2. Preliminaries

In this paper, by a ring we mean an associative ring.

Definition 2.1. A ring with an involution is a pair $(R,*)$ , where $R$ is a ring and $*\;:\;R\to R, a\mapsto a^*$ is a mapping such that

(2.1) \begin{equation} (a+b)^* = a^* + b^*,\;\; (a^*)^* = a, \;\; (ab)^* = b^*a^* \end{equation}

for all $a,b\in R$ . Note that $0^*=(0+0)^* = 0^*+0^*$ implies $0^*=0$ .

Let $(R,*)$ and $(S,\star )$ be rings with involution. A ring homomorphism $f\colon R \to S$ is called a homomorphism of rings with involution, if $f(r^*) = f(r)^\star$ holds for every $r\in R$ .

We recall that a right $R$ module $M_R$ is called unitary if $MR=M$ , and firm if the natural mapping $\mu _M\;:\; M\otimes _R R \to M, m\otimes r\mapsto mr$ is bijective (see [20]). A bimodule is unitary (firm) if it is unitary (firm) as a left and a right module. Note that $MR=M$ if and only if $\mu _M$ is surjective; hence, every firm module is unitary. Recall that a ring $R$ is called idempotent if $R=RR$ .

Rieffel introduced imprimitivity bimodules over $C^*$ -algebras ([Reference Rieffel22, Definition 6.10]) and used them to develop Morita theory for $C^*$ -algebras ([Reference Brown, Green and Rieffel4, Reference Rieffel21]). Inspired by this notion we give the following definition.

In addition to the axioms listed in Definition 2.2, Rieffel contexts have some more calculation rules.

Proof. The proof of RC6 and RC7 is inspired by [Reference Steinberg24, Proposition 2.3], and the proof of RC8 and RC9 by [Reference Rieffel22, Lemma 6.12]. We have

\begin{align*} \left \langle \,x,s^*y\,\right \rangle & = \left \langle \,s^*y,x\,\right \rangle ^* = (s^*\left \langle \,y,x\,\right \rangle )^* = \left \langle \,y,x\,\right \rangle ^* s^{**} = \left \langle \,x,y\,\right \rangle s, \\[5pt] [xt^*,y] & = [y,xt^*]^* = ([y,x]t^*)^* = t^{**}[y,x]^* = t[x,y]. \end{align*}

If the mapping $\left \langle \,,\,\right \rangle$ is full and $s\in S$ , then there exist $n\in \mathbb{N}$ and $x_k,y_k\in X$ such that $s= \sum _{k=1}^n \left \langle \,x_k,y_k\,\right \rangle$ . Now

\begin{align*} [s^*x,y] & = \left [\sum _{k = 1}^n \left \langle \,y_k,x_k\,\right \rangle x,y\right ] \tag*{$({\textbf{RC2}})$}\\[2pt] & = \sum _{k = 1}^n[\left \langle \,y_k,x_k\,\right \rangle x,y] \tag*{$(\text{additivity})$}\\[2pt] & = \sum _{k = 1}^n [y_k[x_k,x],y] \tag*{$({\textbf{RC5}})$} \\[2pt] & = \sum _{k = 1}^n [y_k[x_k,x]^{**},y] \tag*{$(\text{by (2.1)})$} \\[2pt] & = \sum _{k = 1}^n [x_k,x]^{*}[y_k,y] \tag*{$({\textbf{RC7}})$} \\[2pt] & = \sum _{k = 1}^n [x,x_k][y_k,y] \tag*{(${\textbf{RC4}})$} \\[2pt] & = \sum _{k = 1}^n [x,x_k[y_k,y]] \tag*{$({\textbf{RC3}})$} \\[2pt] & = \sum _{k = 1}^n [x,\left \langle \,x_k,y_k\,\right \rangle y] \tag*{$({\textbf{RC5}})$} \\[2pt] & = \left [x,\sum _{k = 1}^n\left \langle \,x_k,y_k\,\right \rangle y\right ] \tag*{$(\text{additivity})$} \\[2pt] & = [x,sy], \end{align*}

so RC9 holds. The proof of RC8 is analogous.

We give some natural examples of Rieffel contexts.

Example 2.4. Consider a ring $S$ with involution as a bimodule $_SS_S$ and define

\begin{equation*} \left \langle \,s,z\,\right \rangle \;:\!=\; sz^*, \;\;\; [s,z]\;:\!=\; s^*z \end{equation*}

for every $s,z\in S$ . We obtain a Rieffel context $(S,S,S,\left \langle \,,\,\right \rangle, [\,,\,])$ . If $S$ is an idempotent ring, then this context is full and unitary.

Example 2.5. Let $R$ be a ring and consider matrix rings $S=\textsf{M}_m(R)$ and $T= \textsf{M}_n(R)$ with transposing as an involution operation. Then $X\;:\!=\;\textsf{M}_{m,n}(R)$ is an $(S,T)$ -bimodule with respect to usual addition of matrices and actions defined by matrix multiplication. Defining

\begin{equation*} \left \langle \,A,B\,\right \rangle \;:\!=\; AB^T,\;\;\; [A,B]\;:\!=\;A^TB \end{equation*}

we see easily that the required axioms are satisfied.

If we consider the relation of having a Rieffel context between two rings with involution, then it is clear that this relation is reflexive and symmetric. Next, we show that the relation of having a full and unitary Rieffel context is also transitive. Therefore, the notion of a Rieffel context allows us to consider an equivalence relation on the class of all idempotent rings with involution. But first let us prove a small, but useful, lemma.

Proof. We know that $M \otimes _R N$ is a right $S$ -module with an obvious $S$ -action. Let $\sum _{k = 1}^t m_k \otimes n_k \in M \otimes _R N$ be arbitrary. Since $N_S$ is unitary, for every $k \in \{0,\ldots, t\}$ , there exist elements $n_{k1},\ldots, n_{ku} \in N$ and $s_{k1},\ldots, s_{ku} \in S$ such that $n_k = n_{k1}s_{k1} + \ldots + n_{ku}s_{ku}$ . Now,

\begin{equation*} \sum _{k = 1}^t m_k \otimes n_k = \sum _{k = 1}^n m_k \otimes \left (\sum _{h = 1}^u n_{kh}s_{kh}\right ) = \sum _{k = 1}^n\sum _{h = 1}^u (m_k \otimes n_{kh})s_{kh} \in (M \otimes _R N)S. \end{equation*}

Hence, $M \otimes _R N$ is a unitary right $S$ -module.

The next result is a ring theoretic analogue of [Reference Steinberg24, Proposition 2.5].

Proof. Let $(S,T,{_SX_T},\langle \,,\,\rangle _1,[\,,\,]_1)$ and $(T,R,{_TY_R},\langle \,,\,\rangle _2,[\,,\,]_2)$ be unitary and full Rieffel contexts. Consider the ( $S,R$ )-bimodule $X \otimes _T Y$ . Since $_SX$ and $Y_R$ are unitary modules, the bimodule $X \otimes _T Y$ is also unitary by Lemma 2.6 and its dual. Define mappings

\begin{align*} \langle \,,\,\rangle \colon (X \!\otimes _T Y) \times (X \!\otimes _T Y) \to S, \ \left (\sum _{k = 1}^{n} x_k \otimes y_k,\sum _{h = 1}^m x_h' \otimes y_h'\right ) \!\mapsto \! \sum _{k = 1}^{n}\sum _{h = 1}^m \langle x_k \langle y_k,y'_h\rangle _2, x_h'\rangle _1, \\[5pt] [\,,\,]\colon (X \!\otimes _T Y) \times (X \!\otimes _T Y) \to R, \ \left (\sum _{k = 1}^{n} x_k \otimes y_k,\sum _{h = 1}^m x_h' \otimes y_h'\right ) \mapsto \sum _{k = 1}^{n}\sum _{h = 1}^m [y_k,[x_k,x_h']_1y_h']_2. \end{align*}

By using the universal property of the tensor product a few times, it can be shown that these mappings are well-defined homomorphisms of groups.

The mappings $\langle \,,\,\rangle$ and $[\,,\,]$ are full because $\langle \,,\,\rangle _1$ , $\langle \,,\,\rangle _2$ , $[\,,\,]_1$ , $[\,,\,]_2$ are full and the modules $X_T$ , $_TY$ are unitary.

It suffices to verify the axioms RC1–RC5 on the generators $x\otimes y$ of $X\otimes _T Y$ . The axioms RC1 and RC3 are clearly satisfied. For every $x,x' \in X$ and $y,y' \in Y$ , we have

\begin{align*} {\langle x \otimes y,x' \otimes y'\rangle }^* &={\langle x{\langle y,y'\rangle }_2, x'\rangle }_1^* \tag*{$(\text{def. of} \left \langle \,,\,\right\rangle)$} \\[5pt] & ={\langle x',x{\langle y,y'\rangle }_2\rangle }_1 \tag*{$({\textbf{RC2}})$} \\[5pt] & ={\langle x'{\langle y,y'\rangle }_2^*,x\rangle }_1 \tag*{$({\textbf{RC8}})$} \\[5pt] &={\langle x'{\langle y',y\rangle }_2,x\rangle }_1 \tag*{$({\textbf{RC2}})$} \\[5pt] &={\langle x' \otimes y', x \otimes y\rangle }. \tag*{$(\textrm{def. of} \left \langle \,,\,\right \rangle)$} \end{align*}

Hence, RC2 holds. RC4 holds for similar reasons. Furthermore, for every $x,x',x'' \in X$ and $y,y',y'' \in Y$ ,

\begin{align*} \langle x \otimes y, x' \otimes y'\rangle (x'' \otimes y'') &= \langle x\langle y,y'\rangle _2, x' \rangle _1x'' \otimes y'' = x\langle y,y'\rangle _2 [x', x'']_1 \otimes y'' \\[5pt] &= x \otimes \langle y,y'\rangle _2 [x', x'']_1 y'' = x \otimes y[y',[x', x'']_1 y'']_2 \\[5pt] & = (x \otimes y)[x'\otimes y',x'' \otimes y'']. \end{align*}

So RC5 also holds. We have constructed a unitary and full Rieffel context $(S,R,X \otimes _T Y,\langle \,,\,\rangle, [\,,\,])$ .

There is also the classical notion of Morita context for rings (see, e.g., [Reference Müller17]).

Let $S$ and $T$ be rings with involution. Given any $(S,T)$ -bimodule $_SX_T$ , one can construct its dual bimodule $_T\overline{X}_S$ as follows (see [Reference Rieffel22, Definition 6.17]). As an abelian group, $\overline{X}$ is the same as $X$ . We write $\overline{x}$ when we consider an element $x\in X$ as a member of $\overline{X}$ . So $\overline x + \overline y = \overline{x+y}$ . The $T$ - and $S$ -actions on $\overline{X}$ are defined by

(2.2) \begin{equation} t\overline{x}\;:\!=\; \overline{xt^*}, \;\;\;\; \overline{x}s\;:\!=\; \overline{s^*x}. \end{equation}

Definition 2.9. By a Morita $*$ -context connecting rings $S$ and $T$ with involution we mean a Morita context $(S,T,{_SX_T},{_T\overline{X}_S},\theta, \phi )$ satisfying

\begin{equation*} \theta (x\otimes \overline y)^* = \theta (y\otimes \overline x) \;\; \mbox { and }\;\; \phi (\overline y\otimes x)^* = \phi (\overline x\otimes y) \end{equation*}

for all $x,y\in X$ .

The following proposition shows that the conditions appearing in Definition 2.9 are natural.

Proof. Recall that the multiplication on $X\otimes _T^\phi \overline{X}$ is given by

\begin{equation*} (x\otimes \overline {y})(x_1\otimes \overline {y_1}) \;:\!=\; x\otimes \phi (\overline {y}\otimes x_1)\overline {y_1} = x\otimes \overline {y_1\phi (\overline {y}\otimes x_1)^*} = x\otimes \overline {y_1 \phi (x_1\otimes y)}. \end{equation*}

By Theorem 2.11 in [Reference Väljako26], the mapping $\theta$ is a homomorphism of rings. We define an involution on the ring $X\otimes _T^\phi \overline{X}$ by

\begin{equation*} (x\otimes \overline {y})^* \;:\!=\; y\otimes \overline {x}. \end{equation*}

For every $x,x_1 \in X$ and $\overline{y},\overline{y_1} \in \overline{X}$ , we have

\begin{align*} ((x\otimes \overline{y})(x_1\otimes \overline{y_1}))^* & = (x\otimes \overline{y_1\phi (\overline{x_1}\otimes y)})^* = y_1\phi (\overline{x_1}\otimes y)\otimes \overline{x} = y_1\otimes \phi (\overline{x_1}\otimes y) \overline{x}\\[5pt] & = (y_1\otimes \overline{x_1})(y\otimes \overline{x}) = (x_1\otimes \overline{y_1})^* (x\otimes \overline{y})^*, \\[5pt] \theta ((x\otimes \overline{y})^*) & = \theta (y\otimes \overline{x}) = \theta (x\otimes \overline{y})^*. \end{align*}

Hence, we see that $X \otimes _T^\phi \overline{X}$ is indeed a ring with involution and $\theta$ is a homomorphism of rings with involution.

Remark 2.11. Note that the bimodules $_SX_T$ and $_T\overline{X}_S$ actually form a natural example of something that might be called a 4-module. Indeed, let $S$ , $R$ , $T$ , $P$ be rings. An abelian group $X$ could be called a $(S,R,T,P)$ 4-module, if there exist four actions

\begin{align*} S \times X \to X,\quad (s,x) &\mapsto sx, & X \times T \to X,\quad (x,t) &\mapsto xt, \\[5pt] R \times X \to X,\quad (r,x) &\mapsto rx, & X \times P \to X,\quad (x,p) &\mapsto xp \end{align*}

such that they induce bimodules $_SX_T$ , $_SX_P$ , $_RX_T$ , $_RX_P$ and the conditions

\begin{equation*} s(rx) = r(sx), \qquad \qquad (xt)p = (xp)t \end{equation*}

hold for every $x \in X$ , $r \in R$ , $s \in S$ , $t \in T$ and $p \in P$ . We believe that such structures merit studying on their own and hope it will be done some day.

Next, we adopt the notion of an enlargement to the case of rings with involution.

Proof. Denote $s\;:\!=\; f^{-1}(u)$ . Then we have $f(s^*) = f(s)^\star = (f(f^{-1}(u)))^\star = u^\star$ , which implies $(f^{-1}(u))^* = s^* = f^{-1}(u^\star )$ . A similar argument works for $g^{-1}$ .

3. Main theorem

Our main theorem is the following one. It can be compared with [Reference Lawson12, Theorem 1.1] for semigroups with local units and with [Reference Laan and Väljako9, Theorem 3.3] for idempotent rings.

Proof. (1) $\implies$ (2). Every firm bimodule is unitary.

(2) $\implies$ (3). Assume that $(S,T, X,\left \langle \,,\,\right \rangle, [\,,\,])$ is a unitary and full Rieffel context. We are going to build a Morita $*$ -context using the bimodules $_SX_T$ and $_T\overline{X}_S$ . Consider the mappings

\begin{align*} \hat \theta \;:\; X\times \overline X \to S,\;\;\; & (x,\overline y)\mapsto \left \langle \,x,y\,\right \rangle, \\[5pt] \hat \phi \;:\; \overline X\times X \to T,\;\;\; & (\overline y,x)\mapsto [y,x]. \end{align*}

We note that the mapping $\hat \theta$ is $T$ -balanced, because it is additive in both arguments and, for all $x,y\in X$ and $t\in T$ ,

\begin{align*} \hat \theta (xt,\overline y) & = \left \langle \,xt,y\,\right \rangle \tag*{$(\text{def. of $\hat \theta $})$} \\[5pt] & = \left \langle \,x,yt^*\,\right \rangle \tag*{$({\textbf{RC8}})$} \\[5pt] & = \hat \theta (x,\overline{yt^*}) \tag*{$(\text{def. of $\hat \theta $})$} \\[5pt] & = \hat \theta (x,t\overline y). \tag*{$\text{by (2.2)}$} \end{align*}

Similarly $\hat \phi$ is $S$ -balanced. By the universal property of tensor product there exist abelian group homomorphisms $\theta \;:\; X\otimes _T \overline X \to S$ and $\phi \;:\; \overline X\otimes _S X\to T$ such that

(3.1) \begin{equation} \theta (x\otimes \overline y) = \left \langle \,x,y\,\right \rangle \;\; \mbox{ and } \;\; \phi (\overline y\otimes x) = [y,x] \end{equation}

for all $x,y\in X$ . Condition RC1 implies that $\theta$ preserves left $S$ -action. It also preserves right $S$ -action, because

\begin{align*} \theta (x\otimes \overline y s) & = \theta (x\otimes \overline{s^*y}) \tag*{$\text{(by (2.2))}$} \\[5pt] & = \left \langle \,x,s^*y\,\right \rangle \tag*{$(\text{by (3.1)})$} \\[5pt] & = \left \langle \,x,y\,\right \rangle s \tag*{$({\textbf{RC6}})$} \\[5pt] & = \theta (x\otimes \overline y)s \tag*{$\text{by (3.1)}$} \end{align*}

for every $x,y\in X$ and $s\in S$ . Thus, $\theta$ (and similarly $\phi$ ) is a homomorphism of bimodules. These mappings satisfy compatibility conditions, because

\begin{align*} \theta (x\otimes \overline y)x' & = \left \langle \,x,y\,\right \rangle x' \tag*{$\text{by (3.1)}$} \\[5pt] & = x[y,x'] \tag*{$({\textbf{RC5}})$} \\[5pt] & = x\phi (\overline y\otimes x') \tag*{$\text{by (3.1)}$} \end{align*}

and

\begin{align*} \phi (\overline y\otimes x)\overline{y'} & = [y,x]\overline{y'} \tag*{$\text{(by (3.1))}$} \\[5pt] & = \overline{y'[y,x]^*} \tag*{$(\text{by (2.2)})$} \\[5pt] & = \overline{y'[x,y]} \tag*{$({\textbf{RC4}})$} \\[5pt] & = \overline{\left \langle \,y',x\,\right \rangle y} \tag*{$({\textbf{RC5}})$} \\[5pt] & = \overline y \left \langle \,y',x\,\right \rangle ^* \tag*{$(\text{by (2.2)})$} \\[5pt] & = \overline y\left \langle \,x,y'\,\right \rangle & & \tag*{$({\textbf{RC2}})$} \\[5pt] & = \overline y\theta (x\otimes \overline{y'}). & & \tag*{$\text{by (3.1)}$} \end{align*}

If $s\in S$ , then there exist $n\in \mathbb{N}$ , $x_1,\ldots, x_n,y_1,\ldots, y_n\in X$ such that

\begin{equation*} s = \sum _{k=1}^n \left \langle \,x_k,y_k\,\right \rangle = \sum _{k=1}^n \theta (x_k\otimes \overline {y_k}) = \theta \left ( \sum _{k=1}^n x_k\otimes y_k\right ). \end{equation*}

Hence, $\theta$ (and analogously $\phi$ ) is surjective. The bimodule $_SX_T$ is unitary by assumption. Let $\overline x\in \overline X$ . Since $_SX$ is unitary, there exist $n\in \mathbb{N}$ , $s_1,\ldots, s_n\in S$ and $x_1,\ldots, x_n\in X$ such that $x= \sum _{k=1}^n s_k x_k$ . Hence,

\begin{equation*} \overline x = \overline {\sum _{k=1}^n s_k x_k } = \sum _{k=1}^n \overline {s_k x_k} = \sum _{k=1}^n \overline {x_k} s_k^*. \end{equation*}

We have shown that $\overline{X}_S$ is unitary. Analogously we see that $_T\overline{X}$ is unitary. Finally, we notice that, for every $x,y\in X$ ,

\begin{align*} \theta (x\otimes \overline y)^* & = \left \langle \,x,y\,\right \rangle ^* = \left \langle \,y,x\,\right \rangle = \theta (y\otimes \overline x), \\[5pt] \phi (\overline y\otimes x)^* & = [y,x]^* = [x,y] = \phi (\overline x\otimes y). \end{align*}

(3) $\implies$ (4). Let $(S,T,{_SX_T},{_T\overline{X}_S}, \theta, \phi )$ be a unitary surjective Morita $*$ -context. We consider the matrix set

\begin{equation*} R= \left \{ \begin {pmatrix} s & x \\[5pt] \overline {y} & t \end {pmatrix} \middle \vert \; s\in S, t\in T, x,y\in X \right \} \end{equation*}

with componentwise addition and with the multiplication

\begin{align*} \begin{pmatrix} s & x \\[5pt] \overline{y} & t \end{pmatrix} \begin{pmatrix} s_1 & x_1 \\[5pt] \overline{y_1} & t_1 \end{pmatrix} & \;:\!=\; \begin{pmatrix} ss_1+ \theta (x\otimes \overline{y_1}) & sx_1+xt_1 \\[5pt] \overline{y}s_1+t\overline{y_1} & \phi (\overline y\otimes x_1) +tt_1 \end{pmatrix} = \begin{pmatrix} ss_1+ \theta (x\otimes \overline{y_1}) & sx_1+xt_1 \\[5pt] \overline{s^*_1y+y_1t^*} & \phi (\overline y\otimes x_1) +tt_1 \end{pmatrix}. \end{align*}

The unary $\star$ -operation on $R$ is defined as

\begin{equation*} \begin {pmatrix} s & x \\[5pt] \overline {y} & t \end {pmatrix}^\star \;:\!=\; \begin {pmatrix} s^* & y \\[5pt] \overline {x} & t^* \end {pmatrix}. \end{equation*}

It is well known (see, e.g., [Reference Nobusawa19, Section 2]) that $R$ with these operations is a ring (called a Morita ring of a Morita context). Two things remain to verify.

a) The operation $\star$ is an involution on $R$ , because

\begin{align*} \!\left (\!\! \begin{pmatrix} s & x \\[5pt] \overline{y} & t \end{pmatrix}+ \begin{pmatrix} s_1 & x_1 \\[5pt] \overline{y_1} & t_1 \end{pmatrix}\!\! \right )^{\!\star} & = \begin{pmatrix} s+s_1 & x+x_1 \\[5pt] \overline{y+y_1} & t+t_1 \end{pmatrix}^{\!\star} = \begin{pmatrix} s^*+s^*_1 & y+y_1 \\[5pt] \overline{x+x_1} & t^*+t^*_1 \end{pmatrix} = \begin{pmatrix} s^* & y \\[5pt] \overline{x} & t^* \end{pmatrix} + \begin{pmatrix} s^*_1 & y_1 \\[5pt] \overline{x_1} & t^*_1 \end{pmatrix} = \begin{pmatrix} s & x \\[5pt] \overline{y} & t \end{pmatrix}^{\!\star} + \begin{pmatrix} s_1 & x_1 \\[5pt] \overline{y_1} & t_1 \end{pmatrix}^{\!\star}\!, \\[5pt] \begin{pmatrix} s & x \\[5pt] \overline{y} & t \end{pmatrix}^{\star \star } & = \begin{pmatrix} s^* & y \\[5pt] \overline{x} & t^* \end{pmatrix}^\star = \begin{pmatrix} s^{**} & x \\[5pt] \overline{y} & t^{**} \end{pmatrix} = \begin{pmatrix} s & x \\[5pt] \overline{y} & t \end{pmatrix}, \\[5pt] \left ( \begin{pmatrix} s & x \\[5pt] \overline{y} & t \end{pmatrix} \begin{pmatrix} s_1 & x_1 \\[5pt] \overline{y_1} & t_1 \end{pmatrix} \right )^\star & = \begin{pmatrix} ss_1+ \theta (x\otimes \overline{y_1}) & sx_1+xt_1 \\[5pt] \overline{s^*_1y+y_1t^*} & \phi (\overline y\otimes x_1)+tt_1 \end{pmatrix}^\star \\[5pt] & = \begin{pmatrix} (ss_1)^*+ \theta (x\otimes \overline{y_1})^* & s^*_1y+y_1t^* \\[5pt] \overline{sx_1+xt_1} & \phi (\overline y\otimes x_1)^*+(tt_1)^* \end{pmatrix} \\[5pt] & = \begin{pmatrix} s_1^*s^*+ \theta (y_1\otimes \overline x) & s^*_1y+y_1t^* \\[5pt] \overline{s^{**}x_1+xt_1^{**}} & \phi (\overline x_1\otimes y)+t_1^*t^* \end{pmatrix} = \begin{pmatrix} s^*_1 & y_1 \\[5pt] \overline{x_1} & t^*_1 \end{pmatrix} \begin{pmatrix} s^* & y \\[5pt] \overline{x} & t^* \end{pmatrix} = \begin{pmatrix} s_1 & x_1 \\[5pt] \overline{y_1} & t_1 \end{pmatrix}^\star \begin{pmatrix} s & x \\[5pt] \overline{y} & t \end{pmatrix}^\star . \end{align*}

b) We will verify the conditions in Definition 2.12, that are related to $S$ . Similar arguments will apply for $T$ . The set

\begin{equation*} S'= \left \{ \begin {pmatrix} s & 0_X \\[5pt] \overline {0_X} & 0_T \end {pmatrix} \middle \vert \, s\in S \right \} \end{equation*}

is a subring of $R$ , which is also closed under $\star$ -operation (for the latter we need that $0^*_T = 0_T$ ). It is easy to see that the mapping

\begin{equation*} f\;:\; S\to S', \;\; s\mapsto \begin {pmatrix} s & 0_X \\[5pt] \overline {0_X} & 0_T \end {pmatrix} \end{equation*}

is an isomorphism of rings. Moreover, it preserves the involution, because

\begin{equation*} f(s^*) = \begin {pmatrix} s^* & 0_X \\[5pt] \overline {0_X} & 0_T \end {pmatrix} = \begin {pmatrix} s^* & 0_X \\[5pt] \overline {0_X} & 0^*_T \end {pmatrix} = \begin {pmatrix} s & 0_X \\[5pt] \overline {0_X} & 0_T \end {pmatrix}^\star = f(s)^\star . \end{equation*}

The equalities $R=RS'R$ and $S'=S'RS'$ hold by [Reference Laan and Väljako9, Proposition 3.5].

(4) $\implies$ (1). Assume that $R$ is a joint enlargement of $S$ and $T$ . It suffices to consider the case, where $S$ and $T$ are subrings of $R$ . In that case $s^\star = s^*$ and $t^\star =t^*$ for every $s\in S$ and $t\in T$ . We consider the bimodule

\begin{equation*} {_SX_T}\;:\!=\; S\otimes _S SRT \otimes _T T \end{equation*}

with natural actions of $S$ and $T$ . By [Reference Marín15, Proposition 2.5], the bimodule $_SX_T$ is firm, because the rings $S$ and $T$ are idempotent. Consider the mapping

\begin{equation*} \hat {\sigma }\;:\; S\times SRT \times T \to R, \;\; (s,\rho, t) \mapsto t^*\rho ^\star s^*. \end{equation*}

Note that

\begin{align*} \hat{\sigma } (ss_1,\rho, t) & = t^*\rho ^\star (ss_1)^* = t^*\rho ^\star s_1^*s^* = t^*\rho ^\star s_1^\star s^* = t^*(s_1 \rho )^\star s^* = \hat{\sigma }(s,s_1 \rho, t), \\[5pt] \hat{\sigma } (s,\rho t_1,t) & = t^*(\rho t_1)^\star s^* = t^*t_1^\star \rho ^\star s^* = t^*t_1^*\rho ^\star s^* = (t_1t)^* \rho ^\star s^* = \hat{\sigma }(s,\rho, t_1t) \end{align*}

and also $\hat{\sigma }$ is additive in all three arguments. Consequently, there exists a group homomorphism $\sigma \;:\; S\otimes _S SRT \otimes _T T \to R$ such that $ \sigma (s\otimes \rho \otimes t)= t^*\rho ^\star s^*$ for every $s\in S$ , $\rho \in SRT$ and $t\in T$ . Note that if $\rho = \sum _{k = 1}^n s_kr_kt_k \in SRT$ , then

\begin{equation*} \rho ^\star = \left (\sum _{k = 1}^n s_kr_kt_k\right )^\star = \sum _{k = 1}^n (s_kr_kt_k)^\star = \sum _{k = 1}^n t_k^*r_k^\star s_k^* \in TRS. \end{equation*}

A little bit shorter argument shows that there also exists a group homomorphism $\tau \;:\; S\otimes _S SRT \otimes _T T\to R$ such that $\tau (s\otimes \rho \otimes t) = s\rho t$ for every $s\in S$ , $\rho \in SRT$ and $t\in T$ . We define

\begin{equation*} \left \langle \,x,y\,\right \rangle \;:\!=\; \tau (x)\sigma (y), \quad [x,y] \;:\!=\; \sigma (x)\tau (y) \end{equation*}

for every $x,y\in X$ . In particular, on elementary tensors we have

\begin{align*} \left \langle \,s\otimes \rho \otimes t, s_1\otimes \rho _1 \otimes t_1\,\right \rangle & \;:\!=\; s\rho tt^*_1\rho ^\star _1s^*_1\in SSRTTTTRSS = SRTRS= SRS = S, \\[5pt] [s\otimes \rho \otimes t, s_1\otimes \rho _1 \otimes t_1] &\;:\!=\; t^*\rho ^\star s^*s_1\rho _1t_1 \in TTRSSSSRTT = TRSRT = TRT = T, \end{align*}

which implies that $\left \langle \,x,y\,\right \rangle \in S$ and $[x,y]\in T$ for every $x,y\in X$ , because every element of $X$ is a finite sum of elementary tensors. Thus we have defined mappings

\begin{equation*} \left \langle \,,\,\right \rangle \;:\; X\times X \to S\; \mbox { and }\; [\,,\,]\;:\; X\times X \to T, \end{equation*}

which are additive in both arguments, because $\tau$ and $\sigma$ preserve addition and $S$ and $T$ have the distributivity law.

To prove that $\left \langle \,\,\right \rangle$ is full, we consider an element $z\in S$ . Since $S=S(SRT)TT(TRS)S$ , $z$ is a finite sum of products of the form $s\rho tt_1\rho _1s_1$ , where $s,s_1\in S$ , $t,t_1\in T$ , $\rho \in SRT$ and $\rho _1 \in TRS$ . It suffices to show that each such product has a preimage with respect to $\left \langle \,,\,\right \rangle$ . Indeed,

\begin{equation*} s\rho tt_1\rho _1s_1 = s\rho t(t_1^*)^*(\rho _1^\star )^\star (s_1^*)^* = \left \langle \,s\otimes \rho \otimes t, s_1^*\otimes \rho _1^\star \otimes t_1^*\,\right \rangle . \end{equation*}

A similar proof shows that $[\,,\,]$ is full. Finally, we verify the conditions RC1–RC5.

RC1. For every $z\in S$ ,

\begin{equation*} \left \langle \,z(s\otimes \rho \otimes t), s_1\otimes \rho _1 \otimes t_1\,\right \rangle = zs\rho tt^*_1\rho ^\star _1s^*_1 = z \left \langle \,s\otimes \rho \otimes t, s_1\otimes \rho _1 \otimes t_1\,\right \rangle . \end{equation*}

RC2. We calculate

\begin{align*} \left \langle \,s\otimes \rho \otimes t, s_1\otimes \rho _1 \otimes t_1\,\right \rangle ^* & = (s\rho tt^*_1\rho ^\star _1s^*_1)^* \tag*{$(\text{def. of $\left \langle \,,\,\right \rangle $})$}\\[5pt] & = (s\rho tt^\star _1\rho ^\star _1s^\star _1)^\star \tag*{$(\text{$t_1^*=t_1^\star, s_1^*=s_1^\star $})$}\\[5pt] & = s_1^{\star \star } \rho _1^{\star \star } t_1^{\star \star } t^\star \rho ^\star s^\star \tag*{$(\text{$(ab)^\star = b^\star a^\star $})$} \\[5pt] & = s_1 \rho _1 t_1 t^\star \rho ^\star s^\star \tag*{$(\text{$a^{\star \star } = a$})$} \\[5pt] & = s_1 \rho _1 t_1 t^* \rho ^\star s^* \tag*{$(\text{$t^\star = t^*, s^\star = s^*$})$} \\[5pt] & = \left \langle \,s_1\otimes \rho _1\otimes t_1, s\otimes \rho \otimes t\,\right \rangle . \tag*{$(\text{def. of $\left \langle \,,\,\right \rangle $})$} \end{align*}

RC3. Analogous to RC1.

RC4. Analogous to RC2.

RC5. To prove this, we need to use that $S$ and $T$ are idempotent rings. We calculate

\begin{align*} \left \langle \,s\otimes \rho \otimes t, s_1\otimes \rho _1 \otimes t_1\,\right \rangle (s_2\otimes \rho _2 \otimes t_2) & = s\rho tt^*_1\rho ^\star _1s^*_1 (s_2\otimes \rho _2 \otimes t_2) \tag*{$(\text{def. of $\left \langle \,,\,\right \rangle $})$} \\[5pt] & = s\rho tt^*_1\rho ^\star _1s^*_1 s_2\otimes \rho _2 \otimes t_2 \tag*{$(\text{def. of $S$-action})$}\\[5pt] & = \left (\sum _{k = 1}^n u_ku'_k\right )\rho tt^*_1\rho ^\star _1s^*_1 s_2\otimes \rho _2 \otimes t_2 \tag*{$(\text{$s=\sum _{k = 1}^n u_ku'_k$})$}\\[5pt] & = \left (\sum _{k = 1}^n u_ku'_k\rho tt^*_1\rho ^\star _1s^*_1 s_2\right )\otimes \rho _2 \otimes t_2 \tag*{$(\text{distributivity})$}\\[5pt] & = \sum _{k = 1}^n (u_ku'_k\rho tt^*_1\rho ^\star _1s^*_1 s_2\otimes \rho _2 \otimes t_2) \tag*{$(\text{property of $\otimes $})$}\\[5pt] & = \sum _{k = 1}^n (u_k\otimes u'_k\rho tt^*_1\rho ^\star _1s^*_1 s_2 \rho _2 \otimes t_2) \tag*{$(\text{$S(SRT)T(TRS)S\!=\! S$})$}\\[5pt] & = \sum _{k = 1}^n (u_ku'_k\otimes \rho tt^*_1\rho ^\star _1s^*_1 s_2 \rho _2 \otimes t_2) \tag*{$(\text{$u'_k\in S$})$}\\[5pt] & = \left (\sum _{k = 1}^n u_ku'_k\right ) \otimes \rho tt^*_1\rho ^\star _1s^*_1 s_2 \rho _2 \otimes t_2 \tag*{$(\text{property of $\otimes $})$}\\[5pt] & = s \otimes \rho tt^*_1\rho ^\star _1s^*_1 s_2 \rho _2 \otimes t_2. \tag*{$(\text{$s=\sum _{k = 1}^n u_ku'_k$})$}\\[5pt] \end{align*}

A similar calculation shows that

\begin{align*} (s\otimes \rho \otimes t)[s_1\otimes \rho _1\otimes t_1,s_2\otimes \rho _2\otimes t_2] & = (s\otimes \rho \otimes t)t_1^*\rho _1^\star s_1^*s_2\rho _2t_2 = s\otimes \rho t t_1^*\rho _1^\star s_1^*s_2\rho _2\otimes t_2. \end{align*}

This completes the proof.

It would be natural to call two idempotent rings with involution Morita equivalent if they satisfy the conditions of Theorem3.1.

Remark 3.4. Consider the implication (4) $\implies$ (2) in Theorem 3.1. Assume that $R$ is a joint enlargement of $S$ and $T$ as in Definition 2.12, with isomorphisms

\begin{equation*} f\;:\; S\to S', s\mapsto \hat {s}, \;\;\;\; g\;:\; T\to T', t\mapsto \hat {t}. \end{equation*}

Then, it turns out that there is a simpler construction yielding a unitary full Rieffel context between $S$ and $T$ . We consider the bimodule

\begin{equation*} {_SX_T}\;:\!=\; S'RT' = \left \{\sum _{k=1}^n \hat {s_k} r_k \hat {t_k} \middle | n\in \mathbb {N}, s_k\in S, r_k\in R, t_k\in T \right \} \subseteq R. \end{equation*}

Addition of $X$ is the restriction of the addition of $R$ . The actions of $S$ and $T$ on $X$ are defined by

\begin{equation*} s\cdot x \;:\!=\; \hat {s}x, \;\;\; x\cdot t \;:\!=\; x\hat {t}, \end{equation*}

where $\hat{s}x$ is the product of $\hat{s}$ and $x$ in $R$ and $x\hat{t}$ is the product of $x$ and $\hat{t}$ in $R$ . It is straightforward to show that $_SX_T$ is an $(S,T)$ -bimodule. Since $S$ and $T$ are idempotent rings, the bimodule $_SX_T$ is unitary.

Note that

\begin{align*} (\hat{s}r\hat{t})(\hat{s_1}r_1\hat{t_1})^\star & = \hat{s}r\hat{t} (\hat{t_1})^\star r_1^\star (\hat{s_1})^\star = \hat{s}r\hat{t} (\widehat{t_1^*}) r_1^\star (\widehat{s_1^*}) \in S'RT'T'RS' = S'RT'RS'=S', \end{align*}

and, analogously, $(\hat{s}r\hat{t})^\star (\hat{s_1}r_1 \hat{t_1}) \in T'$ . Hence, we can define the mappings

\begin{equation*} \left \langle \,,\,\right \rangle \;:\; X\times X \to S\; \mbox { and } \; [\,,\,]\;:\; X\times X \to T, \end{equation*}

by

\begin{equation*} \left \langle \,x,y\,\right \rangle \;:\!=\; f^{-1}(xy^\star ), \;\; [x,y]\;:\!=\; g^{-1}(x^\star y), \end{equation*}

for every $x,y\in X$ . These mappings are additive in both arguments, because $f^{-1}$ and $g^{-1}$ are ring isomorphisms.

To prove that $\left \langle \,,\,\right \rangle$ is full, we consider an element $z\in S$ . Since $S' = S'RT'T'RS'$ , $\hat{z}\in S'$ is a finite sum of products of the form $\hat{s}r\hat{t}\hat{t_1}r_1\hat{s_1}$ , where $s,s_1\in S$ , $t,t_1\in T$ , $r,r_1\in R$ . Now

\begin{align*} f^{-1}(\hat{s}r\hat{t}\hat{t_1}r_1\hat{s_1}) & = f^{-1}(\hat{s}r\hat{t}(\hat{t_1})^{\star \star }r_1^{\star \star } (\hat{s_1})^{\star \star }) = f^{-1}(\hat{s}r\hat{t}((\hat{s_1})^{\star }r_1^{\star } (\hat{t_1})^{\star })^\star ) = \left \langle \,\hat{s}r\hat{t}, (\hat{s_1})^{\star }r_1^{\star } (\hat{t_1})^{\star } \,\right \rangle . \end{align*}

It follows that $z= f^{-1}(\hat{z}) = \sum _{k=1}^n \left \langle \,x_k,y_k\,\right \rangle$ , where $n\in \mathbb{N}$ and $x_k,y_k\in X$ . A similar proof shows that $[\,,\,]$ is full. It is easy to verify the conditions RC1–RC5.

We will say that $(S,T, S'RT',\left \langle \,,\,\right \rangle, [\,,\,])$ is the Rieffel context induced by a joint enlargement $R$ of $S$ and $T$ .

4. Rieffel contexts come from enlargements

In this section, we will show that, up to isomorphism, every unitary Rieffel context is induced by a joint enlargement. Isomorphism of classical Morita contexts is defined in [Reference Müller17]. Its analogue for Rieffel’s contexts is the following.

Definition 4.1. Let $S$ and $T$ be rings with involution. We say that a Rieffel context $(S,T,X, \left \langle \,,\,\right \rangle _X,[\,,\,]_X)$ is isomorphic to a Rieffel context $(S,T,Y, \left \langle \,,\,\right \rangle _Y,[\,,\,]_Y)$ if there exists a bimodule isomorphism $h\;:\; X\to Y$ such that, for every $x_1,x_2\in X$ ,

\begin{equation*} \left \langle \,h(x_1),h(x_2)\,\right \rangle _Y = \left \langle \,x_1,x_2\,\right \rangle _X\;\; \mbox { and } \;\; [h(x_1),h(x_2)]_Y = [x_1,x_2]_Y. \end{equation*}

Proof. Let $(S,T,X, \left \langle \,,\,\right \rangle _X,[\,,\,]_X)$ be a unitary Rieffel context. Putting together implications ( $2 \Longrightarrow 3$ ) and ( $3 \Longrightarrow 4$ ) in Theorem 3.1 we see that it gives rise to a joint enlargement $R$ of $S$ and $T$ . More precisely,

\begin{equation*} R= \left \{ \begin {pmatrix} s & x \\[5pt] \overline {y} & t \end {pmatrix} \middle \vert \; s\in S, t\in T, x,y\in X \right \} \end{equation*}

is a ring with componentwise addition and with the multiplication

\begin{align*} \begin{pmatrix} s & x \\[5pt] \overline{y} & t \end{pmatrix} \begin{pmatrix} s_1 & x_1 \\[5pt] \overline{y_1} & t_1 \end{pmatrix} & \;:\!=\; \begin{pmatrix} ss_1+ \left \langle \,x,y_1\,\right \rangle & sx_1+xt_1 \\[5pt] \overline{y}s_1+t\overline{y_1} & [y,x_1]+tt_1 \end{pmatrix} = \begin{pmatrix} ss_1+ \left \langle \,x,y_1\,\right \rangle & sx_1+xt_1 \\[5pt] \overline{s^*_1y+y_1t^*} & [y,x_1]+tt_1 \end{pmatrix}. \end{align*}

The unary $\star$ -operation on $R$ is defined by

\begin{equation*} \begin {pmatrix} s & x \\[5pt] \overline {y} & t \end {pmatrix}^\star \;:\!=\; \begin {pmatrix} s^* & y \\[5pt] \overline {x} & t^* \end {pmatrix}. \end{equation*}

The mappings

\begin{equation*} f\;:\; S\to R, \;\;\; s\mapsto \begin {pmatrix} s & 0_X\\[5pt] \overline {0_X} & 0_T \end {pmatrix} \;\;\; \mbox { and }\;\;\; g\;:\; T\to R, \;\;\; t\mapsto \begin {pmatrix} 0_S & 0_X\\[5pt] \overline {0_X} & t \end {pmatrix} \end{equation*}

are injective ring homomorphisms and the subrings

\begin{equation*} S'= f(S)= \left \{ \begin {pmatrix} s & 0_X\\[5pt] \overline {0_X} & 0_T \end {pmatrix} \middle \vert \;s\in S\right \} \;\;\; \mbox { and }\;\;\; T'= g(T)= \left \{ \begin {pmatrix} 0_S & 0_X\\[5pt] \overline {0_X} & t \end {pmatrix} \middle \vert \;t\in T\right \} \end{equation*}

of $R$ satisfy the equalities $S'=S'RS'$ , $R = RS'R$ , $T'=T'RT'$ and $R = RT'R$ . Let the Rieffel context induced by the joint enlargement $R$ be $(S,T,Y,\left \langle \,,\,\right \rangle _Y,[\,,\,]_Y)$ . By Remark 3.4, this means that ${_SY_T} = S'RT'$ , $\left \langle \,y_1,y_2\,\right \rangle _Y = f^{-1}(y_1y_2^\star )$ and $[y_1,y_2]_Y = g^{-1}(y_1^\star y_2)$ . Observe that

\begin{equation*} \begin {pmatrix} s' & 0_X\\[5pt] \overline {0_X} & 0_T \end {pmatrix} \begin {pmatrix} s & x\\[5pt] \overline {y} & t \end {pmatrix} \begin {pmatrix} 0_S & 0_X\\[5pt] \overline {0_X} & t' \end {pmatrix} = \begin {pmatrix} s's & s'x\\[5pt] \overline {0_X} & 0_T \end {pmatrix} \begin {pmatrix} 0_S & 0_X\\[5pt] \overline {0_X} & t' \end {pmatrix} = \begin {pmatrix} 0_S & s'xt'\\[5pt] \overline {0_X} & 0_T \end {pmatrix} \end{equation*}

for every $s,s'\in S$ , $t,t'\in T$ and $x\in X$ . Hence,

\begin{equation*} {_SY_T} = \left \{ \sum _{k=1}^n \begin {pmatrix} 0_S & s_kx_kt_k\\[5pt] \overline {0_X} & 0_T \end {pmatrix} \middle |\; n\in \mathbb {N}, s_k\in S, x_k\in X, t_k\in T \right \} = \left \{ \begin {pmatrix} 0_S & x\\[5pt] \overline {0_X} & 0_T \end {pmatrix} \middle \vert \;x\in X\right \}, \end{equation*}

where the last equality holds because $_SX_T$ is a unitary bimodule.

We define a mapping $h\;:\; X\to Y$ by

\begin{equation*} h(x) \;:\!=\; \begin {pmatrix} 0_S & x\\[5pt] \overline {0_X} & 0_T \end {pmatrix}. \end{equation*}

Then, $h$ is clearly an isomorphism of abelian groups. Note that

\begin{align*} s\cdot h(x) & = s\cdot \begin{pmatrix} 0_S & x\\[5pt] \overline{0_X} & 0_T \end{pmatrix} = \begin{pmatrix} s & 0_X\\[5pt] \overline{0_X} & 0_T \end{pmatrix} \begin{pmatrix} 0_S & x\\[5pt] \overline{0_X} & 0_T \end{pmatrix} = \begin{pmatrix} 0_S & sx\\[5pt] \overline{0_X} & 0_T \end{pmatrix} = h(sx), \\[5pt] h(x)\cdot t & = \begin{pmatrix} 0_S & x\\[5pt] \overline{0_X} & 0_T \end{pmatrix} \begin{pmatrix} 0_S & 0_X\\[5pt] \overline{0_X} & t \end{pmatrix} = \begin{pmatrix} 0_S & xt\\[5pt] \overline{0_X} & 0_T \end{pmatrix} = h(xt) \end{align*}

for every $x\in X$ , $s\in S$ and $t\in T$ . Hence, $h$ is an isomorphism of $(S,T)$ -bimodules.

Finally, if $x_1,x_2\in X$ , then

\begin{align*} \left \langle \,h(x_1),h(x_2)\,\right \rangle _Y & = f^{-1}(h(x_1)h(x_2)^\star ) = f^{-1} \left ( \begin{pmatrix} 0_S & x_1\\[5pt] \overline{0_X} & 0_T \end{pmatrix} \begin{pmatrix} 0_S & 0_X\\[5pt] \overline{x_2} & 0_T \end{pmatrix} \right )\\[5pt] & = f^{-1} \left ( \begin{pmatrix} \left \langle \,x_1,x_2\,\right \rangle _X & 0_X\\[5pt] \overline{0_X} & 0_T \end{pmatrix} \right ) = \left \langle \,x_1,x_2\,\right \rangle _X, \end{align*}

and, analogously, $[h(x_1),h(x_2)]_Y = [x_1,x_2]_Y$ .

5. From a Rieffel context to an equivalence of module categories

In this section, we study relationships between Rieffel contexts and equivalences between module categories. When moving from unital rings to bigger classes of rings, one has to choose which categories of modules to use in order to develop Morita theory (see the Introduction of [Reference García and Marín5]). There are three natural choices proposed in the literature: the categories of a) unitary torsion-free modules (see, e.g., [Reference Ara3, Reference García and Simón7, Reference Nobusawa18]), b) closed modules, c) firm modules ([Reference García and Marín5, Reference García and Marín6]). In the case of an idempotent ring, all three categories are equivalent. While [Reference Ara3] uses unitary torsion-free (= nondegenerate) modules, we prefer to work with firm modules.

In [Reference Ara3, Theorem 4.1(i)] it is shown that if there exists a full nondegenerate Rieffel context between two nondegenerate idempotent rings with involution, then there is a $*$ -equivalence between the categories of unitary torsion-free modules over these rings. In this section we will prove an analogue of this result by a) replacing unitary torsion-free modules by firm modules and b) abandoning the requirement that the rings and Rieffel contexts be nondegenerate. We start by modifying the necessary definitions for our purposes and by proving some lemmas that will be needed later.

Compared to the definition given on page 229 of [Reference Ara3], we have dropped the condition of nondegeneracy (condition (4)). Note that if $S$ is a ring with involution and $\delta \;:\; P\times Q \to S$ is a dual $S$ -pair, then there exists another dual $S$ -pair

(5.1) \begin{equation} \overleftarrow{\delta }\colon \overline{Q} \times \overline{P} \to S, \qquad \overleftarrow{\delta }(\overline{q},\overline{p}) \;:\!=\; \delta (p,q)^*. \end{equation}

We also use a modification of Definition 1.2 in [Reference Ara3]. Instead of categories of unitary nondegenerate modules, we use the category $_S\textsf{FMod}$ of firm left $S$ -modules and the category $\textsf{FMod}_S$ of firm right $S$ -modules.

For a ring $S$ with involution, we have the dualization functors

\begin{align*} H_S\;:\; \textsf{FMod}_S\to{_S\textsf{FMod}}, \;\; H_S(X_S) &\;:\!=\;{_S\overline{X}}, \end{align*}

and

\begin{equation*} H'_S\;:\; {_S\textsf {FMod}}\to \textsf {FMod}_S, \;\; H'_S({_SX})\;:\!=\; {\overline {X}_S}. \end{equation*}

If $f\;:\; X_S\to Y_S$ is a morphism in $\textsf{FMod}_S$ , then $H_S(f)\;:\;{_S\overline{X}} \to{_S\overline{Y}}$ is defined by $H_S(f)(\overline{x}) \;:\!=\; \overline{f(x)}$ , and similarly for $H'_S$ . Note that $H_S \circ H'_S$ and $H'_S \circ H_S$ are identity functors.

Definition 5.3 (Cf. [Reference Ara3, page 234]). Let $S$ and $T$ be idempotent rings with involution. A $*$ -functor from $S$ to $T$ is a triple $(F,G,\alpha )$ , where $(F,G)$ is a functor-multiplier from $S$ to $T$ and $\alpha \;:\; F\Rightarrow H_T \circ G \circ H'_S$ a natural isomorphism such that, for every dual pair $\delta \;:\; V\times W \to S$ , every $a\in F(V)$ and $b\in (H'_T \circ F \circ H_S)(W)$ , we have

(5.2) \begin{equation} \overleftarrow{\overleftarrow{\delta }^{FG}}\left (\alpha _V(a),b \right ) = \delta ^{FG}(a,H'_T(\alpha _{H_S(W)})(b)). \end{equation}

Note that here $(\alpha _V(a),b) \in (H_T \circ G \circ H'_S)(V) \times (H'_T \circ F \circ H_S)(W) = \overline{G(\overline{V})} \times \overline{F(\overline{W})}$ ,

\begin{align*} \alpha _{H_S(W)}\colon &(F \circ H_S)(W) \to (H_T \circ G \circ H'_S \circ H_S)(W) = (H_T \circ G)(W), \\[5pt] H'_T(\alpha _{H_S(W)}) \colon &(H'_T \circ F \circ H_S)(W) \to (H'_T \circ H_T \circ G)(W) = G(W), \end{align*}

and $(a,H'_T(\alpha _{H_S(W)})(b)) \in F(V)\times G(W)$ .

Lemma 5.4. If $(F,G,\alpha )$ is a $*$ -functor from $S$ to $T$ , $\delta \;:\; V\times W \to S$ is a dual pair, $x\in G(W)$ and $y\in (H_T \circ G \circ H'_S)(V)$ , then

\begin{equation*} \overleftarrow {\overleftarrow {\delta }^{FG}}\left ( y, H'_T(\alpha _{H_S(W)}^{-1})(x) \right ) = \delta ^{FG}\left ( \alpha _V^{-1}(y),x\right ). \end{equation*}

Proof. Denote $a\;:\!=\; \alpha ^{-1}_V(y)$ and $b\;:\!=\; H'_T(\alpha _{H_S(W)}^{-1})(x)$ . Then

\begin{align*} \overleftarrow{\overleftarrow{\delta }^{FG}}\left ( y, H'_T(\alpha _{H_S(W)}^{-1})(x) \right ) & = \overleftarrow{\overleftarrow{\delta }^{FG}}\left ( \alpha _V(a), b \right ) \\[5pt] & = \delta ^{FG}\left (a,H'_T(\alpha _{H_S(W)})(b) \right ) \tag*{$(\text{$(F,G,\alpha )$ is a $*$-functor})$}\\[5pt] & = \delta ^{FG}\left (a,H'_T(\alpha _{H_S(W)} \circ \alpha _{H_S(W)}^{-1})(x) \right ) \tag*{$(\text{$H'_T$ is a functor})$}\\[5pt] & = \delta ^{FG}\left ( a,{\mathrm{id}}_{G(W)}(x) \right )\\[5pt] & = \delta ^{FG}\left ( \alpha _V^{-1}(y),x\right ), \end{align*}

which proves the lemma.

Note that $({\mathrm{id}}_{{_S\textsf{FMod}}},{\mathrm{id}}_{\textsf{FMod}_S},{\mathrm{id}}_{{\mathrm{id}}_{{_S\textsf{FMod}}}})$ is a $*$ -functor from $S$ to $S$ . By Lemma 2.1 in [Reference Ara3], we know that if $(F,G,\alpha )$ is a $*$ -functor from $S$ to $T$ and $(F',G',\beta )$ a $*$ -functor from $T$ to $R$ , then the triple $(F' \circ F, G' \circ G', \beta * \alpha )$ is a $*$ -functor from $S$ to $R$ , where $\beta * \alpha$ is the horizontal composition of natural transformations $\alpha$ and $\beta$ . Also, for every dual $S$ -pair $\delta \colon V \times W \to S$ we have the formula

(5.3) \begin{equation} \delta ^{F'F,G'G} = (\delta ^{FG})^{F'G'}\colon \ F'(F(V)) \times G'(G(W)) \to R. \end{equation}

Definition 5.5. [Cf. [Reference Ara3, page 235]] A $*$ -functor $(F,G,\alpha )$ from $S$ to $T$ is called a $*$ -equivalence if there exists a $*$ -functor $(F',G',\beta )$ from $T$ to $S$ such that there exist natural isomorphisms $\varphi \colon F' \circ F \Rightarrow{\mathrm{id}}_{{_S\textsf{FMod}}}$ , $\psi \colon F \circ F' \Rightarrow{\mathrm{id}}_{{_T\textsf{FMod}}}$ , for every dual pair $\delta _S\colon V \times W \to S$ we have

(5.4) \begin{equation} \delta _S(\varphi _V(a),b) = \delta _S^{F'F,G'G}\left (a,H_S'\left ((\beta *\alpha )_{H_S(W)} \circ \varphi _{H_S(W)}^{-1} \right )(w)\right ) \end{equation}

for every $a\in (F' \circ F)(V)$ and $b\in W$ and for every dual pair $\delta _T\colon V' \times W' \to T$ we have

(5.5) \begin{equation} \delta _T(\psi _V(x),y) = \delta ^{FF',GG'}_T\left (x,H_T'\left ((\alpha * \beta )_{H_T(W)} \circ \psi _{H_T(W)}^{-1}\right )(y)\right ) \end{equation}

for every $x \in (F' \circ F)(V')$ and $y \in W'$ . Here $\beta *\alpha \;:\; F' \circ F\Rightarrow H_S \circ G' \circ G \circ H'_S$ is the horizontal composition of $\alpha$ and $\beta$ and

\begin{equation*} H_S(W) \to ^{\varphi ^{-1}_{H_S(W)}} (F' \!\circ F \circ H_S)(W) \to ^{(\beta *\alpha )_{H_S(W)}} (H_S \circ G' \!\circ G \circ H'_S \circ H_S)(W) \!=\! (H_S \circ G' \circ G)(W). \end{equation*}

Proof. Let the assumptions of this lemma hold with natural isomophisms $\varphi \colon F' \circ F \Rightarrow{\mathrm{id}}_{{_S\textsf{FMod}}}$ and $\psi \colon F \circ F' \Rightarrow{\mathrm{id}}_{{_T\textsf{FMod}}}$ . The required natural isomorphisms from the claim of the lemma have components $\sigma _W = H_S'((\beta *\alpha )_{H_S(W)} \circ \varphi _{H_S(W)}^{-1} )\colon W\to (G'\circ G)(W)$ (for every $W \in \textsf{FMod}_S$ ) and $\rho _U = H_T'((\alpha * \beta )_{H_T(U)} \circ \psi _{H_T(U)}^{-1})\colon U\to (G\circ G')(U)$ (for every $U \in \textsf{FMod}_T$ ).

Now we are ready to prove the main result of this section. It is an analogue of the classical Morita I theorem (see, e.g., [Reference Lam10, Theorem 18.24]).

Proof. We construct a functor-multiplier from $S$ to $T$ . If $M_S\in \textsf{FMod}_S$ , then $M\otimes _S X_T\in \textsf{FMod}_T$ , because $\mu _{M\otimes X} ={\mathrm{id}}_M\otimes \mu _X$ , where ${\mathrm{id}}_M$ and $\mu _X$ are isomorphisms. Similarly, ${_T\overline{X}}\otimes _S N \in{_T\textsf{FMod}}$ if ${_SN} \in{_S\textsf{FMod}}$ . This allows us to consider the tensor functors

\begin{align*} & F\;:\!=\; \overline{X}\otimes _S \underline{\hspace{0.8em}}:{_S\textsf{FMod}}\to{_T\textsf{FMod}},\\[5pt] & G\;:\!=\; \underline{\hspace{0.8em}}\otimes _S X\;:\; \textsf{FMod}_S\to \textsf{FMod}_T. \end{align*}

The following proof is divided into several claims, which are proved separately.

Claim 1. The pair $(F,G)$ is a functor-multiplier.

Conditions (1) and (2) in Definition 5.2 are obvious.

(3) Let $\delta \;:\; V\times W\to S$ be a dual $S$ -pair. Define a mapping

\begin{equation*} \delta ^{FG}\;:\; (\overline {X}\otimes _S V) \times (W\otimes _S X) \to T \end{equation*}

by

(5.6) \begin{equation} \delta ^{FG}(\overline y\otimes v, w\otimes x) \;:\!=\; [y,\delta (v,w) x]\in T. \end{equation}

We will show that $\delta ^{FG}$ is well-defined. Note that for every pair $(w',x') \in W \times X$ , there is a well-defined mapping

\begin{equation*} d_{w',x'}\colon \ \overline {X} \times V \to T, \qquad (\overline {y},v) \mapsto [y,\delta (v,w')x']. \end{equation*}

The mapping $d_{w',x'}$ is $S$ -balanced, because, for every $y,y' \in X$ , $v,v' \in V$ and $s \in S$ , we have

\begin{align*} d_{w',x'}(\overline{y} + \overline{y'},v) &= d_{w',x'}(\overline{y + y'},v) = [y + y',\delta (v,w')x'] = d_{w',x'}(\overline{y},v) + d_{w',x'}(\overline{y'},v), \\[5pt] d_{w',x'}(\overline{y}, v + v') &= [y,\delta (v + v',w')x'] = [y,(\delta (v,w') + \delta (v',w'))x'] \\[5pt] & = [y,\delta (v,w')x'] + [y,\delta (v',w')x'] = d_{w',x'}(\overline{y},v) + d_{w',x'}(\overline{y},v'), \\[5pt] d_{w',x'}(\overline{y}s,v) &= d_{w',x'}(\overline{s^*y},v) = [s^*y,\delta (v,w')x'] = [y,s\delta (v,w')x'] = [y,\delta (sv,w')x'] \\[5pt] &= d_{w',x'}(y,sv). \end{align*}

By the universal property of the tensor product, we have a well-defined group homomorphism $\overline{d_{w',x'}}\colon \overline{X} \otimes _S V \to T$ such that

\begin{equation*} \overline {d_{w',x'}}(\overline {y}\otimes v) = [y,\delta (v,w')x'] \end{equation*}

for every $y\in X$ and $v\in V$ . Also, for every $t \in T$ , $\overline{y} \in \overline{X}$ and $v \in V$ , using RC7 we have

\begin{equation*} \overline {d_{w',x'}}(t\overline {y} \otimes v) = \overline {d_{w',x'}}(\overline {yt^*} \otimes v) = [yt^*,\delta (v,w')x'] = t[y,\delta (v,w')x'] = t\overline {d_{w',x'}}(\overline {y} \otimes v). \end{equation*}

Hence, $\overline{d_{w',x'}}$ is a homomorphism of left $T$ -modules. Now we may define a mapping

\begin{equation*} \tau \colon \ W \times X \to \textrm {Hom}\left (\overline {X} \otimes _S V,T\right ), \qquad (w,x) \mapsto \overline {d_{w,x}}. \end{equation*}

Note, that for every $w,w' \in W$ , $x,x',y \in X$ , $v\in V$ and $s \in S$ we have

\begin{align*} \tau (w + w',x)(\overline{y} \otimes v) &= \overline{d_{w + w',x}}(\overline{y} \otimes v) = [y,\delta (v,w + w')x] = [y,\delta (v,w)x] + [y,\delta (v,w')x] \\[5pt] &= \overline{d_{w,x}}(\overline{y} \otimes v) + \overline{d_{w',x}}(\overline{y} \otimes v) = (\tau (w,x) + \tau (w',x))(\overline{y} \otimes v), \\[5pt] \tau (ws,x)(\overline{y} \otimes v) &= \overline{d_{ws,x}}(\overline{y} \otimes v) = [y,\delta (v,ws)x] = [y,\delta (v,w)sx] = \tau (w,sx)(\overline y\otimes v) \end{align*}

and $\tau (w,x + x') = \tau (w,x) + \tau (w,x')$ . Hence, $\tau$ is $S$ -balanced and by the universal property of the tensor product, we may consider a well-defined group homomorphism $\overline{\tau }\colon W \otimes _S X \to \textrm{Hom}\left (\overline{X} \otimes _S V,T\right )$ such that $\overline{\tau }(w\otimes x) = \overline{d_{w,x}}$ for every $w\in W$ and $x\in X$ . Putting

\begin{equation*} \delta ^{FG}(a,b) \;:\!=\; \overline {\tau }(b)(a)\in T \end{equation*}

we have a well-defined mapping $\delta ^{FG} \;:\; (\overline{X}\otimes _S V) \times (W\otimes _S X)\to T$ such that

\begin{equation*} \delta ^{FG}\left (\sum _{k = 1}^n \overline {y}_k \otimes v_k,\sum _{h = 1}^m w_h \otimes x_h\right ) \!=\! \overline {\tau }\left (\sum _{h = 1}^m w_h \otimes x_h\right )\!\!\left (\sum _{k = 1}^n \overline {y}_k \otimes v_k\right ) \!=\! \sum _{k = 1}^n \sum _{h = 1}^m [y_k,\delta (v_k,w_h)x_h]. \end{equation*}

It easily follows that $\delta ^{FG}$ is additive in both arguments. We also have

\begin{align*} \delta ^{FG}(t(\overline y\otimes v), w\otimes x) & = \delta ^{FG}(\overline{yt^*}\otimes v, w\otimes x) = [yt^*,\delta (v,w)x] = t[y,\delta (v,w)x] \\[5pt] & = t \delta ^{FG}(\overline y\otimes v, w\otimes x), \\[5pt] \delta ^{FG}(\overline y\otimes v, (w\otimes x)t) & = [y,\delta (v,w) xt] = [y,\delta (v,w) x]t = \delta ^{FG}(\overline y\otimes v, w\otimes x) t \end{align*}

for every $t\in T$ , $x,y\in X$ and $v\in V$ . So $\delta ^{FG}$ is a dual $T$ -pair.

(4) Let $\delta _1\;:\; V_1\times W_1\to S$ and $\delta _2\;:\; V_2\times W_2\to S$ be dual pairs and let $f\;:\; V_1\to V_2$ , $f^\sharp \;:\; W_2\to W_1$ be module homomorphisms such that

\begin{equation*} \delta _2(f(v_1), w_2) = \delta _1\left (v_1,f^\sharp (w_2)\right ) \end{equation*}

in $S$ for all $v_1\in V_1$ and $w_2\in W_2$ . Then, for every $x,y\in X$ ,

\begin{align*} & [y,\delta _2(f(v_1), w_2) x] = [y,\delta _1(v_1,f^\sharp (w_2)) x] \\[5pt] & \implies \delta _2^{FG}(\overline{y}\otimes f(v_1),w_2\otimes x) = \delta _1^{FG}(\overline{y}\otimes v_1, f^\sharp (w_2)\otimes x) \\[5pt] & \implies \delta _2^{FG}(({\mathrm{id}}_{\overline X}\otimes f)(\overline y\otimes v_1), w_2\otimes x) = \delta _1^{FG}(\overline y\otimes v_1, (f^\sharp \otimes{\mathrm{id}}_X)(w_2\otimes x)) \\[5pt] & \implies \delta _2^{FG}(F(f)(\overline y\otimes v_1), w_2\otimes x) = \delta _1^{FG}(\overline y\otimes v_1, G(f^\sharp )(w_2\otimes x)), \end{align*}

as needed. Thus, the pair $(F,G)$ is a functor-multiplier.

Claim 2. The functor-multiplier $(F,G)$ is a $*$ -functor.

We need to define a natural transformation $\alpha \;:\; F\Rightarrow H_T\circ G\circ H'_S$ . For every ${_SM}\in{_S\textsf{FMod}}$ , we define a mapping

\begin{equation*} \alpha _M\;:\; F(M) = \overline {X}\otimes _S M \to \overline {\overline {M}\otimes _S X} = (H_T\circ G\circ H'_S)(M) \end{equation*}

by

\begin{equation*} \alpha _M(\overline y\otimes m)\;:\!=\; \overline {\overline m\otimes y}. \end{equation*}

We will show that $\alpha \;:\!=\; (\alpha _M)_{M \in{_S\textsf{FMod}}}\colon \ F \Rightarrow H_T\circ G\circ H_S$ is a natural isomorphism.

Well-defined. Let ${_SM} \in{_S\textsf{FMod}}$ . Consider the mapping

\begin{equation*} \hat {\alpha }_M\colon \overline {X} \times M \to \overline {\overline {M} \otimes _S X}, \qquad (\overline {y},m) \mapsto \overline {\overline {m} \otimes y}. \end{equation*}

Note that, for every $y,y' \in X$ , $m,m' \in M$ and $s \in S$ , we have

\begin{align*} \hat{\alpha }_M(\overline{y} + \overline{y'},m) &= \hat{\alpha }_M(\overline{y + y'},m) = \overline{\overline{m} \otimes (y + y')} = \overline{\overline{m} \otimes y + \overline{m} \otimes y'} \\[5pt] &= \overline{\overline{m} \otimes y} + \overline{\overline{m} \otimes y'} = \hat{\alpha }_M(\overline{y},m) + \hat{\alpha }_M(\overline{y'},m), \\[5pt] \hat{\alpha }_M(\overline{y}s,m) &= \hat{\alpha }_M(\overline{s^*y},m) = \overline{\overline{m} \otimes s^*y} = \overline{\overline{m}s^* \otimes y} = \overline{\overline{sm} \otimes y} = \hat{\alpha }_M(\overline{y},sm) \end{align*}

and $\hat{\alpha }_M(\overline{y},m + m') = \hat{\alpha }_M(\overline{y},m) + \hat{\alpha }_M(\overline{y},m')$ . By the universal property of the tensor product, we see that $\alpha _M$ is a well-defined group homomorphism.

Homomorphism of $T$ -modules. Let ${_SM} \in{_S\textsf{FMod}}$ . Note that, for every $t \in T$ , $y \in X$ and $m \in M$ , we have

\begin{equation*} \alpha _M(t\overline {y} \otimes m) = \alpha _M(\overline {yt^*} \otimes m) = \overline {\overline {m} \otimes yt^*} = \overline {(\overline {m} \otimes y)t^*} = t\overline {\overline {m} \otimes y} = t\alpha _M(\overline {y} \otimes m). \end{equation*}

Hence, we see that $\alpha _M$ is a homomorphism of left $T$ -modules.

Bijective. Let $M_S \in \textsf{FMod}_S$ . Analogously to the first point, we see that there exists a well-defined group homomorphism

\begin{equation*} \beta _M'\colon \ \overline {M} \otimes _S X \to \overline {X} \otimes _S M, \qquad \overline {m} \otimes y \mapsto \overline {y} \otimes m. \end{equation*}

We may also consider a mapping $\beta _M\colon \overline{\overline{M} \otimes _S X} \to \overline{X} \otimes _S M$ , $\overline{\overline{m} \otimes y} \mapsto \overline{y} \otimes m$ , because $\beta _M = \beta _M'\circ u_{\overline{M} \otimes X}$ , where $u_{\overline{M} \otimes X}\colon \overline{\overline{M} \otimes _S X} \to \overline{M} \otimes _S X$ , $\overline{a} \mapsto a$ . Clearly, $\beta _M$ is the inverse of $\alpha _M$ in $\textsf{Set}$ , which proves that $\alpha _M$ is bijective.

Natural transformation. Let ${_SM},{_SM'} \in{_S\textsf{FMod}}$ and $f\colon{_SM} \to{_SM'}$ be a homomorphism of left $S$ -modules. We need to show that the diagram

commutes. If $y \in X$ and $m \in M$ , then

\begin{align*} ((H_T\circ G\circ H_S)(f) \circ \alpha _M)(\overline{y} \otimes m) &= (H_T\circ G\circ H_S)(f) (\overline{\overline{m} \otimes y}) \\[5pt] & = \overline{(G\circ H_S)(f)(\overline{m})\otimes y} = \overline{(H_S(f)\otimes{\mathrm{id}}_X)(\overline{m}\otimes y)} \\[5pt] & = \overline{H_S(f)(\overline{m})\otimes y} = \overline{\overline{f(m)} \otimes y}\\[5pt] & = \alpha _{M'}(\overline{y} \otimes f(m)) = (\alpha _{M'} \circ ({\mathrm{id}}_X \otimes f))(\overline{y} \otimes m). \end{align*}

Therefore we see that $\alpha$ is indeed a natural transformation.

In conclusion, we have shown that $\alpha$ is a natural isomorphism.

To verify the equality (5.2), let $\delta \;:\; V\times W\to S$ be a dual pair and consider $a\in \overline{X}\otimes _S V = F(V)$ and $b\in \overline{\overline{X}\otimes _S \overline{W}} = (H'_T \circ F \circ H_S)(W)$ . Then $a= \sum _{k = 1}^n\overline y _k\otimes v_k$ , $b= \sum _{h = 1}^m\overline{\overline{x}_h\otimes \overline{w}_h}$ and

\begin{align*} \overleftarrow{\overleftarrow{\delta }^{FG}}(\alpha _V(a),b) &= \overleftarrow{\overleftarrow{\delta }^{FG}}\left (\alpha _V\left (\sum\limits_{k = 1}^n\overline{y}_k \otimes v_k\right ), \sum\limits_{h = 1}^m\overline{\overline{x}_h \otimes \overline{w}_h}\right ) \\[5pt] &= \overleftarrow{\overleftarrow{\delta }^{FG}}\left (\sum\limits_{k = 1}^n \overline{\overline{v}_k \otimes y_k}, \sum\limits_{h = 1}^m\overline{\overline{x}_h \otimes \overline{w}_h}\right ) \tag*{$(\text{def. of $\alpha _V$})$} \\[5pt] & = \overleftarrow{\delta }^{FG}\left (\sum\limits_{h = 1}^m \overline{x}_h \otimes \overline{w}_h, \sum\limits_{k = 1}^n\overline{v}_k \otimes y_k\right )^* \tag*{$(\text{by (5.1)})$} \\[5pt] &= \sum\limits_{h = 1}^m \sum\limits_{k = 1}^n [x_h,\overleftarrow{\delta }(\overline{w}_h,\overline{v}_k)y_k]^* \tag*{$(\text{by (5.6)})$} \\[5pt] & = \sum\limits_{h = 1}^m \sum\limits_{k = 1}^n [x_h,\delta (v_k,w_h)^*y_k]^* \tag*{$(\text{by (5.1)})$} \\[5pt] &= \sum\limits_{h = 1}^m \sum\limits_{k = 1}^n [\delta (v_k,w_h)^*y_k,x_h] \tag*{$({\textbf{RC4}})$} \\[5pt] & = \sum\limits_{h = 1}^m \sum\limits_{k = 1}^n [y_k,\delta (v_k,w_h)x_h] \tag*{$({\textbf{RC9}})$} \\[5pt] &= \delta ^{FG}\left (\sum\limits_{k = 1}^n \overline{y}_k \otimes v_k,\sum\limits_{h = 1}^m w_h \otimes x_h\right )\tag*{$(\text{by (5.6)})$} \\[5pt] &= \delta ^{FG}\left ( a,\;\sum\limits_{h = 1}^m \overline{\overline{w_h \otimes x_h}}\right ) \\[5pt] &= \delta ^{FG}\left ( a, \;\overline{\alpha _{\overline{W}}\left (\sum\limits_{h = 1}^m \overline{x}_h \otimes \overline{w}_h\right )}\right ) \tag*{$(\text{def. of $\alpha _{\overline{W}}$})$} \\[5pt] &= \delta ^{FG}\left ( a, \; H_T'(\alpha _{\overline{W}})\left (\sum\limits_{h = 1}^m \overline{\overline{x}_h \otimes \overline{w}_h}\right )\right ) \tag*{$(\text{def. of $H'_T$})$} \\[5pt] & = \delta ^{FG}(a,H_T'(\alpha _{H_S(W)})(b)). \tag*{$(\text{def. of $H_S$})$} \end{align*}

Hence, we see that $(F,G,\alpha )$ is indeed a $*$ -functor from $S$ to $T$ .

Claim 3. The $*$ -functor $(F,G)$ is a $*$ -equivalence.

Analogously, we see that the pair $(F',G',\beta )$ is a $*$ -functor from $T$ to $S$ , where

\begin{align*} & F' \;:\!=\; X \otimes _T \underline{\hspace{0.8em}}\colon \ {_T\textsf{FMod}} \to{_S\textsf{FMod}}, \\[5pt] & G' \;:\!=\; \underline{\hspace{0.8em}} \otimes _T \overline{X}\colon \ \textsf{FMod}_T \to \textsf{FMod}_S; \\[5pt] & \beta = (\beta _M)_{M \in{_T\textsf{FMod}}}\colon \ F'\Rightarrow H_S\circ G'\circ H_T', \\[5pt] & \qquad \beta _M(x \otimes m) \;:\!=\; \overline{\overline{m} \otimes \overline{x}}. \end{align*}

Also, for any dual $T$ -pair $\delta _T\colon V \times W \to T$ , we have

(5.7) \begin{equation} \delta _T^{F'G'}\colon F'(V) \times G'(W) \to S, \qquad (x \otimes v, w \otimes \overline{y}) \mapsto \langle x \, \delta _T(v,w),y\rangle . \end{equation}

To complete the proof we need to show that the requirements of Definition 5.5 are met. For every ${_SM} \in{_S\textsf{FMod}}$ , consider the mapping

\begin{equation*} \tau _M\colon (F' \circ F)({_SM}) = X \otimes _T \overline {X} \otimes _S M \to S \otimes _S M, \qquad x \otimes \overline {y} \otimes m \mapsto \langle x,y\rangle \otimes m. \end{equation*}

As in (3.1), we have a well-defined group homomorphism $\theta \colon X \otimes _T \overline{X} \to S$ , $x \otimes \overline{y} \mapsto \langle x,y\rangle$ , which is a homomorphism of ( $S,S$ )-bimodules. This implies that $\tau _M = \theta \otimes{\mathrm{id}}_M$ is a well-defined homomorphism of left $S$ -modules. The homomorphism $\tau _M$ is surjective, because $\langle \,,\,\rangle$ is full. We will show that $\tau _M$ is also injective. Consider the short exact sequence of ( $S,S$ )-bimodules

\begin{equation*} \{0\} \to {\mathrm {Ker}}\theta \overset {\iota }{\to } X \otimes _T \overline {X} \overset {\theta }{\to } S \to \{0\}, \end{equation*}

where $\iota \colon{\mathrm{Ker}}\theta \to X \otimes _T \overline{X}$ is the inclusion. By [Reference Wisbauer28, Result 12.8(1)], the sequence of left $S$ -modules

(5.8) \begin{equation} {\mathrm{Ker}}\theta \otimes _S M \overset{\iota \otimes{\mathrm{id}}_M}{\to } X \otimes _T \overline{X} \otimes _S M \overset{\tau _M}{\to } S \otimes _S M \to \{0\} \end{equation}

is also exact. We show that $({\mathrm{Ker}}\theta )S = \{0\}$ . Let $\sum _{k = 1}^n x_k \otimes \overline{y}_k \in{\mathrm{Ker}}\theta$ and $s \in S$ . Since $\theta$ is surjective, there exists $\sum _{h = 1}^m \xi _h \otimes \overline{\zeta }_h \in X \otimes _T \overline{X}$ such that $s = \theta (\sum _{h = 1}^m \xi _h \otimes \overline{\zeta }_h)$ . Now

\begin{align*} \left (\sum _{k = 1}^n x_k \otimes \overline{y}_k\right )&s = \sum _{k = 1}^n x_k \otimes \overline{y}_k \theta \left (\sum _{h = 1}^m \xi _h \otimes \overline{\zeta }_h\right ) = \sum _{k = 1}^n\sum _{h = 1}^m x_k \otimes \overline{y}_k\langle \xi _h, \zeta _h\rangle \\[5pt] &= \sum _{k = 1}^n\sum _{h = 1}^m x_k \otimes \overline{\langle \xi _h, \zeta _h\rangle ^*y_k} = \sum _{k = 1}^n\sum _{h = 1}^m x_k \otimes \overline{\langle \zeta _h,\xi _h\rangle y_k} = \sum _{k = 1}^n\sum _{h = 1}^m x_k \otimes \overline{\zeta _h[\xi _h, y_k]} \\[5pt] & = \sum _{k = 1}^n\sum _{h = 1}^m x_k \otimes [\xi _h, y_k]^*\overline{\zeta _h} = \sum _{k = 1}^n\sum _{h = 1}^m x_k[\xi _h, y_k]^* \otimes \overline{\zeta _h} = \sum _{k = 1}^n\sum _{h = 1}^m x_k[y_k,\xi _h] \otimes \overline{\zeta _h} \\[5pt] & = \sum _{k = 1}^n\sum _{h = 1}^m \langle x_k,y_k\rangle \xi _h \otimes \overline{\zeta _h} = \sum _{h = 1}^m \theta \left (\sum _{k = 1}^n x_k\otimes y_k\right ) \xi _h \otimes \overline{\zeta _h} = \sum _{h = 1}^m 0 \xi _h \otimes \overline{\zeta _h} = 0. \end{align*}

Hence, $({\mathrm{Ker}}\theta )S = \{0\}$ .

Now let $\xi \otimes m$ be an elementary tensor in ${\mathrm{Ker}}\theta \otimes _S M$ (then $\xi \otimes m$ is also a generator of ${\mathrm{Ker}}\theta \otimes _S M$ ). Since $_SM$ is unitary, there exist $s_{1},\ldots, s_{j} \in S$ and $m_{1},\ldots, m_{j} \in M$ such that $m = s_{1}m_{1} + \ldots + s_{j}m_{j}$ . Then $\xi s_h=0$ for every $h\in \{1,\ldots, j\}$ and

\begin{equation*} \xi \otimes m = \xi \otimes \left (\sum _{h = 1}^j s_{h}m_{h}\right ) = \sum _{h = 1}^j \xi \otimes s_{h}m_{h} = \sum _{h = 1}^j \xi s_{h} \otimes m_{h} = 0. \end{equation*}

It follows that ${\mathrm{Ker}}\theta \otimes _S M = \{0\}$ . Hence, $\tau _M$ is bijective, due to the exactness of the sequence (5.8).

Let ${_SM},{_SM'} \in{_S\textsf{FMod}}$ and $f\colon{_SM} \to{_SM'}$ a morphism in $_S\textsf{FMod}$ . The diagram

commutes, because

\begin{align*} (({\mathrm{id}}_S\otimes f)\circ \tau _M)(x\otimes \overline{y}\otimes m) & = ({\mathrm{id}}_S\otimes f)(\left \langle \,x,y\,\right \rangle \otimes m) = \left \langle \,x,y\,\right \rangle \otimes f(m) \\[5pt] & = \tau _{M'}(x\otimes \overline{y}\otimes f(m)) = (\tau _{M'}\circ ({\mathrm{id}}_{X\otimes \overline{X}}\otimes f)) (x\otimes \overline{y}\otimes m). \end{align*}

We have shown that $\tau = (\tau _M)_{M \in{_S\textsf{FMod}}}\colon F' \circ F \Rightarrow S \otimes _S \underline{\hspace{0.8em}}$ is a natural isomorphism. By the definition of the category $_S\textsf{FMod}$ , there exists a natural isomorphism $\nu = (\nu _M)_{M\in{_S\textsf{FMod}}} \colon S \otimes _S \underline{\hspace{0.8em}} \Rightarrow{\mathrm{id}}_{{_S\textsf{FMod}}}$ , where

\begin{equation*} \nu _M\colon S\otimes _S M\to M, \;\; x\otimes m\mapsto sm. \end{equation*}

Composing $\tau$ and $\nu$ , we obtain a natural isomorphism

\begin{align*} \varphi = (\varphi _M)&\colon \ F' \circ F \Rightarrow{\mathrm{id}}_{{_S\textsf{FMod}}}, \\[5pt] \varphi _M&\colon X \otimes _T \overline{X} \otimes _S M \to{_SM}, \qquad x \otimes \overline{y} \otimes m \mapsto \langle x,y\rangle m. \end{align*}

We will show that $\varphi$ satisfies condition (5.4). Similar reasoning will lead to a natural transformation $\psi \;:\; F\circ F'\Rightarrow{\mathrm{id}}_{{_T\textsf{FMod}}}$ satisfying (5.5). First, we calculate the natural isomorphism

\begin{equation*} \beta * \alpha \colon F' \circ F = X \otimes _T \overline {X} \otimes _S \underline {\hspace {0.8em}} \Rightarrow (H_S \circ G' \circ H_T') \circ (H_T \circ G \circ H_S') = H_S \circ (\underline {\hspace {0.8em}} \otimes _S X \otimes _T \overline {X}) \circ H'_S. \end{equation*}

For every ${_SM} \in{_S\textsf{FMod}}$ ,

\begin{align*} &(\beta * \alpha )_{M} = \beta _{H_TGH_S'(M)} \circ F'(\alpha _M)\colon X\otimes _T \overline{X}\otimes _S M \to \overline{\overline{M}\otimes _S X\otimes _T \overline{X}}, \\[5pt] & x \otimes \overline{y} \otimes m \mapsto \beta _{H_TGH_S'(M)}(x \otimes \overline{\overline{m} \otimes y}) = \overline{\overline{m} \otimes y \otimes \overline{x}}. \end{align*}

Let $\delta \colon V \times W \to S$ be a dual pair. We will prove (5.4). Let $a \in (F' \circ F)(V)$ and $b \in W$ . Then $a = \sum _{k = 1}^n x_k \otimes \overline{y}_k \otimes v_k$ for some $x_k,y_k\in X$ and $v_k\in V$ . Since $W_S$ is firm (thus unitary) and $\langle \,,\,\rangle$ is full, we can write $b = \sum _{h = 1}^j w_h\langle \xi _h,\zeta _h\rangle$ , where $w_h\in W$ and $\xi _h,\zeta _h\in X$ . Now

\begin{equation*} \overline {b} = \sum _{h=1}^j \overline {w_h \left \langle \,\xi _h,\zeta _h\,\right \rangle } = \sum _{h=1}^j \left \langle \,\xi _h,\zeta _h\,\right \rangle ^*\overline {w_h} = \sum _{h=1}^j \left \langle \,\zeta _h,\xi _h\,\right \rangle \overline {w_h} = \varphi _{H(W)} \left ( \sum _{h=1}^j \zeta _h\otimes \overline {\xi _h} \otimes \overline {w_h} \right ), \end{equation*}

so we have

\begin{align*} H_S'((\beta * \alpha )_{H(W)} \circ \varphi _{H(W)}^{-1})(b) &= \left ( H_S'((\beta * \alpha )_{H(W)}) \circ H'_S(\varphi _{H(W)}^{-1})\right ) \left (\overline{\overline{b}}\right ) \tag*{($\text{$H'_S$ is a functor}$)} \\[5pt] &= H_S'((\beta * \alpha )_{H(W)}) \left (\overline{\varphi _{H(W)}^{-1} \left (\overline{b}\right )} \right ) \tag*{($\text{def. of $H'_S$}$)}\\[5pt] &= H_S'((\beta * \alpha )_{H(W)})\left (\overline{\sum _{h = 1}^j \zeta _h \otimes \overline{\xi }_h \otimes \overline{w_h}}\right ) \\[5pt] &= \overline{(\beta * \alpha )_{H(W)}\left (\sum _{h = 1}^j \zeta _h \otimes \overline{\xi }_h \otimes \overline{w_h}\right )} \tag*{($\text{def. of $H'_S$}$)} \\[5pt] &= \overline{\sum _{h = 1}^j \overline{\overline{\overline{w_h}} \otimes \xi _h \otimes \overline{\zeta }_h}} \tag*{($\text{def. of $(\beta *\alpha )_{H(W)}$})$} \\[5pt] & = \sum _{h = 1}^j w_h \otimes \xi _h \otimes \overline{\zeta }_h, \end{align*}

and therefore

\begin{align*} \delta ^{F'F,G'G}&(a,H_S'((\beta * \alpha )_{H(W)} \circ \varphi _{H(W)}^{-1})(b)) \\[5pt] &= \delta ^{F'F,G'G}\left (\sum _{k = 1}^n x_k \otimes \overline{y}_k \otimes v_k,\sum _{h = 1}^j w_h \otimes \xi _h \otimes \overline{\zeta }_h\right ) \\[5pt] &= \sum _{k = 1}^n\sum _{h = 1}^j\left \langle x_k\delta ^{FG}(\overline{y}_k \otimes v_k, w_h \otimes \xi _h),\zeta _h\right \rangle \tag*{$(\text{by (5.7)})$} \\[5pt] & = \sum _{k = 1}^n\sum _{h = 1}^j\left \langle x_k[y_k, \delta (v_k, w_h) \xi _h],\zeta _h\right \rangle \tag*{$(\text{by (5.6)})$} \\[5pt] &= \sum _{k = 1}^n\sum _{h = 1}^j\left \langle x_k,\zeta _h[y_k, \delta (v_k, w_h) \xi _h]^*\right \rangle \tag*{$({\textbf{RC8}})$}\\[5pt] & = \sum _{k = 1}^n\sum _{h = 1}^j\langle x_k,\zeta _h[\delta (v_k, w_h) \xi _h, y_k]\rangle \tag*{$({\textbf{RC4}})$} \\[5pt] &= \sum _{k = 1}^n\sum _{h = 1}^j \langle x_k,\langle \zeta _h,\delta (v_k, w_h) \xi _h\rangle y_k\rangle \tag*{$({\textbf{RC5}})$}\\[5pt] & = \sum _{k = 1}^n\sum _{h = 1}^j \langle x_k, y_k\rangle \langle \zeta _h,\delta (v_k, w_h) \xi _h\rangle ^* \tag*{$({\textbf{RC6}})$}\\[5pt] &= \sum _{k = 1}^n\sum _{h = 1}^j \langle x_k, y_k\rangle \langle \delta (v_k, w_h) \xi _h, \zeta _h\rangle \tag*{$({\textbf{RC2}})$}\\[5pt] & = \sum _{k = 1}^n\sum _{h = 1}^j \langle x_k, y_k\rangle \delta (v_k, w_h) \langle \xi _h, \zeta _h\rangle \tag*{$({\textbf{RC1}})$}\\[5pt] &= \delta \left (\sum _{k = 1}^n \langle x_k, y_k\rangle v_k, \sum _{h = 1}^j w_h\langle \xi _h,\zeta _h\rangle \right ) \tag*{$(\text{additivity})$} \\[5pt] &= \delta \left (\varphi _V\left (\sum _{k = 1}^n x_k \otimes \overline{y}_k \otimes v_k\right ),b\right ) \tag*{$(\text{def. of $\varphi _V$})$} \\[5pt] &= \delta (\varphi _V(a),b). \end{align*}

The proof is complete.

6. From a category equivalence to a Rieffel context

In this section, we will give a necessary and sufficient condition for the existence of a firm and full Rieffel context between two idempotent rings with involution in terms of equivalence of categories of firm modules.

Lemma 6.1. Let $S$ and $T$ be rings with involution. For every bimodule $_SX_T$ , there exists a group homomorphism $\sigma _X\;:\; X\otimes _T\overline{X} \to X\otimes _T\overline{X}$ such that

\begin{equation*} \sigma _X(x\otimes \overline {y}) = y\otimes \overline {x} \end{equation*}

for every $x,y\in X$ .

Proof. The mapping $ \hat{\sigma }\;:\; X\times \overline{X} \to X\otimes _T\overline{X}, \;\; (x,\overline{y})\mapsto y\otimes \overline{x}$ , is $T$ -balanced, because

\begin{equation*} \hat {\sigma }(xt,\overline {y}) = y\otimes \overline {xt} = y\otimes t^*\overline {x} = yt^*\otimes \overline {x} = \hat {\sigma }(x,\overline {yt^*}) = \hat {\sigma }(x,t\overline {y}), \end{equation*}

and it is obviously additive in both arguments. Now the claim follows from the universal property of tensor product.

We recall that there exist two natural transformations $\mu = (\mu _M)_{M\in \textsf{Mod}_S}\colon \underline{\hspace{0.8em}} \otimes _S S \Rightarrow{\mathrm{id}}_{\textsf{Mod}_S}$ and $\nu = (\nu _N)_{N\in{_S\textsf{Mod}}}\colon S \otimes _S \underline{\hspace{0.8em}} \Rightarrow{\mathrm{id}}_{{_S\textsf{Mod}}}$ with components

\begin{align*} \mu _M\colon M\otimes _S S \to M, \;\;& m\otimes s \mapsto ms, \\[5pt] \nu _N\colon S\otimes _S N \to N, \;\;& s\otimes n\mapsto sn. \end{align*}

Proof. The mappings

\begin{align*} \hat{\kappa }\;:\; X\times T \to T\otimes _T \overline{X},\;\;\; & (x,t)\mapsto t^*\otimes \overline{x}, \\[5pt] \hat{\omega }\;:\; T\times \overline{X}\to X\otimes _T T,\;\;\; & (t,\overline{x})\mapsto x\otimes t^* \end{align*}

are $T$ -balanced, because

\begin{align*} \hat{\kappa }(xt_1, t) & = t^*\otimes \overline{xt_1} = t^*\otimes t_1^* \overline{x} = t^*t_1^*\otimes \overline{x} = (t_1t)^* \otimes \overline{x} = \hat{\kappa }(x,t_1t), \\[5pt] \hat{\omega } (tt_1,\overline{x}) & = x\otimes (tt_1)^* = x\otimes t_1^*t^* = xt_1^*\otimes t^* = \hat \omega (t,\overline{xt_1^*}) = \hat \omega (t,t_1\overline{x}), \end{align*}

and additivity in both arguments is clear. Hence, there exist group homomorphisms $\kappa \;:\; X\otimes _T T \to T\otimes _T \overline{X}$ and $\omega \;:\; T\otimes _T \overline{X}\to X\otimes _T T$ such that

\begin{align*} \kappa (x\otimes t) & = t^*\otimes \overline{x}, \\[5pt] \omega (t\otimes \overline{x}) & = x\otimes t^*. \end{align*}

It is easy to see that $\kappa =\omega ^{-1}$ , so $\kappa$ is a bijective mapping. Moreover, the square

commutes, because

\begin{equation*} (\overline {(\;\;)}\circ \mu _X)(x\otimes t) = \overline {xt} = t^*\overline {x} = \nu _{\overline {X}}(t^*\otimes \overline {x}) = (\nu _{\overline {X}}\circ \kappa )(x\otimes t). \end{equation*}

Since $\kappa$ and $\overline{(\;\;)}$ are bijections, $\mu$ is a bijection if and only if $\nu$ is a bijection.

Proof. (1) $\implies$ (2). Let $(S,T,{_SX_T}, \left \langle \,,\,\right \rangle, [\,,\,])$ be a firm and full Rieffel context. In the proof of Theorem5.7, we have seen that

\begin{align*} & F= \overline{X} \otimes _S - \colon{_S\textsf{FMod}}\to{_T\textsf{FMod}}, & G= -\otimes _S X \colon \textsf{FMod}_S\to \textsf{FMod}_T, \\[5pt] & F'= X \otimes _T - \colon{_T\textsf{FMod}}\to{_S\textsf{FMod}}, & G'= -\otimes _T \overline{X} \colon \textsf{FMod}_T\to \textsf{FMod}_S \end{align*}

are equivalence functors and $\varphi = (\varphi _{_SM})\;:\; F'\circ F \Rightarrow{\mathrm{id}}_{{_S\textsf{FMod}}}$ , where

\begin{equation*} \varphi _{_SM}\;:\; X\otimes _T \overline {X}\otimes _S M \to {_SM}, \;\; x\otimes \overline {y} \otimes m \mapsto \left \langle \,x,y\,\right \rangle m, \end{equation*}

is a natural isomorphism. Analogously we can construct natural isomorphisms $\psi$ , $\xi$ and $\zeta$ by defining

\begin{align*} \psi _{_TN}\;:\; \overline{X}\otimes _S X \otimes _T N \to{_TN},\hspace{1cm} & \overline{y}\otimes x \otimes n \mapsto [y,x]n, \\[5pt] \xi _{N_S}\;:\; N\otimes _S X\otimes _T \overline{X} \to N_S, \hspace{1cm} & n\otimes x\otimes \overline{y} \mapsto n\left \langle \,x,y\,\right \rangle, \\[5pt] \zeta _{M_T}\;:\; M\otimes _T \overline{X} \otimes _S X \to M_T, \hspace{1cm} & m\otimes \overline{y} \otimes x \mapsto m[y,x]. \end{align*}

Now, for every $x,x',y,y'\in X$ ,

\begin{align*} \zeta _{X_T} (x'\otimes \overline{y}\otimes x) & = x'[y,x] = \left \langle \,x',y\,\right \rangle x = \varphi _{_SX}(x'\otimes \overline{y} \otimes x), \\[5pt] \psi _{_T\overline{X}} (\overline{y}\otimes x \otimes \overline{y'}) & = [y,x]\overline{y'} = \overline{y'[y,x]^*} = \overline{y'[x,y]} = \overline{\left \langle \,y',x\,\right \rangle y} \\[5pt] & = \overline{\left \langle \,x,y'\,\right \rangle ^*y} = \overline{y}\left \langle \,x,y'\,\right \rangle = \xi _{\overline{X}_S} (\overline{y}\otimes x \otimes \overline{y'}). \end{align*}

So $\zeta _{X_T}= \varphi _{_SX}$ and $\psi _{_T\overline{X}} = \xi _{\overline{X}_S}$ .

We will prove the commutativity of rectangle (R1) (for rectangle (R2) the proof will be similar). Take $x\otimes \overline{y}\in X\otimes _T \overline{X}$ . Since $_SX$ is unitary, we can write

(6.1) \begin{equation} x= \sum _{i = 1}^{\hat{\imath }} s_ix_i, \;\; x_i= \sum _{j = 1}^{\hat{\jmath }} s_{ij}x_{ij}, \;\; y= \sum _{k = 1}^{\hat{k}} u_ky_k, \;\; y_k = \sum _{l = 1}^{\hat{l}} u_{kl}y_{kl} \end{equation}

for some $x_i,x_{ij}, y_k, y_{kl}\in X$ and $s_i,s_{ij}, u_k, u_{kl}\in S$ . Then

\begin{align*} & \left ( ({\mathrm{id}}_X\otimes \mu _{\overline{X}})\circ ({\mathrm{id}}_X \otimes \mu _{\overline{X}}\otimes{\mathrm{id}}_S)\right ) \left ( \sum _{k = 1}^{\hat{k}} \sum _{l = 1}^{\hat{l}} x\otimes \overline{y_{kl}}\otimes u_{kl}^* \otimes u_k^*\right ) \\[5pt] & = \left ({\mathrm{id}}_X\otimes \mu _{\overline{X}}\right ) \left ( \sum _{k = 1}^{\hat{k}} \sum _{l = 1}^{\hat{l}} x\otimes \overline{y_{kl}} u_{kl}^* \otimes u_k^*\right ) \tag*{$(\text{def. of $\mu _{\overline{X}}$})$} \\[5pt] & = \left ({\mathrm{id}}_X\otimes \mu _{\overline{X}}\right ) \left ( \sum _{k = 1}^{\hat{k}} \sum _{l = 1}^{\hat{l}} x\otimes \overline{u_{kl}y_{kl}} \otimes u_k^*\right ) \tag*{$(\text{by (2.2)})$} \\[5pt] & = \left ({\mathrm{id}}_X\otimes \mu _{\overline{X}}\right ) \left ( \sum _{k = 1}^{\hat{k}} x\otimes \overline{y_k} \otimes u_k^*\right ) \tag*{$(\text{by (6.1)})$} \\[5pt] & = \sum _{k = 1}^{\hat{k}} x\otimes \overline{y_k}u_k^* \tag*{$(\text{def. of $\mu _{\overline{X}}$})$} \\[5pt] & = \sum _{k = 1}^{\hat{k}} x\otimes \overline{u_ky_k} \tag*{$(\text{by (2.2)})$} \\[5pt] & = x\otimes \overline{y} \tag*{$(\text{by (6.1)})$} \end{align*}

and, analogously,

(6.2) \begin{equation} \left ( ({\mathrm{id}}_X\otimes \mu _{\overline{X}})\circ ({\mathrm{id}}_X \otimes \mu _{\overline{X}}\otimes{\mathrm{id}}_S)\right ) \left ( \sum _{i = 1}^{\hat{\imath }} \sum _{j = 1}^{\hat{\jmath }} y\otimes \overline{x_{ij}}\otimes s_{ij}^* \otimes s_i^*\right ) = y\otimes \overline{x}. \end{equation}

Hence,

\begin{align*} &\left ( (\;)^* \circ \mu _S \circ \varphi _{S\otimes S} \circ ({\mathrm{id}}_X \otimes \mu _{\overline{X}}^{-1}\otimes{\mathrm{id}}_S)\circ ({\mathrm{id}}_X \otimes \mu _{\overline{X}}^{-1}) \right ) (x\otimes \overline{y}) \\[5pt] & = \left ( (\;)^* \circ \mu _S \circ \varphi _{S\otimes S} \right ) \left ( \sum _{k = 1}^{\hat{k}} \sum _{l = 1}^{\hat{l}} x\otimes \overline{y_{kl}}\otimes u_{kl}^* \otimes u_k^*\right ) \\[5pt] & = \sum _{k = 1}^{\hat{k}} \sum _{l = 1}^{\hat{l}} \left ( (\;)^* \circ \mu _S \circ \varphi _{S\otimes S} \right ) \left ( x\otimes \overline{y_{kl}}\otimes u_{kl}^* \otimes u_k^*\right ) \tag*{$(\text{additivity})$} \\[5pt] & = \sum _{k = 1}^{\hat{k}} \sum _{l = 1}^{\hat{l}} \left ( (\;)^* \circ \mu _S \right ) \left ( \left \langle \,x,y_{kl}\,\right \rangle (u_{kl}^* \otimes u_k^*) \right ) \tag*{$(\text{def. of $\varphi _{S\otimes S}$})$} \\[5pt] & = \sum _{k = 1}^{\hat{k}} \sum _{l = 1}^{\hat{l}} \left ( (\;)^* \circ \mu _S \right ) \left ( \left \langle \,x,y_{kl}\,\right \rangle u_{kl}^* \otimes u_k^* \right ) \tag*{$(\text{$S$-action of $S\otimes S$})$} \end{align*}

\begin{align*} & = \sum _{k = 1}^{\hat{k}} \sum _{l = 1}^{\hat{l}} \left ( \left \langle \,x,y_{kl}\,\right \rangle u_{kl}^*u_k^* \right )^* \tag*{$(\text{def. of $\mu _S$ and $(\;)^*$})$}\\[5pt] & = \left ( \sum _{k = 1}^{\hat{k}} \sum _{l = 1}^{\hat{l}} \left \langle \,x,y_{kl}\,\right \rangle u_{kl}^*u_k^* \right )^* \tag*{$(\text{by (2.1)})$} \\[5pt] & = \left ( \sum _{k = 1}^{\hat{k}} \sum _{l = 1}^{\hat{l}} \left \langle \,x,u_{kl}y_{kl}\,\right \rangle u_k^* \right )^* \tag*{$({\textbf{RC6}})$}\\[5pt] & = \left ( \sum _{k = 1}^{\hat{k}} \left ( \left \langle \,x,\sum _{l = 1}^{\hat{l}} u_{kl}y_{kl}\,\right \rangle \right ) u_k^* \right )^* \tag*{$(\text{additivity})$} \\[5pt] & = \left ( \sum _{k = 1}^{\hat{k}} \left \langle \,x, y_k\,\right \rangle u_k^* \right )^* \tag*{$(\text{by (6.1)})$} \\[5pt] & = \left ( \sum _{k = 1}^{\hat{k}} \left \langle \,x, u_ky_k\,\right \rangle \right )^* \tag*{$({\textbf{RC6}})$} \\[5pt] & = \left \langle \,x,y\,\right \rangle ^* \tag*{$(\text{by (6.1)})$} \\[5pt] & = \left \langle \,y,x\,\right \rangle \tag*{$({\textbf{RC2}})$} \end{align*}

and

\begin{align*} & \left ( \mu _S \circ \varphi _{S\otimes S}\circ ({\mathrm{id}}_X \otimes \mu _{\overline{X}}^{-1}\otimes{\mathrm{id}}_S)\circ ({\mathrm{id}}_X \otimes \mu _{\overline{X}}^{-1}) \circ \sigma _X\right ) (x\otimes \overline{y}) \\[5pt] & = \left ( \mu _S \circ \varphi _{S\otimes S}\circ ({\mathrm{id}}_X \otimes \mu _{\overline{X}}^{-1}\otimes{\mathrm{id}}_S)\circ ({\mathrm{id}}_X \otimes \mu _{\overline{X}}^{-1}) \right ) (y\otimes \overline{x}) \tag*{$(\text{def. of $\sigma _X$})$} \\[5pt] & = (\mu _S \circ \varphi _{S\otimes S}) \left ( \sum _{i = 1}^{\hat{\imath }}\sum _{j = 1}^{\hat{\jmath }} y\otimes \overline{x_{ij}} \otimes s_{ij}^* \otimes s_i^* \right ) \tag*{$(\text{by (6.2)})$} \\[5pt] & = \sum _{i = 1}^{\hat{\imath }}\sum _{j = 1}^{\hat{\jmath }} (\mu _S \circ \varphi _{S\otimes S}) \left ( y\otimes \overline{x_{ij}} \otimes s_{ij}^* \otimes s_i^* \right ) \tag*{$(\text{additivity})$} \\[5pt] & = \sum _{i = 1}^{\hat{\imath }}\sum _{j = 1}^{\hat{\jmath }} \mu _S \left ( \left \langle \,y, x_{ij}\,\right \rangle ( s_{ij}^* \otimes s_i^*) \right ) \tag*{$(\text{def. of $\varphi _{S\otimes S}$})$}\\[5pt] & = \sum _{i = 1}^{\hat{\imath }}\sum _{j = 1}^{\hat{\jmath }} \mu _S \left ( \left \langle \,y, x_{ij}\,\right \rangle s_{ij}^* \otimes s_i^* \right ) \tag*{$(\text{$S$-action of $S\otimes S$})$} \\[5pt] & = \sum _{i = 1}^{\hat{\imath }}\sum _{j = 1}^{\hat{\jmath }} \left \langle \,y, x_{ij}\,\right \rangle s_{ij}^* s_i^* \tag*{$(\text{def. of $\mu _S$})$} \\[5pt] & = \sum _{i = 1}^{\hat{\imath }}\sum _{j = 1}^{\hat{\jmath }} \left \langle \,y, s_{ij}x_{ij}\,\right \rangle s_i^* \tag*{$({\textbf{RC6}})$} \end{align*}

\begin{align*} & = \sum _{i = 1}^{\hat{\imath }} \left \langle \,y, x_i\,\right \rangle s_i^* \tag*{$(\text{by (6.1)})$} \\[5pt] & = \sum _{i = 1}^{\hat{\imath }} \left \langle \,y, s_ix_i\,\right \rangle \tag*{$({\textbf{RC6}})$}\\[5pt] & = \left \langle \,y,x\,\right \rangle . \tag*{$(\text{by (6.1)})$} \end{align*}

Thus, the rectangle (R1) commutes.

(2) $\implies$ (1). By Theorem3.1, it suffices to prove that there exists a firm and surjective Morita $*$ -context $(S,T,{_SX_T},{_T\overline{X}_S}, \theta, \phi )$ . We will use the mappings $\varphi _{S\otimes S}\;:\; X\otimes \overline{X} \otimes S\otimes S\to S\otimes S$ and $\zeta _{T\otimes T}\;:\; T\otimes T\otimes \overline{X}\otimes X \to T\otimes T$ for this. Since $_T\overline{X}_S$ is a firm bimodule by Lemma 6.2, the mappings

\begin{align*} \mu _{\overline{X}}\;:\; \overline{X}\otimes _S S\to \overline{X},\;\;\; & \overline{x}\otimes s \mapsto \overline{x}s = \overline{s^*x}, \\[5pt] \nu _{\overline{X}}\;:\; T\otimes _T \overline{X} \to \overline{X}, \;\;\; & t\otimes \overline{x} \mapsto t\overline{x} = \overline{xt^*} \end{align*}

are bijective. Idempotency of $S$ and $T$ means that the mappings

\begin{align*} \mu _S\;:\; S\otimes _S S\to S,\;\;\; & s\otimes u \mapsto su, \\[5pt] \mu _T\;:\; T\otimes _T T \to T, \;\;\; & t\otimes v \mapsto tv \end{align*}

are surjective. We define $\theta$ as the composite

and $\phi$ as the composite

From here we see immediately that $\theta$ and $\phi$ are surjective homomorphisms of left modules. Take $x,y\in X$ . Using twice the fact that $_SX_T$ is unitary, we can write $y$ as

(6.3) \begin{equation} y= \sum _{k=1}^n u_ks_ky_k = \sum _{l=1}^m z_lt_lv_l \end{equation}

for some $m,n\in \mathbb{N}$ , $u_k,s_k\in S$ , $t_l,v_l\in T$ and $y_k,z_l\in X$ . Then,

\begin{align*} x\otimes \overline{y} & = x\otimes \overline{\sum _{k=1}^n u_ks_ky_k} = \sum _{k=1}^n x\otimes \overline{u_ks_ky_k} = \sum _{k=1}^n x\otimes \overline{y_k}s_k^* u_k^* \\[5pt] & = \sum _{k=1}^n \left ({\mathrm{id}}_X\otimes \mu _{\overline{X}} \right ) (x\otimes \overline{y_k} s_k^* \otimes u_k^*) \tag*{$(\text{def. of $\mu _{\overline{X}}$})$}\\[5pt] & = \sum _{k=1}^n \left ( ({\mathrm{id}}_X\otimes \mu _{\overline{X}})\!\circ \! ({\mathrm{id}}_X\otimes \mu _{\overline{X}}\otimes{\mathrm{id}}_S)\right ) (x\otimes \overline{y_k}\otimes s_k^* \otimes u_k^*). \tag*{$(\text{def. of $\mu _{\overline{X}}$})$} \end{align*}

Hence,

(6.4) \begin{equation} \theta (x\otimes \overline{y}) = \sum _{k=1}^n (\mu _S\circ \varphi _{S\otimes S}) (x\otimes \overline{y_k}\otimes s_k^*\otimes u_k^*) \end{equation}

and, analogously,

(6.5) \begin{equation} \phi (\overline{y}\otimes x) = \sum _{l=1}^m (\mu _T \circ \zeta _{T\otimes T}) (v_l^* \otimes t_l^* \otimes \overline{z_l} \otimes x). \end{equation}

Note that $\rho _s\;:\;{_SS}\otimes _S S\to{_SS}\otimes _S S, \;\; u\otimes v\mapsto u\otimes vs$ is a homomorphism of left $S$ -modules for every $s\in S$ . Since $\varphi$ is a natural transformation, the square

commutes. Hence,

\begin{align*} \varphi _{S\otimes S}((x\otimes \overline{y}\otimes u\otimes v)s) & = \varphi _{S\otimes S}(x\otimes \overline{y}\otimes u\otimes vs) \tag*{$(\text{$S$-action of $X \otimes \overline{X} \otimes S \otimes S$})$}\\[5pt] & = \left ( \varphi _{S\otimes S} \circ ({\mathrm{id}}_{X\otimes \overline{X}}\otimes \rho _s)\right ) (x\otimes \overline{y}\otimes u\otimes v) \tag*{$(\text{def. of $\rho _s$})$}\\[5pt] & = (\rho _s\circ \varphi _{S\otimes S})(x\otimes \overline{y} \otimes u\otimes v) \tag*{$(\text{naturality of $\varphi $})$}\\[5pt] & = \varphi _{S\otimes S}(x\otimes \overline{y}\otimes u\otimes v)s, \tag*{$(\text{def. of $\rho _s$})$} \end{align*}

which means that $\varphi _{S \otimes S}$ is a homomorphism of $(S,S)$ -bimodules. Analogously, $\zeta _{T\otimes T}$ is a homomorphism of $(T,T)$ -bimodules. Using this and the equality $s^*y= \sum _{k=1}^n (s^*u_k)s_ky_k$ , we see that

\begin{align*} \theta ((x\otimes \overline{y})s) & = \theta (x\otimes \overline{s^*y}) \tag*{$(\text{by (2.2)})$} \\[5pt] & = \sum _{k=1}^n (\mu _S\circ \varphi _{S\otimes S})(x\otimes \overline{y_k} \otimes s_k^*\otimes (s^*u_k)^*) \tag*{$(\text{by (6.4)})$}\\[5pt] & = \sum _{k=1}^n (\mu _S\circ \varphi _{S\otimes S}) (x\otimes \overline{y_k} \otimes s_k^*\otimes u_k^*s) \tag*{$(\text{by (2.1)})$}\\[5pt] & = \sum _{k=1}^n (\mu _S\circ \varphi _{S\otimes S}) ((x\otimes \overline{y_k} \otimes s_k^*\otimes u_k^*)s) \tag*{$(\text{$S$-action of $X \otimes \overline{X} \otimes S \otimes S$})$}\\[5pt] & = \sum _{k=1}^n (\mu _S\circ \varphi _{S\otimes S}) (x\otimes \overline{y_k} \otimes s_k^* \otimes u_k^*)s \tag*{$(\text{$\mu _S, \varphi _{S \otimes S}$ are homomorphisms})$}\\[5pt] & = \theta (x\otimes \overline{y})s, \tag*{$(\text{by (6.4)})$} \end{align*}

so $\theta$ is a homomorphism of $(S,S)$ -bimodules, and, analogously, $\phi$ is a homomorphism of $(T,T)$ -bimodules.

Next we verify the compatibility conditions. Let $x,y$ be as above and take also $x'\in X$ . Consider the module homomorphisms

(6.6) \begin{align} r_{x'}\;:\;{_SS}\to{_SX},\;\;\; & s \mapsto sx', \end{align}

(6.7) \begin{align} l_x\;:\; T_T \to X_T, \;\;\; & t \mapsto xt \end{align}

and the composites $\rho _{x'}\;:\!=\; r_{x'}\circ \mu _S\;:\; S\otimes _S S\to X$ and $\lambda _x\;:\!=\; l_x\circ \mu _T\;:\; T\otimes _T T \to X_T$ . Since $\varphi$ and $\zeta$ are natural transformations, the squares

commute. Therefore,

\begin{align*} \theta (x\otimes \overline{y})x' & = \sum _{k=1}^n (\mu _S\circ \varphi _{S\otimes S}) (x\otimes \overline{y_k} \otimes s_k^*\otimes u_k^*) x' \tag*{$(\text{by (6.4)})$} \\[5pt] & = \sum _{k=1}^n (r_{x'}\circ \mu _S\circ \varphi _{S\otimes S}) (x\otimes \overline{y_k}\otimes s_k^*\otimes u_k^*) \tag*{$(\text{by (6.6)})$} \\[5pt] & = \sum _{k=1}^n (\rho _{x'}\circ \varphi _{S\otimes S}) (x\otimes \overline{y_k}\otimes s_k^*\otimes u_k^*) \tag*{$(\text{def. of $\rho _{x'}$})$}\\[5pt] & = \sum _{k=1}^n (\varphi _X\circ ({\mathrm{id}}_{X\otimes \overline{X}}\otimes \rho _{x'})) (x\otimes \overline{y_k}\otimes s_k^*\otimes u_k^*) \tag*{$(\text{naturality of $\varphi $})$}\\[5pt] & = \sum _{k=1}^n\varphi _X(x\otimes \overline{y_k} \otimes s_k^*u_k^*x') \tag*{$(\text{def. of $\rho _{x'}$})$}\\[5pt] & = \sum _{k=1}^n\varphi _X(x\otimes \overline{y_k}s_k^*u_k^*\otimes x') \tag*{$(\text{property of $\otimes $})$}\\[5pt] & = \sum _{k=1}^n\varphi _X(x\otimes \overline{u_ks_ky_k}\otimes x') \tag*{$(\text{by (2.2)})$}\\[5pt] & = \varphi _X(x\otimes \overline{y}\otimes x') \tag*{$(\text{by (6.3)})$}\\[5pt] & = \zeta _X(x\otimes \overline{y}\otimes x') \tag*{$(\text{$\varphi _X= \zeta _X$})$}\\[5pt] & = \sum _{l=1}^m \zeta _X(x\otimes \overline{z_lt_lv_l}\otimes x') \tag*{$(\text{by (6.3)})$} \\[5pt] & = \sum _{l=1}^m \zeta _X(x\otimes v_l^*t_l^*\overline{z_l}\otimes x') \tag*{$(\text{by (2.2)})$} \\[5pt] & = \sum _{l=1}^m \zeta _X(xv_l^*t_l^*\otimes \overline{z_l}\otimes x') \tag*{$(\text{property of $\otimes $})$} \end{align*}

\begin{align*} & = \sum _{l=1}^m (\zeta _X\circ (\lambda _x\otimes{\mathrm{id}}_{\overline{X}\otimes X}))(v_l^*\otimes t_l^*\otimes \overline{z_l}\otimes x') \tag*{$(\text{def. of $\lambda _x$})$}\\[5pt] & = \sum _{l=1}^m (\lambda _x\circ \zeta _{T\otimes T})(v_l^*\otimes t_l^*\otimes \overline{z_l}\otimes x') \tag*{$(\text{naturality of $\zeta $})$} \\[5pt] & = \sum _{l=1}^m (l_x\circ \mu _T\circ \zeta _{T\otimes T})(v_l^*\otimes t_l^*\otimes \overline{z_l}\otimes x') \tag*{$(\text{def. of $\lambda _x$})$}\\[5pt] & = \sum _{l=1}^m x(\mu _T\circ \zeta _{T\otimes T})(v_l^*\otimes t_l^*\otimes \overline{z_l} \otimes x') \tag*{$(\text{by (6.7)})$}\\[5pt] & = x\phi (\overline{y}\otimes x'). \tag*{$(\text{by (6.5)})$} \end{align*}

Similarly, we can establish the compatibility condition $\phi (\overline{y}\otimes x)\overline{y'} = \overline{y}\theta (x\otimes \overline{y'})$ .

The equalities $\theta (x\otimes \overline{y})^* = \theta (y\otimes \overline{x})$ and $\phi (\overline{y}\otimes x)^* = \phi (\overline{x}\otimes y)$ follow from the commutativity of the rectangles (R1) and (R2).

7. Comparison with earlier results

In this section, we compare our results with those of Ara [Reference Ara3].

A module $M_R$ is called nondegenerate ([Reference Ara3]) or torsion-free ([Reference Marín15, p. 221]) if, for every $m\in M$ , $mR=0$ implies $m=0$ . A ring $R$ is called nondegenerate if it is nondegenerate as a right and left module over itself.

Recall that a module $M_R$ is called s-unital ([Reference Tominaga25]) if, for every $m\in M$ , there exists $r\in R$ such that $m=mr$ . A ring $R$ is called s-unital if it is s-unital as a right and left module over itself. It is easy to see that every s-unital module is nondegenerate.

Definition 7.2. We say that a Rieffel context $(S,T, X,\left \langle \,,\,\right \rangle, [\,,\,])$ is nondegenerate if, for every $x\in X$ ,

\begin{equation*} (\left \langle \,X,x\,\right \rangle = \{0_S\} \implies x=0_X) \; \mbox { and }\; ([x,X]= \{0_T\} \implies x=0_X), \end{equation*}

where $\left \langle \,X,x\,\right \rangle \;:\!=\; \{\left \langle \,\xi, x\,\right \rangle \mid \xi \in X\}$ and $[x,X] \;:\!=\; \{[x,\xi ]\mid \xi \in X\}$ .

Proof. Let $(S,T, X,\left \langle \,,\,\right \rangle, [\,,\,])$ be a unitary and full Rieffel context and assume that $\left \langle \,X,x\,\right \rangle =\{0_S\}$ . By Lemma 7.1, $X_T$ is s-unital. Hence, $x=xt$ for some $t\in T$ . Since $[\,,\,]$ is full, there exist $y_k,y'_k\in X$ such that $t =\sum _{k=1}^n [y_k,y'_k]$ . Now

\begin{align*} x & = xt = x\left ( \sum _{k=1}^n [y_k,y'_k]\right ) = \sum _{k=1}^n x[y_k,y'_k] = \sum _{k=1}^n \left \langle \,x,y_k\,\right \rangle y'_k = \sum _{k=1}^n \left \langle \,x,y_k\,\right \rangle ^{**} y'_k \\[5pt] & = \sum _{k=1}^n \left \langle \,y_k,x\,\right \rangle ^* y'_k = \sum _{k=1}^n 0^*_Sy'_k = \sum _{k=1}^n 0_Sy'_k = 0_X. \end{align*}

A similar proof shows that $[x,X]=\{0_T\}$ implies $x=0_X$ .

If $(S,T, X,\left \langle \,,\,\right \rangle, [\,,\,])$ is a nondegenerate Rieffel context, then according to the terminology of [Reference Ara3, page 243], $_SX_T$ is called an inner product $(S,T)$ -bimodule.

In [Reference Ara3, page 235], two nondegenerate idempotent rings with involution $R$ and $S$ are called Morita $*$ -equivalent if there exists a $*$ -equivalence $(F,G)$ (which is a pair of certain functors between categories of unitary torsion-free modules) from $R$ to $S$ .

Proof. Necessity. As explained in the second paragraph of [Reference Ara3, page 243], by Theorem3.1 of that article, there exists a full nondegenerate unitary Rieffel context.

Sufficiency. This is shown in [Reference Ara3, Theorem 4.1(i)].

As a consequence, we obtain the following result.

Proof. Necessity. Let $S$ and $T$ be Morita $*$ -equivalent. By s-unitality, rings $S$ and $T$ are both idempotent and nondegenerate. Hence Theorem7.4 applies.

Sufficiency. Assume that $S$ and $T$ are connected by a unitary and full Rieffel context. By Proposition 7.3, this context is nondegenerate. By Theorem7.4, $S$ and $T$ are Morita $*$ -equivalent.

So for s-unital rings with involution, the theory developed in this paper is compatible with that of [Reference Ara3].