Proof. Let \((f,g): (M,N) \rightarrow (M_1,N_1)\), with \(M,M_1 \in {\mathcal{M}}_{Q}\), \(N,N_1 \in {}_{Q}{\mathcal{M}}\). We shall define \(f\otimes g: M\otimes N \rightarrow M_1 \otimes N_1\) e \(h\otimes k: M_1\otimes N_1 \rightarrow M_2 \otimes N_2\), and prove that \(- \otimes-\) preserves the composition.

Let \(\varphi: (x,y) \in M\times N \mapsto f(x)\otimes g(y) \in M_1\otimes N_1\); we have that \[\begin{array}{l} \varphi(xq,y) = f(xq)\otimes g(y) = f(x)q\otimes g(y) = \\ = f(x)\otimes qg(y) = f(x)\otimes g(qy) = \varphi(x,qy). \end{array}\] We also have \[\begin{array}{l} \varphi\left(\bigvee X,y\right) = f\left(\bigvee X\right)\otimes g(y) = \left(\bigvee_{x \in X} f(x)\right)\otimes g(y) = \\ = \bigvee_{x \in X} (f(x)\otimes g(y)) = \bigvee_{x \in X} \varphi(x,y), \end{array}\] and, similarly, \(\varphi \left( x,\bigvee Y\right) = \bigvee_{y \in Y} \varphi(x,y)\). So, \(\varphi\) is a \(Q\)-bimorphism and, therefore, it canonically induces a sup-lattice morphism \(f\otimes g: M \otimes N \to M_1 \otimes N_1\). Analogously, we can define sup-lattice morphisms \(h \otimes k: M_1\otimes N_1 \rightarrow M_2 \otimes N_2\) and \((h\circ f)\otimes(k\circ g): M \otimes N \to M_2 \otimes N_2\). Furthermore, we have \((h\circ f)\otimes(k\circ g) = (h\otimes k)\circ(f\otimes g)\) because \(h(f(x))\otimes k(g(y)) = (h\otimes k)(f(x)\otimes g(y)) = ((h\otimes k)\circ(f\otimes g))(x\otimes y)\).

It follows that \(- \otimes-\) is a bifunctor. The second part of the proposition is obvious. ◻

Proof. For all \(x \in M\), let us define \[\alpha_x: (y,z) \in N \times O \mapsto (x\otimes y)\otimes z \in (M\otimes N)\otimes O,\] and show that it is a bimorphism. Indeed, we have: \[\begin{array}{l} \alpha_x(yr,z) = (x\otimes yr)\otimes z = (x\otimes y)r\otimes z = (x\otimes y)\otimes rz = \alpha_x(y,rz), \\ \alpha_x\left(\bigvee Y,z\right) = \left( x\otimes\bigvee Y\right)\otimes z = \left(x\otimes\bigvee\limits_{y \in Y} y\right)\otimes z = \left(\bigvee\limits_{y \in Y} (x\otimes y)\right)\otimes z = \\ = \bigvee\limits_{y \in Y} ((x\otimes y)\otimes z) = \bigvee\limits_{y \in Y} \alpha_x(y,z), \;\text{ and, similarly,} \\ \alpha_x\left(y,\bigvee Z\right) = \bigvee\limits_{z \in Z} \alpha_x(y,z). \end{array}\] So, there exists a unique sup-lattice morphism \(\overline{\alpha}_x: N \otimes O \rightarrow (M\otimes N)\otimes O\) extending \(\alpha_x\); observe that, for all \(x \in M\), \(y \in N\), and \(z \in O\), \(\overline{\alpha}_x(y\otimes z) = (x\otimes y)\otimes z\).

Now, let \[\alpha: \left(x,\bigvee_{i \in I}y_i\otimes z_i\right) \in M \times (N \otimes O) \mapsto \overline{\alpha}_x\left(\bigvee_{i \in I}y_i \otimes z_i\right) \in (M\otimes N)\otimes O,\] which is a bimorphism too. Indeed, we have \[\begin{array}{l} \alpha\left(\bigvee X,\bigvee\limits_{i \in I}y_i \otimes z_i\right) = \overline{\alpha}_{\bigvee X}\left(\bigvee\limits_{i \in I}y_i \otimes z_i\right) = \bigvee\limits_{i \in I}\overline{\alpha}_{\bigvee X}(y_i \otimes z_i) =\\ =\bigvee\limits_{i \in I}\alpha_{\bigvee X}(y_i , z_i) = \bigvee\limits_{i \in I}\left( \bigvee X\otimes y_i\right)\otimes z_i = \bigvee\limits_{i \in I}\bigvee\limits_{x \in X}(x\otimes y_i)\otimes z_i = \\ = \bigvee\limits_{x \in X}\bigvee\limits_{i \in I}(x\otimes y_i)\otimes z_i = \bigvee\limits_{x \in X}\bigvee\limits_{i \in I}\overline{\alpha}_x(y_i\otimes z_i)= \bigvee\limits_{x \in X}\overline{\alpha}_x\left(\bigvee\limits_{i \in I}y_i\otimes z_i\right) = \\ = \bigvee\limits_{x\in X}\alpha\left(x,\bigvee\limits_{i \in I}y_i \otimes z_i\right). \end{array}\]

So, by the universal property of the tensor product \(M \otimes_Q (N \otimes_R O)\), there exists a unique sup-lattice morphism \(\overline{\alpha}: M \otimes_Q (N \otimes_R O) \longrightarrow (M \otimes_Q N)\otimes_R O\)

such that, for all \(x \in M\), \(y \in N\), and \(z \in O\), we have \(\overline{\alpha}\bigl(x \otimes (y \otimes z)\bigr) = (x \otimes y) \otimes z\).

Symmetrically, for each \(z \in O\) let us define \(\beta_z : M \times N \longrightarrow M \otimes_Q (N \otimes_R O)\),

\(\beta_z(x,y) = x \otimes (y \otimes z)\).

As before, \(\beta_z\) is a bimorphism, so there exists a unique sup-lattice morphism

\(\overline{\beta}_z : M \otimes_Q N \longrightarrow M \otimes_Q (N \otimes_R O)\)

such that \(\overline{\beta}_z(x \otimes y) = x \otimes (y \otimes z)\) for all \(x \in M\), \(y \in N\). Now, define

\(\beta : (M \otimes_Q N) \times O \longrightarrow M \otimes_Q (N \otimes_R O), \quad \beta\!\left(\bigvee_j x_j \otimes y_j, z\right) = \bigvee_j \overline{\beta}_z(x_j \otimes y_j)\).

Again, \(\beta\) is a bimorphism, so by the universal property of the tensor product \((M \otimes_Q N)\otimes_R O\), there exists a unique sup-lattice morphism

\(\overline{\beta} : (M \otimes_Q N)\otimes_R O \longrightarrow M \otimes_Q (N \otimes_R O)\) such that, for all \(x \in M\), \(y \in N\),

\(z \in O\), \(\overline{\beta}\bigl((x \otimes y) \otimes z\bigr) = x \otimes (y \otimes z)\).

So, \(\overline{\beta} \circ \overline{\alpha}\bigl(x \otimes (y \otimes z)\bigr) = \overline{\beta}\bigl((x \otimes y)\otimes z\bigr) = x \otimes (y \otimes z)\), and

\(\overline{\alpha} \circ \overline{\beta}\bigl((x \otimes y)\otimes z\bigr) = \overline{\alpha}\bigl(x \otimes (y \otimes z)\bigr) = (x \otimes y) \otimes z\).

Since both morphisms preserve arbitrary joins and the tensors generate the respective tensor products, we conclude that

\(\overline{\beta} \circ \overline{\alpha} = \mathrm{id}_{M \otimes_Q (N \otimes_R O)} \quad\text{and}\quad \overline{\alpha} \circ \overline{\beta} = \mathrm{id}_{(M \otimes_Q N)\otimes_R O}\).

Therefore, \(\overline{\alpha}\) is an isomorphism with inverse \(\overline{\beta}\). ◻

Proof. We note that \(\left\langle\operatorname{End}_Q(M), \bigvee \right\rangle\) is a sup-lattice, and \(\left\langle\operatorname{End}_Q(M), \cdot, \operatorname{id}_M \right\rangle\) is a monoid, where the product \(h\cdot g\) is the composition \(g\circ h\).

Let \(h \in \operatorname{End}_Q(M)\) and \(\{g_i\}_{i \in I} \subseteq \operatorname{End}_Q(M)\). Note that \[\begin{array}{l} \left(\left(\bigvee\limits_{i \in I} g_i\right)\cdot h\right)(x) = \\ = \left(h \circ \left(\bigvee\limits_{i \in I} g_i\right)\right) (x) = h\left(\left(\bigvee\limits_{i \in I} g_i\right)(x)\right) = \\ = h\left(\bigvee\limits_{i \in I} g_i(x)\right) = \bigvee\limits_{i \in I} h(g_i(x)) = \\ = \bigvee\limits_{i \in I} (h\circ g_i)(x) = \bigvee\limits_{i \in I} (g_i\cdot h)(x). \end{array}\] Similarly, we have \(\left( h\cdot\left(\bigvee_{i \in I} g_i\right)\right)(x) = \bigvee_{i \in I} (h\cdot g_i )(x)\), whence \(E=\operatorname{End}_Q(M)\) is a quantale.

Furthermore, we have that \(M\) is a right \(E\)-module. Indeed, taking \(xh = h(x)\), \(x \in M\), \(h \in E\), \[x(f\cdot g) = (f\cdot g)(x) = (g\circ f)(x) = g(f(x)) = f(x)g= (xf)g.\] We have also, for \(x \in M\), \(f \in E\), \(F \subseteq E\) and \(N \subseteq M\), \[x\left(\bigvee F\right) = \left(\bigvee F\right)(x) = \left(\bigvee_{f \in F} f \right)(x) = \bigvee_{f \in F} f(x) = \bigvee_{f \in F} xf\] and \[\left(\bigvee N\right)f = f\left(\bigvee N\right) = f\left(\bigvee_{x \in N} f \right)(x) = \bigvee_{x \in N} f(x) = \bigvee_{x \in N} xf.\] It is clear that \(x\operatorname{id}_E = \operatorname{id}_E(x) = x\). ◻

Proof. Indeed, if \(\alpha(1) := r \in Q\), then \(\alpha(x) = \alpha(x1) = x\alpha(1) = xr\), for any \(x \in Q\). ◻

Proof. Follows readily from [15]. ◻

Proof. Let us define the operations

\[\begin{array}{ll} (h,f) \in E \times M^* \mapsto hf = f \circ h \in M^*, & \text{ and}\\ (f,q) \in M^* \times Q \mapsto fq \in M^*, & \text{ where }fq :x \in M \mapsto f(x)q \in Q. \end{array}\]

Next, we set \[\alpha': (m,f) \in M \times M^* \mapsto f(m) \in Q, \text{and show that it is a bimorphism.}\]

Indeed, since \(M\) is a right \(E\)-module, taking \(mh = h(m)\), \(m \in M\), \(h \in E\), we have \[\begin{array}{l} \alpha'(mh,f) = \alpha'(h(m),f) = f(h(m)) = f \circ h(m) = hf(m) = \alpha'(m,hf). \end{array}\] Furthermore, \[\begin{array}{l} \alpha'\left(\bigvee X,f\right) = f\left( \bigvee X\right) = \bigvee\limits_{x \in X} f(x) =\bigvee\limits_{x \in X} (\alpha'(x,f)) \text{ and }\\ \alpha'\left(x,\bigvee F\right) = \bigvee F (x) = \bigvee\limits_{f \in F} f(x) = \bigvee\limits_{f \in F} (\alpha'(x,f)). \end{array}\]

So, there exists a unique sup-lattice morphism \(\alpha:M \otimes M^* \to Q\) extending \(\alpha'\), where, for all \(m \in M\) and \(f \in M^*\), \((m\otimes f) \in M \otimes M^* \mapsto f(m) \in Q\).

Note also that \(\alpha\) is a \(Q\)-\(Q\)-homomorphism, because

\[\begin{array}{l} \alpha\left(q\left(\bigvee\limits_{i \in I}x_i\otimes f_i\right)\right) = \alpha\left(\bigvee\limits_{i \in I}qx_i\otimes f_i\right) = \bigvee\limits_{i \in I}\alpha(qx_i\otimes f_i) = \\ = \bigvee\limits_{i \in I}f_i(qx_i) = \bigvee\limits_{i \in I}qf_i(x_i) = q\bigvee\limits_{i \in I}f_i(x_i) = q \bigvee\limits_{i \in I}\alpha(x_i\otimes f_i) = \\ = q\alpha \left(\bigvee\limits_{i \in I}x_i\otimes f_i\right), \\ \text{and} \\ \alpha\left(\left(\bigvee\limits_{i \in I}x_i\otimes f_i\right)q\right) = \alpha\left(\bigvee\limits_{i \in I}x_i\otimes f_iq\right) = \bigvee\limits_{i \in I}\alpha(x_i\otimes f_iq) = \\ = \bigvee\limits_{i \in I}f_iq(x_i) = \bigvee\limits_{i \in I}f_i(x_i)q = \left(\bigvee\limits_{i \in I}f_i(x_i)\right)q = \left(\bigvee\limits_{i \in I}\alpha(x_i\otimes f_i)\right)q = \\ = \alpha \left(\bigvee\limits_{i \in I}(x_i\otimes f_i)\right)q. \end{array}\]

Now consider the map

\[\beta': (f,m) \in M^* \times M \mapsto fm \in E, \text{and show that it is a bimorphism.}\]

Indeed, since \(M^*\) is a right \(Q\)-module, taking \(fq = f(x)q\), \(x \in M\), \(q \in Q\), we have \[\begin{array}{l} \beta'(fq,m) = (fq)m = (f(x)q)m = f(x)(qm) = f\cdot(qm) = \beta'(f,qm). \end{array}\] Furthermore, \[\begin{array}{l} \beta'\left(\bigvee F,m\right) = \left(\bigvee F\right)m =\left(\bigvee\limits_{f \in F} f\right) m = \left(\bigvee\limits_{f \in F} fm\right) =\bigvee\limits_{f \in F} \beta'(f,m), \\ \text{and }\\ \\ \beta'\left(f,\bigvee Y\right) = f \bigvee Y = \bigvee\limits_{m \in Y} fm =\bigvee\limits_{m \in Y} \beta'(f,m). \end{array}\]

So, there exists a unique sup-lattice morphism \(\beta:M^* \otimes M \to E\) extending \(\beta'\), where, for all \(m \in M\) and \(f \in M^*\), \((f\otimes m) \in M^* \otimes M \mapsto fm \in E\).

We have also that \(\beta\) is a \(E\)-\(E\)-homomorphism, because

\[\begin{array}{l} \beta\left(h\left(\bigvee\limits_{i \in I}f_i\otimes m_i\right)\right)(x) = \beta\left(\bigvee\limits_{i \in I}hf_i\otimes m_i\right)(x) = \bigvee\limits_{i \in I}\beta(hf_i\otimes m_i)(x) =\\ = \bigvee\limits_{i \in I}(hf_i)(x)m_i =\bigvee\limits_{i \in I}(f_i\circ h)(x)m_i = \bigvee\limits_{i \in I}f_i (h(x))m_i = \\ = \bigvee\limits_{i \in I}\beta(f_i\otimes m_i)(h(x)) = \bigvee\limits_{i \in I}(\beta(f_i\otimes m_i)\circ h)(x) = \bigvee\limits_{i \in I}h\beta(f_i\otimes m_i)(x) = \\ = h \left(\bigvee\limits_{i \in I}\beta(f_i\otimes m_i)\right)(x) = h\beta \left(\bigvee\limits_{i \in I}f_i\otimes m_i\right)(x)\\ \text{and} \\ \beta\left(\left(\bigvee\limits_{i \in I}f_i\otimes m_i\right)h\right)(x) = \beta\left(\bigvee\limits_{i \in I}f_i\otimes m_ih\right)(x) = \bigvee\limits_{i \in I}\beta(f_i\otimes m_ih)(x) =\\ = \bigvee\limits_{i \in I}\beta(f_i\otimes h(m_i))(x) = \bigvee\limits_{i \in I}f_i(x)h(m_i) = \bigvee\limits_{i \in I}f_i(x)h(m_i) = \\ = \bigvee\limits_{i \in I}h(f_i(x)(m_i)) = \left(\bigvee\limits_{i \in I}h(\beta(f_i\otimes m_i))(x)\right) = \left(\bigvee\limits_{i \in I}h \circ (\beta f_i\otimes m_i))(x)\right) = \\ = \left(\bigvee\limits_{i \in I}(\beta(f_i\otimes m_i)h)(x)\right) = \left(\beta\left(\bigvee\limits_{i \in I}(f_i\otimes m_i)\right)h\right)(x). \end{array}\] ◻

Proof. \((a) \Rightarrow (b)\)

Let \(M\) be a projective module, and let us identify \(M\) with its isomorphic submodule of some free module \(\mm[Q^I][Q]{}\), according to Proposition 7. Then there exists an endomorphism \(e: \mm[Q^I][Q]{} \rightarrow \mm[Q^I][Q]{}\) such that the restriction of \(e\) to the submodule \(M\) is the identity, i. e. \(me=m, \forall m\). Let \(\overline{e_i^*}\) be the restriction to \(M\) of the composition \(\mm[Q^I][Q]{} \stackrel{e}{\longrightarrow} \mm[Q^I][Q]{} \stackrel{e_i^*}{\longrightarrow} \mm[Q][Q]{}\), for each \(i \in I\). It is clear that \(x=\bigvee(xe_i^*)e_i\) for any element \(x=\bigvee x_ie_i \in \mm[Q^I][Q]{}\), hence, applying the “MEM-associativity”, one has \(m=me=\bigvee(mee_i^*)e_ie=\bigvee(m\overline{e_i^*})(e_ie)=\bigvee m\overline{e_i^*}(e_ie)=m \bigvee\overline{e_i^*}(e_ie)\) \(\forall m \in M\), with \(e_ie \in M\). Therefore \(\bigvee\overline{e_i^*}(e_ie)=\beta(\bigvee\overline{e_i^*}\otimes(e_ie))=1_M \in E\), and, since \(\beta\) in an \((E,E)\)-homomorphism, \(\beta\) is a surjection.

Suppose now that \[\beta\left(\bigvee\limits_{i,j \in I} e_i^*a_i\otimes b_je_j\right) = \beta\left(\bigvee\limits_{i,j \in I} e_i^*a_i'\otimes b_j'e_j\right),a_i, a_i',b_j, b_j' \in Q.\] Then, for any \(e_i\), we have \[\begin{array}{l} a_i\bigvee\limits_{j \in I} b_je_j = \bigvee\limits_{j \in I} a_i b_je_j = \bigvee\limits_{j \in I} e_ie_i^*a_i b_je_j = e_i\beta\left(\bigvee\limits_{i,j \in I} e_i^*\otimes a_i b_je_j\right) = \\ e_i\beta\left(\bigvee\limits_{i,j \in I} e_i^*a_i\otimes b_je_j\right) = e_i\beta\left(\bigvee\limits_{i,j \in I} e_i^*a_i'\otimes b_j'e_j\right) = \\ = e_i\beta\left(\bigvee\limits_{i,j \in I} e_i^*\otimes a_i' b_j'e_j\right) =\bigvee\limits_{j \in I} e_ie_i^*a_i' b_j'e_j = \bigvee\limits_{j \in I} a_i' b_j'e_j = a_i'\bigvee\limits_{j \in I} b_j'e_j. \end{array}\] Therefore, \[\begin{array}{l} \bigvee\limits_{i,j \in I} e_i^*a_i\otimes b_je_j = \bigvee\limits_{i,j \in I} e_i^*\otimes a_i b_je_j = \bigvee\limits_{i\in I} e_i^*\left(a_i \bigvee_{j \in I} b_je_j\right) = \\ = \bigvee\limits_{i\in I} e_i^*\left(a_i' \bigvee_{j \in I} b_j'e_j\right) = \bigvee\limits_{i,j \in I} e_i^*\otimes a_i' b_j'e_j = \bigvee\limits_{i,j \in I} e_i^*a_i'\otimes b_j'e_j, \end{array}\] i.e. \(\beta\) is injective. Therefore, \(\beta\) is an isomorphism.

\((b) \Rightarrow (c)\) is obvious.

\((c) \Rightarrow (a)\)

Suppose that there are \(n_i \in M^*\) and \(m_j \in M\) \(i \in I\) such that \(\bigvee n_im_i = \beta\left(\bigvee\limits_{i \in I} n_i\otimes m_i\right) = 1_M \in E\).

Then, for the surjection

and the injection \(\mu: \mm{}\longrightarrow \mm[Q^I][Q]{}\), defined by the assignments \(e_i \mapsto m_i\), and \(m \mapsto \bigvee mn_ie_i\), respectively, we get that \(\theta \mu = 1_M\). So \(M\) is a retract of a free module, therefore projective. ◻

Proof.

1. Let \(\operatorname{tr}(M) = Q\), i.e., \(Q=\left\langle f M: f \in M^* \right\rangle\), and let \(g: M^{M^*} \to Q\) be the morphism which makes, for each \(f \in M^*\), the following diagram commute: \[\begin{tikzcd} {M^{M^*}} \arrow[rrd, "g"] & & \\ & & {Q} \\ {M} \arrow[rru, "f"'] \arrow[uu, "\alpha_f", hook'] & & \end{tikzcd}.\] Note that \(g((x_i)_{i\in I}) = \bigvee_{i\in I}g_i(x_i)\), for all \((x_i)_{i \in I} \in M^I\).

   Since \(\operatorname{tr}(M) = Q\), \(\forall q \in Q\), \(\exists \{x_i\}_{i\in I} \in M^I\) and \(\exists \{q_i\}_{i\in I} \in Q^I\) such that \(q_ig_i(x_i)=q\). Furthermore, \(\forall q \in Q\), \(\exists S \subseteq \bigcup\limits_{i\in I} g_i M\) and \(\exists \{q_x\}_{x\in S} \subseteq Q\) such that \(\bigvee\limits_{x\in S}q_xx=q\). We also have that \(\forall x \in S\), \(\exists y_x \in M\) such that \(x=g_i(y_x)\).

   So \(q=\bigvee\limits_{x\in S}q_xx=\bigvee\limits_{x\in S}q_xg_i(y_x)=\bigvee\limits_{x\in S}q_xg\alpha_i(y_x)= g\left(\bigvee\limits_{x\in S}q_x\alpha_i(y_x)\right)\). Therefore \(g\) is a surjection.

   Let \(f:q\in Q \mapsto q.g_*(1) \in M^I\). Thus, \(\forall q\in Q, \;gf(q)=qgf(1)=qgg_*(1)=q.1=q\), i. e., \(f\) is a right inverse for \(g\) and, therefore, \(Q\) is a retract of \(M^I\).

2. On the other hand, suppose \(Q\) is a retract of \(M^I\), with

   being the retraction. We already know that \(\operatorname{tr}(M) \leq Q\). We have that, with \(\beta_i=\beta \circ \mu_i\), for each \(i \in I\), the following diagram commutes: \[\begin{tikzcd} {M^{I}} \arrow[rrd, "\beta", two heads] & & \\ & & {Q} \\ {M} \arrow[rru, "\beta_i"'] \arrow[uu, "\mu_i", hook'] & & \end{tikzcd}.\]

   So \(\forall (x_i)_{i \in I}, \, \beta ((x_i)_{i \in I}) = \bigvee \limits_{i \in I} \beta_i (x_i)\), and we have that \(\beta\) is a surjection. Since \(\bigcup\limits_{i \in I}\beta_i[M] \subseteq \bigcup\limits_{\alpha \in M^*}\alpha[M]\), then \(Q=\left\langle\bigcup\limits_{i \in I}\beta_i[M]\right\rangle \subseteq \operatorname{tr}(M)\). Therefore \(\operatorname{tr}(M) = Q\).

 ◻

Proof. Since \(M\) is a generator, by Proposition 9, \(Q\) is a retract of \(M^I\), for some set \(I\). Let

be the retraction, and consider the canonical embedding \(\alpha_i: M \hookrightarrow M^I\). Let \(N\) and \(P\) be left \(Q\)-modules, and let \(f\) and \(g\) be different morphisms from \(N\) to \(P\). Then there exists \(y \in N\) such that \(f(y)\neq g(y)\); let us define the \(Q\)-module morphism \(h: q \in Q \mapsto qy \in N\). Note that \(fh\neq gh: Q \rightarrow P\), since \(fh(1)=f(y)\neq g(y)=gh(1)\). Hence, \(fh\beta ((x_i)_{i \in I})=fh(1)=f(y)\neq g(y)=gh(1)=gh\beta ((x_i)_{i \in I})\) for every family \((x_i)_{i\in I} \in \beta^{-1}(1)\), and therefore \(fh\beta \neq gh\beta: M^I\rightarrow P\).

On the other hand, for all \((x_i)_{i \in I} \in M^I\), \((x_i)_{i \in I} = \bigvee_{i \in I} \alpha_i(x_i)\), so \[\bigvee_{i\in I}fh\beta \alpha_i(x_i)\neq \bigvee_{i\in I}gh\beta \alpha_i(x_i),\] which implies that there exists \(i \in I\) such that \(fh\beta \alpha_i(x_i)\neq gh\beta \alpha_i(x_i)\). Then, \(fh\beta \alpha_i \neq gh\beta \alpha_i: M \rightarrow P\). ◻

Proof. The first affirmation follows from the definition of \(\alpha\) and the Proposition 9. For the second statement, suppose that \(\alpha\) is a surjection, and let us show that it is an injection too. So, let \[\bigvee_{i \in I}f_i(x_i) = \alpha \left(\bigvee_{i \in I}x_i\otimes f_i\right) = \alpha \left(\bigvee_{j \in j}x_j'\otimes f_j'\right) = \bigvee_{j \in J}f_j'(x_j'),\] and \[\alpha \left(\bigvee_{k \in K}x_k''\otimes f_k''\right) = \bigvee_{k \in K}f_k(x_k) = \operatorname{id}_Q,\] for some sets \(I\), \(J\) and \(K\).

So, using the \(MNM\)-associativity and the \(NMN\)-associativity, we have \[\begin{array}{l} \bigvee\limits_{i \in I}x_i\otimes f_i = \bigvee\limits_{i \in I}\left(\bigvee\limits_{k \in K}(x''_kf''_k)x_i\right)\otimes f_i = \bigvee\limits_{k \in K}\left(\bigvee\limits_{i \in I}(x''_kf''_k)x_i\otimes f_i\right) = \\ = \bigvee\limits_{k \in K}\left(\bigvee\limits_{i \in I}x''_k(f''_kx_i)\otimes f_i\right) = \bigvee\limits_{k \in K}\left(\bigvee\limits_{i \in I}x''_k\otimes(f''_kx_i)f_i\right) = \\ = \bigvee\limits_{k \in K}\left(\bigvee\limits_{i \in I}x''_k\otimes f''_k\cdot(x_if_i)\right) = \bigvee\limits_{k \in K}\left(x''_k\otimes f''_k\bigvee\limits_{i \in I}x_if_i\right) = \\ = \bigvee\limits_{k \in K}\left(x''_k\otimes f''_k\bigvee\limits_{j \in J}x'_jf'_j\right) = \bigvee\limits_{k \in K}\left(\bigvee\limits_{j \in J}x''_k\otimes f''_k\cdot(x'_jf'_j)\right) = \\ =\bigvee\limits_{k \in K}\left(\bigvee\limits_{j \in J}x''_k\otimes(f''_kx'_j)f'_j\right) = \bigvee\limits_{k \in K}\left(\bigvee\limits_{j \in J}x''_k(f''_kx'_j)\otimes f'_j)\right) = \\ = \bigvee\limits_{k \in K}\left(\bigvee\limits_{j \in J}(x''_kf''_k)x'_j\otimes f'_j\right) = \bigvee\limits_{j \in J}\left(\bigvee\limits_{k \in K}\left((x''_kf''_k)x'_j\right)\otimes f'_j\right) = \\ = \bigvee\limits_{j \in J}x'_j\otimes f'_j. \end{array}\] ◻

Proof.

(i) By associating, to each \(f \in M^*\), the \(E\)-homomorphism \(\lambda(f) \in {\mathcal{M}}_{E}(M_E,E_E)\), defined by \(\lambda(f)(m) = fm \in E\), \(\forall m \in M\), we get the map \(\lambda: M^* \rightarrow {\mathcal{M}}_{E}(M_E,E_E)\). We shall prove that \(\lambda\) is an \(E\)-\(Q\)-module isomorphism.

First, we note that NME-associativity implies that
$\lambda(f) \in {\mathcal{M}}_{E}(M_E,E_E)$, while ENM-associativity
and NMQ-associativity guarantee that $\lambda$ is an
$E$-$Q$-bimodule homomorphism.

Since $M$ is a generator for ${}_{Q}{\mathcal{M}}$, like in
Proposition [11](#alphaiso){reference-type="ref"
reference="alphaiso"}, there exists a set $K$,
$\{x_k\}_{k \in K} \subseteq M$, and $\{f_k\}_{k \in K} \subseteq N$
such that $\bigvee_{k \in K}x_kf_k = \operatorname{id}_Q$. So,
suppose $\lambda(f)(m) = fm = f'm = \lambda(f')(m)$ for some
$f, f' \in Q$ and all $m \in M$. Then, using the NMN-associativity,
we have
$$\begin{array}{l} f = f\operatorname{id}_Q = f\bigvee\limits_{k \in K}x_kf_k = \bigvee\limits_{k \in K}f\cdot (x_kf_k) = \bigvee\limits_{k \in K}(fx_k)f_k = \\
= \bigvee\limits_{k \in K}(f'x_k)f_k= \bigvee\limits_{k \in K}f'\cdot(x_kf_k) = f'\bigvee\limits_{k \in K}x_kf_k = f',
\end{array}$$ which proves that $\lambda$ is injective. In order to
show that it is also surjective, let us consider a homomorphism
$f \in {\mathcal{M}}_{E}(M_E,E_E)$ and an $m \in M$. Using the
MNM-associativity and the ENM-associativity, we have
$$\begin{array}{l}
f(m) = f\left(\left(\bigvee\limits_{k \in K}x_kf_k\right)m\right) = f\left(\bigvee\limits_{k \in K}(x_kf_k)m\right) = f\left(\bigvee\limits_{k \in K}x_k(f_km)\right) =\\
= \bigvee\limits_{k \in K}f(x_k)\cdot(f_km) = \bigvee\limits_{k \in K}(f(x_k)f_k)m= \lambda\left(\bigvee\limits_{k \in K}(f(x_k)f_k)\right)(m).
\end{array}$$

(ii) It can be proved in a similar way to the previous one.

(iii) Consider the quantale homomorphisms \(\sigma: Q \rightarrow \operatorname{End}(M_E)\) and \(\tau: Q \rightarrow \operatorname{End}({}_{E}N)\), defined by \(\sigma(q)(m) = qm\) and \(\tau(q)(f) = fq\) for any \(q \in Q\), \(m \in M\) and \(f \in N\). In the same way as in (i), we show that \(\sigma\) is injective. Indeed, using the QMN-associativity, suppose that \(qm = \sigma(q)(m) = \sigma(q')(m) = q'm\), for some \(q, q' \in Q\), \(m \in M\). Then \[\begin{array}{l} q = q\operatorname{id}_Q = q\bigvee\limits_{k \in K}x_kf_k = \bigvee\limits_{k \in K}q\cdot(x_kf_k) = \bigvee\limits_{k \in K}(qx_k)f_k = \bigvee\limits_{k \in K}(q'x_k)f_k = \\ = \bigvee\limits_{k \in K}q'(x_kf_k) = q'\bigvee\limits_{k \in K}x_kf_k = q'.\end{array}\] Furthermore, just like in (i), we can show that \(\sigma\left(\bigvee_{k \in K}f(x_k)f_k\right) = f\) for any \(f \in \operatorname{End}(M_E)\), so \(\sigma\) is also surjective and, therefore, an isomorphism. The proof of the fact that also \(\tau\) is an isomorphism is completely analogous.

The proofs of (iv) and (v) are totally similar to ones of (i) and (iii), respectively, using only the fact that by Proposition 8 there exist a finite set \(K\), \(\{x_k\}_{k \in K} \subseteq M\), and \(\{f_k\}_{k \in K} \subseteq N\), such that \(\bigvee_{k \in K}f_kx_k = \operatorname{id}_R\). ◻

Proof. For \({}_{E}N\), by (ii) and (iii) of the previous proposition, we have \(M \cong {}_{E}{\mathcal{M}}({}_{E}N,{}_{E}E)\) and \(Q \cong \operatorname{End}({}_{E}N)\). Therefore, considering that \(\alpha\) and \(\beta\) are surjections and applying Propositions 8 and 11 to the module \({}_{E}N\), it follows that \({}_{E}N\) is a progenerator for \({}_{E}{\mathcal{M}}\). The proofs for the other cases are analogous. ◻

Proof. By [16], \({\mathcal{M}}_{Q}{}\) has products and coproducts, hence, by Proposition 14, it is enough show that the category it also has equalizers and coequalizers.

So, let \(f\) and \(g\) be morphisms from \(M\) to another module \(N\) and let \(A = \{x \in M: f(x)=g(x)\}\), which is easily seen to be a submodule of \(M\). The diagram

shows that the inclusion morphism \(e\) is equalizer of \((f,g)\). Indeed, by the definition of \(A\), we have \(f e = g e\). If \(e':A' \rightarrow M\) is a homomorphism such that \(f e' = g e'\), then the image of \(e'\) is contained in \(A\), hence the same morphism \(e'': y \in A' \mapsto e'(y) \in A\) is the unique homomorphism such that \(e' = e e''\).

For the coequalizer, in the diagram

let us consider the congruence \(\vartheta = \langle(f(x),g(x)): x \in M\rangle\), and let \(B=N/\vartheta\) and \(c\) be the canonical projection. Given any homomorphism \(c':N \rightarrow B'\) such that \(c' f = c' g\), we get \(\vartheta = \ker c \subseteq \ker c'\), hence there exists a unique homomorphism \(d:B \rightarrow B'\) such that \(c' = d c\). ◻

Proof. The implication \((a) \Rightarrow (b)\) follows from [32], and \((c) \Rightarrow (a)\) follows from Proposition 4.

For \((b) \Rightarrow (c)\), let \(P := F(Q) \in {\mathcal{M}}_{R}{}\), then the functor \(F\) induces a quantale homomorphism \(Q \cong {\mathcal{M}}_{Q}{}(Q,Q) \rightarrow {\mathcal{M}}_{R}{}(F(Q),F(Q))={\mathcal{M}}_{R}{}(P,P)\), which turns \(P\) a \(Q\)-\(R\)-bimodule, i.e. \(P \in {}_{Q}{\mathcal{M}}{}_R\). Let \(M \in {}_{Q}{\mathcal{M}}{}\) be an arbitrary module and a surjection \(\gamma: Q^I \twoheadrightarrow M\) for some free module \(Q^I \in {\mathcal{M}}_{Q}{}\). Since any surjective homomorphism in \({\mathcal{M}}_{Q}\) is a coequalizer of some pair of homomorphisms, let \(\alpha,\beta:A \rightarrow Q^I\) be such that \(\gamma=\operatorname{coeq}(\alpha,\beta)\), and therefore, for some surjection \(\theta:Q^J \twoheadrightarrow A\) there is a right exact diagram \[\begin{tikzcd} {Q^J} \arrow[r, "\alpha\theta", shift right=-1] \arrow[r, "\beta\theta"', shift right=1] & {Q^I} \stackrel{\gamma}{\twoheadrightarrow} M. \end{tikzcd}\] Applying the functors \(-\otimes_Q P\) and \(F\) to this diagram, we obtain the following commutative diagram in \({\mathcal{M}}_{R}{}\) \[\begin{tikzcd} {P^J}\arrow[transform canvas={xshift=0.3ex},-]{d} \arrow[transform canvas={xshift=-0.4ex},-]{d} \arrow[r, shift right=-1] \arrow[r, shift right=1] & {P^I} \arrow[transform canvas={xshift=0.3ex},-]{d} \arrow[transform canvas={xshift=-0.4ex},-]{d}\arrow[r, shift right=0] & M\otimes_Q P \\ {(F(Q))^J} \arrow[r, shift right=-1] \arrow[r, shift right=1] & (F(Q))^I \arrow[r, shift right=0] & {F(M)}\\ \end{tikzcd},\] in which both rows are right exact diagrams. Then there exists an isomorphism between \(M\otimes_Q P\) and \(F(M)\) which completes the diagram above, and it can easily be verified to be a natural one in \(M \in {}_{Q}{\mathcal{M}}\). Hence, \(F \cong -\otimes_Q P\). ◻

Proof. This follows from the observations that in \({\mathcal{M}}_{Q}{}\) any equalizer is a surjection and any surjection is coequalizer of some pair of homomorphisms, and from [33]. ◻

Proof. By [32], \(F \dashv G\) and \(G \dashv F\), i.e. the functor \(F\) is both left and right adjoint to \(G\). Then, by of [32] and its dual, \(F\) preserves limits and colimits, and hence the (i) follows from the previous lemma.

Items (ii) and (iii) are obvious because being a generator and projectivity are categorical properties, hence they are preserved by equivalences. ◻

Proof. Let \(\alpha: R \rightarrow \operatorname{End}(\mm[P]{})\) be an isomorphism between the \(R\)-\(R\)-bimodules \(R\) and \(\operatorname{End}(\mm[P]{})\).

Then there exist \(r_1 \in R\) and \(f_1 \in \operatorname{End}(\mm[P]{})\) such that \(\alpha(r_1) = \operatorname{id}_P\) and \(\alpha(1_R) = f_1\), and they are uniquely determined. So, for all \(r \in R\), we have \[\label{alphar} \alpha(r) = \alpha(1_R\cdot r) = \alpha(1_R) \cdot r = f_1\cdot r = f_1 \circ \hat{r} = \hat{r} f_1,\tag{3}\] where \(\hat{r}: x \mapsto x\cdot r\) is an element of \(\operatorname{End}(\mm[P]{})\). Let \(h: r \in R \mapsto \hat{r} \in \operatorname{End}(\mm[P]{})\). \(h\) is obviously a surjective quantale homomorphism. In order to prove injectivity, let \(r, s \in R\) be such that \(h(r) = h(s)\). Then, by (3 ), we have \(\alpha(r) = \hat{r} f_1 = h(r) f_1 = h(s) f_1 = \hat{s} f_1 = \alpha(s)\), but \(\alpha\) is bijective, and therefore such equalities imply \(r = s\). ◻

Proof. Applying \[{}_{Q}{\mathcal{M}}_Q(P \otimes_R G, Q) \cong {}_{R}{\mathcal{M}}_Q(G,P^*)\] from proposition 4, we have the \(R\)-\(Q\)-module homomorphism \(\gamma : {}_RG_Q \to {}_RP^*_Q\), defined by \(p(\gamma(g)) = \alpha'(p \otimes g).\)

Then, applying Proposition 11 to the commutative diagram

\[\begin{tikzcd}[column sep=large] P \otimes_R G \arrow[d, "\alpha'"'] \arrow[r, "1_P \otimes \gamma"] & P \otimes_R P^* \arrow[d, "\alpha"] \\ Q \arrow[r, phantom, "\scriptstyle ="] & Q \end{tikzcd}\]

we obtain that \(\mm[P]{}\) is a generator for \({}_{Q}{\mathcal{M}}\) and \(\alpha\) is an isomorphism. Furthermore, applying the functor \(G \otimes_Q -\) to this diagram and taking into consideration Proposition 5 and the fact that \(\beta' : G \otimes_Q P \to R\) is an isomorphism, we get that \(\gamma\) is an isomorphism too. Applying again the functor \(G \otimes_Q -\) to the commutative diagram

\[\begin{tikzcd} (P \otimes_R P^*) \otimes_Q P \arrow[r, "1_P \otimes \beta"] \arrow[d, "\alpha \otimes 1_P"'] & P \otimes_R R \arrow[d, "\cong"] \\ Q \otimes_Q P \arrow[r, "\scalebox{0.9}{\cong}"', shorten <=14pt, shorten >=14pt] & P \end{tikzcd}\] and using Proposition 5 and the fact that \(\beta' : G \otimes_Q P \to R\) is an isomorphism, we get that also \(\beta : P^* \otimes_Q P \to R\) is an isomorphism, and therefore, by Theorem 12, \(\mm[P]{}\) is a progenerator for \({}_{Q}{\mathcal{M}}\). Now, applying the functor \(P \otimes_R -\) to the \(R\)-\(R\)-module \(\operatorname{End}(G \otimes_Q P)\), and using Proposition 5 and the isomorphism \(\beta' : G \otimes_Q P \to R\), as well as applying the functor \(P \otimes_R (G \otimes_Q -)\) to the \(R\)-\(R\)-module \(\operatorname{End}(\mm[P]{})\) and again using Proposition 5 and the isomorphism \(\alpha' : P \otimes_R G \to Q\), one readily sees that the \(R\)-\(R\)-modules \((G \otimes_Q P)\) and \(\operatorname{End}(\mm[P]{})\) are isomorphic. From the latter, applying Lemma 4 and the \(R\)-\(R\)-isomorphism \(\beta': G \otimes_Q P \to R\), we have that \(R \cong \operatorname{End}(\mm[P]{})\) as quantales, and therefore, identifying the quantales \(R\) and \(\operatorname{End}(\mm[P]{})\), and using Proposition 13, we conclude that \(Q \cong \operatorname{End}(P_R)\) as quantales, too. ◻

Proof. By Theorem 12, there are the bimodule isomorphisms

\[\alpha_P: P \otimes_R P^* \rightarrow Q \text{ and }\beta_P: P^* \otimes_Q P \rightarrow R\] and \[\alpha_M: M \otimes_S M^* \rightarrow R\text{ and }\beta_M: M^* \otimes_R M \rightarrow S.\]

Using such isomorphisms and Proposition 5, we obtain the bimodule isomorphisms

\[(P \otimes_R M) \otimes_S (M^* \otimes_R P^*) \cong Q\text{ and }(M^* \otimes_R P^*) \otimes_S (P \otimes_R M) \cong S.\] Therefore, the result follows from the previous proposition. ◻

Proof. For the left-to-right implication, suppose that \(Q\) and \(R\) are Morita equivalent. By Theorem 16 we have that \(F\cong -\otimes_Q P\) for the \(Q\)-\(R\)-bimodule \(P=F(Q) \in {\mathcal{M}}_{R}{}\), and \(G\cong -\otimes_R T\) for the \(R\)-\(Q\)-bimodule \(T=G(R) \in {\mathcal{M}}_{Q}{}\). First notice that by Proposition 2 the modules \(P=F(Q) \in {\mathcal{M}}_{R}{}\) and \(T=G(R) \in {\mathcal{M}}_{Q}{}\) are progenerators for \({\mathcal{M}}_{R}{}\) and \({\mathcal{M}}_{Q}{}\) respectively. Then, taking into consideration the natural isomorphisms, we have: \[\begin{array}{l} {\mathcal{M}}_{R}{}(P,P) = {\mathcal{M}}_{R}{}(F(Q),P) \cong {\mathcal{M}}_{Q}{}(Q,G(P)) \cong \\ \cong {\mathcal{M}}_{Q}{}(Q,G(F(Q))) \cong {\mathcal{M}}_{Q}{}(Q,Q) \cong Q, \end{array}\] i.e., \(\operatorname{End}(Y_{R}) \cong Q\), that is equivalent to the affirmed.

In order to prove the right-to-left implication, let \(\mm[P]{}\) be a progenerator for \({}_{Q}{\mathcal{M}}{}\) such that the quantales \(R\) and \(\operatorname{End}(\mm[P]{})\) are isomorphic, and \(P \in {}_{Q}{\mathcal{M}}_R\), and let \(N = P^* := {}_{Q}{\mathcal{M}}{}(\mm[P]{}, \mm[Q]{})\) be the dual \(R\)-\(Q\)-module \(P^*\) of the \(Q\)-\(R\)-module P. Considering the composition of the functors \(-\otimes_Q P: {\mathcal{M}}_{Q}{} \rightarrow {\mathcal{M}}_{R}{}\) and \(-\otimes_R N = -\otimes_R P^*: {\mathcal{M}}_{R}{} \rightarrow {\mathcal{M}}_{Q}{}\) and using Proposition 4 and Theorem 12, we have the commutative diagram in \({\mathcal{M}}_{Q}{}\): \[\begin{tikzcd} X \otimes_Q P \otimes_R P^* \arrow[d, "\gamma \otimes 1_P \otimes 1_{P^*}"'] & {} \arrow[r, "1_X\otimes\alpha"] & {} & X \otimes_Q Q \arrow[d, "\gamma \otimes 1_Q"] & {} \arrow[r, "\cong"] & {} & X \arrow[d, "\gamma"'] \\ Y \otimes_Q P \otimes_R P^* & {} \arrow[r, "1_Y\otimes\alpha"] & {} & Y \otimes_Q Q & {} \arrow[r, "\cong"] & {} & Y \end{tikzcd}\] for any \(\gamma: X \rightarrow Y \in {\mathcal{M}}_{Q}{}\). From this diagram one can easily see that the functors \(-\otimes_R P^* \circ -\otimes_Q P: {\mathcal{M}}_{Q}{} \rightarrow {\mathcal{M}}_{Q}{}\) and \(\operatorname{Id}_{{\mathcal{M}}_{Q}{}}: {\mathcal{M}}_{Q}{} \rightarrow {\mathcal{M}}_{Q}{}\) are naturally isomorphic, i.e. \(-\otimes_R P^* \circ -\otimes_Q P \cong \operatorname{Id}_{{\mathcal{M}}_{Q}{}}\). Similarly, by Proposition 13 and Corollary 1, one obtains the functor natural isomorphism \(-\otimes_Q P\) \(\circ\) \(-\otimes_R P^* \cong \operatorname{Id}_{{\mathcal{M}}_{R}{}}\), and the equivalence of categories \(-\otimes_Q P: {\mathcal{M}}_{Q}{} \rightleftarrows {\mathcal{M}}_{R}{}: -\otimes_R P^*\). ◻