h)=\tau(h)a , \quad E_1^{\sigma, v}(a\rtimes_{\sigma, v}h)=\tau(h)a$$ for any $a\in A$, $h\in H$, respectively. Furthermore, we suppose that $A' \cap C={\mathbf C}1$. Then there is a $C^*$-Hopf algebra automorphism $\lambda^0$ of $H^0$ such that $(\rho, u)$ is strongly Morita equivalent to the twisted coaction $(({{\rm{id}}}_A \otimes \lambda^0 )\circ \sigma,\, ({{\rm{id}}}_A \otimes \lambda^0 \otimes \lambda^0 )(v))$.

By [@KT2:coaction Theorem 3.3], there are isomorphism of $\Phi_{\sigma}$ of $A\otimes M_N ({\mathbf C})$ onto $D_1$ and a unitary element $U_{\sigma}\in D_1 \otimes H^0$ such that $$\begin{aligned}
{{\rm{Ad}}}(U_{\sigma})\circ\widehat{\widehat{\sigma}} & =(\Phi_{\sigma}\otimes{{\rm{id}}}_{H^0})
\circ(\sigma\otimes{{\rm{id}}}_{M_N ({\mathbf C})})\circ\Phi_{\sigma}^{-1} , \\
(\Phi_{\sigma}\otimes{{\rm{id}}}_{H^0}\otimes{{\rm{id}}}_{H^0})(v\otimes I_N) & =(U_{\sigma}\otimes 1^0 )
(\widehat{\widehat{\sigma}}\otimes{{\rm{id}}}_{H^0})(U_{\sigma})({{\rm{id}}}\otimes\Delta^0 )(U_{\sigma}^* ) ,\end{aligned}$$ where we identify $M_N ({\mathbf C})\otimes H^0$ with $H^0 \otimes M_N ({\mathbf C})$. Thus by Lemma \[lem:exterior\], $({{\rm{id}}}_{D_1}\otimes\lambda^0 )\circ\widehat{\widehat{\sigma}}$ is exterior equivalent to the twisted coaction $$\begin{aligned}
(({{\rm{id}}}_{D_1}\otimes\lambda^0 )& \circ(\Phi_{\sigma}\otimes{{\rm{id}}}_{H^0 })
\circ(\sigma\otimes{{\rm{id}}}_{M_N ({\mathbf C})})\circ\Phi_{\sigma}^{-1} \, ,  \\
& ({{\rm{id}}}_{D_1}\otimes\lambda^0 \otimes\lambda^0 )((\Phi_{\sigma}\otimes{{\rm{id}}}_{H^0}\otimes{{\rm{id}}}_{H^0})(v\otimes I_N ))) ,\end{aligned}$$ where $\lambda^0 $ is the $C^*$-Hopf automorphism of $H^0$ defined before Lemma \[lem:Ciso\]. Since $$\begin{aligned}
& ({{\rm{id}}}_{D_1}\otimes\lambda^0 ) \circ(\Phi_{\sigma}\otimes{{\rm{id}}}_{H^0 })
\circ(\sigma\otimes{{\rm{id}}}_{M_N ({\mathbf C})})\circ\Phi_{\sigma}^{-1} \\
& =(\Phi_{\sigma}\otimes{{\rm{id}}}_{H^0})\circ({{\rm{id}}}_{A\otimes M_N ({\mathbf C})}\otimes\lambda^0 )
\circ(\sigma\otimes{{\rm{id}}}_{M_N ({\mathbf C})})\circ\Phi_{\sigma}^{-1},\end{aligned}$$ and $$\begin{aligned}
& ({{\rm{id}}}_{D_1}\otimes\lambda^0 \otimes\lambda^0 )
((\Phi_{\sigma}\otimes{{\rm{id}}}_{H^0}\otimes{{\rm{id}}}_{H^0})(v\otimes I_N ))) \\
& =(\Phi_{\sigma}\otimes{{\rm{id}}}_{H^0}\otimes{{\rm{id}}}_{H^0})
(({{\rm{id}}}_{A\otimes M_N ({\mathbf C})}\otimes\lambda^0 \otimes\lambda^0 )(v\otimes I_N )) ,\end{aligned}$$ by Lemma \[lem:iso\], the twisted coaction $$(({{\rm{id}}}_{A\otimes M_N ({\mathbf C})}\otimes\lambda^0 )
\circ(\sigma\otimes{{\rm{id}}}_{M_N ({\mathbf C})}) \, , \,
({{\rm{id}}}_{A\otimes M_N ({\mathbf C})}\otimes\lambda^0 \otimes\lambda^0 )(v\otimes I_N ))$$ is strongly Morita equivalent to the twisted coaction $$\begin{aligned}
(({{\rm{id}}}_{D_1}\otimes\lambda^0 )& \circ(\Phi_{\sigma}\otimes{{\rm{id}}}_{H^0 })
\circ(\sigma\otimes{{\rm{id}}}_{M_N ({\mathbf C})})\circ\Phi_{\sigma}^{-1} \, ,  \\
& ({{\rm{id}}}_{D_1}\otimes\lambda^0 \otimes\lambda^0 )((\Phi_{\sigma}\otimes{{\rm{id}}}_{H^0}\otimes{{\rm{id}}}_{H^0})(v\otimes I_N ))) ,\end{aligned}$$ Hence $({{\rm{id}}}_{D_1}\otimes\lambda^0 )\circ\widehat{\widehat{\sigma}}$ is strongly Morita equivalent to $$(({{\rm{id}}}_{A\otimes M_N ({\mathbf C})}\otimes\lambda^0 )
\circ(\sigma\otimes{{\rm{id}}}_{M_N ({\mathbf C})}) \, , \,
({{\rm{id}}}_{A\otimes M_N ({\mathbf C})}\otimes\lambda^0 \otimes\lambda^0 )(v\otimes I_N )) .$$ Thus by [@Kodaka:equivariance Lemma 4.10], $({{\rm{id}}}_{D_1}\otimes\lambda^0 )\circ\widehat{\widehat{\sigma}}$ is strongly Morita equivalent to $$(({{\rm{id}}}_A \otimes\lambda^0 )\circ\sigma \, , \, 
({{\rm{id}}}_A \otimes\lambda^0 \otimes \lambda^0 )(v)) .$$ On the other hand, by Lemmas \[lem:exterior\] and \[lem:conjugate\], $\widehat{\widehat{\rho}}$ is strongly Moriat equivalent to $({{\rm{id}}}_{D_1}\otimes\lambda^0 )\circ\widehat{\widehat{\sigma}}$. Since $\widehat{\widehat{\rho}}$ is strongly Morita equivalent to $(\rho, u)$ by Remark \[remark:review\], $(\rho, u)$ is strongly Morita equivalent to $$(({{\rm{id}}}_A \otimes\lambda^0 )\circ\sigma \, , \, 
({{\rm{id}}}_A \otimes\lambda^0 \otimes \lambda^0 )(v)) .$$

\[cor:key3\]Let $(\rho, u)$ and $(\sigma, v)$ be twisted coactions of $H^0$ on $A$. Let $C=A\rtimes_{\rho, u}H$ and $D=A\rtimes_{\sigma, v}H$. We regard $A$ as an $A-A$equivalent bimodule in the usual way. We suppose that the unital inclusions $A\subset C$ and $A\subset D$ are strongly Morita equivalent with respect to a $C-D$-equivalence bimodule $Y$ and its closed subspace $A$, that is, we assume that the $A-A$-equivalence bimodule $A$ is included in $Y$. Furthermore, we suppose that $A' \cap C={\mathbf C}1$. Then there is a $C^*$-Hopf algebra automorphism $\lambda^0$ of $H^0$ such that $(\rho, u)$ is strongly Morita equivalent to the twisted coaction $$(({{\rm{id}}}_A \otimes \lambda^0 )\circ\sigma ,\, ({{\rm{id}}}_A \otimes \lambda^0 \otimes \lambda^0 )(v)) .$$

This is immediate by Proposition \[prop:key2\] and the discussions before Lemma \[lem:equation\].

The main result {#sec:main}
===============

In this section, we present the main result in the paper. We recall the previous discussions: Let $A$ and $B$ be unital $C^*$-algebras and $H$ a finite dimensional $C^*$-Hopf algebra with its dual $C^*$-Hopf algebra $H^0$. Let $(\rho, u)$ and $(\sigma, v)$ be twisted coactions of $H^0$ on $A$ and $B$, respectively. Le $C=A\rtimes_{\rho, u}H$ and $D=B\rtimes_{\sigma, v}H$ and let $C_1 =C\rtimes_{\widehat{\rho}}H^0$ and $D_1 =D\rtimes_{\widehat{\sigma}}H^0$, where $\widehat{\rho}$ and $\widehat{\sigma}$ are the dual coactions of $H$ on $C$ and $D$ induced by $(\rho, u)$ and $(\sigma, v)$, respectively. W suppose that the inclusions $A\subset C$ and $B\subset D$ are strongly Morita equivalent with respect to a $C-D$-equivalence bimodule $Y$ and its closed subspace $X$ in the sense of [@KT4:morita Definition 2.1]. Furthermore, we suppose that $A' \cap C={\mathbf C}1$. Then by Proposition \[prop:leftcoaction\] and Lemma \[lem:action\] there are a coaction $\beta$ of $H$ on $D$ and a coaction $\mu$ of $H$ on $Y$ such that $(C, D, Y, \widehat{\rho}, \beta, \mu, H)$ is a covariant system. Hence the coactions $\widehat{\rho}$ and $\beta$ of $H$ on $D$ are strongly Morita equivalent. Moreover, by the discussions at the end of Section \[sec:coaction\], the unital inclusions $D\subset D_1 (=D\rtimes_{\widehat{\sigma}}H^0 )$ and $D\subset D\rtimes_{\beta}H^0 $ are strongly Morita equivalent with respect to a $D_1 -D\rtimes_{\beta}H^0$-equivalence bimodule $W(=\widetilde{Y_1}\otimes_{C_1}(Y\rtimes_{\mu}H^0 ))$ and its closed subspace $D$, where we regard $D$ as a $D-D$-equivalence bimodule in the usual way and we can also see that $D$ is included in $W$ by the discussions at the end of Section \[sec:coaction\]. Since $A' \cap C={\mathbf C}1$, by [@KT4:morita Lemma 10.3] and $D' \cap D_1 ={\mathbf C}1$. Thus by Corollary \[cor:key3\], there is a $C^*$-Hopf algebra automorphism $\lambda$ of $H$ such that $\beta$ is strongly Morita equivalent to $({{\rm{id}}}_D \otimes\lambda)\circ\widehat{\sigma}$. Hence $\widehat{\rho}$ is strongly Morita equivalent to $({{\rm{id}}}_D \otimes\lambda)\circ\widehat{\sigma}$. Let $\lambda^0$ be the $C^*$-Hopf algebra automorphism of $H^0$ induced by $\lambda$, that is, we define $\lambda^0$ as follows: For any $\psi\in H^0$, $h\in H$, $$\lambda^0 (\psi)(h)=\psi(\lambda^{-1}(h)) .$$ Let $$\sigma_{\lambda^0}=({{\rm{id}}}_B \otimes\lambda^0 )\circ\sigma , \quad
v_{\lambda^0}=({{\rm{id}}}_B \otimes\lambda^0 \otimes\lambda^0 )(v) .$$ Then by Lemma \[lem:induced\], $(\sigma_{\lambda^0 }, v_{\lambda^0})$ is a twisted coaction of $H^0$ on $B$. Let $D_{\lambda^0} =B\rtimes_{\sigma_{\lambda^0}, v_{\lambda^0}}H$.

\[lem:iso3\]With the above notations, let $\pi$ be the map from $D$ to $D_{\lambda^0}$ defined by $$\pi(b\rtimes_{\sigma, v}h)=b\rtimes_{\sigma_{\lambda^0}, v_{\lambda^0}}\lambda(h)$$ for any $b\in B$, $h\in H$. Then $\pi$ is an isomorphism of $D$ onto $D_{\lambda^0}$ such that $$\widehat{\sigma_{\lambda^0}}\circ\pi=(\pi\otimes{{\rm{id}}}_{H})\circ({{\rm{id}}}_D \otimes\lambda)\circ\widehat{\sigma} ,$$ where $\widehat{\sigma_{\lambda^0}}$ is the dual coaction of $(\sigma_{\lambda^0}, v_{\lambda^0})$, a coaction of $H$ on $D_{\lambda^0}$.

This is immediate by routine computations.

\[thm:main\]Let $A$ and $B$ be unital $C^*$-algebras and $H$ a finite dimensional $C^*$-Hopf algebra with its dual $C^*$-Hopf algebra $H^0$. Let $(\rho, u)$ and $(\sigma, v)$ be twisted coactions of $H^0$ on $A$ and $B$, respectively. Let $C=A\rtimes_{\rho, u}H$ and $D=B\rtimes_{\sigma ,v}H$. We suppose that the unital inclusions $A\subset C$ and $B\subset D$ are strongly Morita equivalent. Also, we suppose that $A' \cap C={\mathbf C}1$. Then there is a $C^*$-Hopf algebra automorphism $\lambda^0 $ of $H^0$ such that the twisted coaction $(\rho, u)$ is strongly Morita equivalent to the twisted coaction $$(({{\rm{id}}}_B \otimes \lambda^0 )\circ\sigma , \, ({{\rm{id}}}_B \otimes\lambda^0 \otimes\lambda^0 )(v)) ,$$ induced by $(\sigma, v)$ and $\lambda^0$.

By the discussions at the beginning of this section, the dual coaction $\widehat{\rho}$ of $(\rho, u)$ is strongly Morita equivalent to $({{\rm{id}}}_D \otimes\lambda)\circ\widehat{\sigma}$. Also, by Lemmas \[lem:iso\] and \[lem:iso3\], $({{\rm{id}}}_D \otimes\lambda)\circ\widehat{\sigma}$ is strongly Morita equivalent to $(({{\rm{id}}}_B \otimes \lambda^0 )\circ\sigma)\,\widehat{}$, the dual coaction of the twisted coaction $$(({{\rm{id}}}_B \otimes\lambda^0 )\circ\sigma , \, ({{\rm{id}}}\otimes\lambda^0 \otimes\lambda^0 )(v)) ,$$ where $\lambda^0$ is the $C^*$-Hopf algebra automorphism of $H^0$ induced by $\lambda$. Hence by [@KT3:equivalence Corollary 4.8], $(\rho, u)$ is strongly Morita equivalent to $$(({{\rm{id}}}_B \otimes\lambda^0 )\circ\sigma , \, ({{\rm{id}}}\otimes\lambda^0 \otimes\lambda^0 )(v)) ,$$ induced by $(\sigma, v)$ and $\lambda^0$.

In the rest of the paper, we show the inverse implication of Theorem \[thm:main\].

\[lem:inverse1\]Let $A\subset C$ and $B\subset D$ be unital inclusions of unital $C^*$-algebras. We suppose that there is an isomorphism $\pi$ of $D$ onto $C$ such that $\pi|_B$ is an isomorphism of $B$ onto $A$. Then $A\subset C$ and $B\subset D$ are strongly Morita equivalent.

Let$Y_{\pi}$ be the $D-C$-equivalence bimodule induced by $\pi$, that is, $Y_{\pi}=D$ as vector spaces and the left $D$-action on $Y_{\pi}$ and the left $D$-valued inner product are defined in the evident way. We define the right $C$-action and the right $C$-valued inner product as follows: For any $c\in C$ and $y, z\in Y_{\pi}$, $$y\cdot c=y\pi^{-1}(c), \quad {\langle}y, z {\rangle}_C=\pi(y^* z) .$$ Then $B$ is a closed subset of $Y_{\pi}$ and $Y_{\pi}$ and $B$ satisfy Conditions (1), (2) in [@KT4:morita Definition 2.1] by easy computations. Therefore, $A\subset C$ and $B\subset D$ are strongly Morita equivalent.

Let $(\rho, u)$ be a twisted coaction of $H^0$ on $A$ and $\lambda^0$ a $C^*$-Hopf algebra automorphism of $H^0$. Let $(\rho_{\lambda^0}, \, u_{\lambda^0})$ be the twisted coaction of $H^0$ on $A$ induced by $(\rho, u)$ and $\lambda^0$, that is $$\rho_{\lambda^0}=({{\rm{id}}}\otimes\lambda^0 )\circ\rho, \quad u_{\lambda^0}=({{\rm{id}}}\otimes\lambda^0 \otimes\lambda^0 )(u) .$$

\[lem:inverse2\]With the above notations, the unital inclusions $A\subset A\rtimes_{\rho, u}H$ and $A\subset A\rtimes_{\rho_{\lambda^0}, u_{\lambda^0}}H$ are strongly Morita equivalent.

Let $\pi$ be the map from $A\rtimes_{\rho, u}H$ to $A\rtimes_{\rho_{\lambda^0}, u_{\lambda^0}}H$ defined by $$\pi(a\rtimes_{\rho, u}h)=a\rtimes_{\rho_{\lambda^0}, u_{\lambda^0}}\lambda(h)$$ for any $a\in A$, $h\in H$, where $\lambda$ is the $C^*$-Hopf algebra automorphism of $H$ induced by $\lambda^0$, that is, $$\lambda^0 (\psi)(h)=\psi(\lambda^{-1}(h))$$ for any $\psi\in H^0$ and $h\in H$. By easy computations, $\pi$ is an isomorphism of $A\rtimes_{\rho, u}H$ onto $A\rtimes_{\rho_{\lambda^0}, u_{\lambda^0}}H$. Also, for any $a\in A$, $$\pi(a\rtimes_{\rho, u}1) =a\rtimes_{\rho_{\lambda^0}, u_{\lambda^0 }}\lambda(1)
=a\rtimes_{\rho_{\lambda^0}, u_{\lambda^0}}\lambda(1)=a\rtimes_{\rho_{\lambda^0}, u_{\lambda^0}}1 .$$ Hence $\pi_A ={{\rm{id}}}_A$. Thus, by Lemma \[lem:inverse1\], we obtain the conclusion.

\[cor:iff\]Let $A$ and $B$ be unital $C^*$-algebras and $H$ a finite dimensional $C^*$-Hopf algebra wit hits dual $C^*$-Hopf algebra $H^0$. Let $(\rho, u)$ and $(\sigma, u)$ be twisted coaction of $H^0$ on $A$ and $B$, respectively. Let $C=A\rtimes_{\rho, u}H$ and $D=B\rtimes_{\sigma, v}H$. We suppose that $A' \cap C={\mathbf C}1$. Then the following conditions are equivalent: $(1)$ The unital inclusions $A\subset C$ and $B\subset D$ are strongly Morita equivalent, $(2)$ There is a $C^*$-Hopf algebra automorphism $\lambda^0$ of $H^0$ such that the twisted coaction $(\rho, u)$ is strongly Morita equivalent to the twisted coaction $$(({{\rm{id}}}_B \otimes\lambda^0 )\circ\sigma \, , \, ({{\rm{id}}}_B \otimes\lambda^0 \otimes\lambda^0 )(v))$$ induced by $(\sigma, v)$ and $\lambda^0$.

This is immediate by Theorem \[thm:main\] and Lemmas \[lem:inverse1\], \[lem:inverse2\].

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