Proof. (i) Assume that there is another $X^{\prime }\in \mathscr{P}(\text{}_{R}A_{S})$ with right inverse $Y^{\prime }\in \mathscr{P}(\text{}_{S}A_{R})$ . Then, for any pair of morphisms $(f,g)\in \text{Hom}_{\mathscr{P}\left(\text{}_{R}A_{S}\right)}\left(X^{\prime },X\right)\times \text{Hom}_{\mathscr{P}\left(mbox_{S}A_{R}\right)}\left(Y,Y^{\prime }\right)$ , we define

(10)

Explicitly, $\pmb{\unicode[STIX]{x1D719}}$ and $\pmb{\unicode[STIX]{x1D713}}$ are given by

(11) $$\begin{eqnarray}\displaystyle \pmb{\unicode[STIX]{x1D719}}_{Y,Y^{\prime }}(f) & := & \displaystyle l_{Y^{\prime }}^{S}\circ \left(m_{Y,X}\otimes _{S}Y^{\prime }\right)\circ (Y\otimes _{R}f\otimes _{S}Y^{\prime })\circ \left(Y\otimes _{R}\left(m_{X^{\prime },Y^{\prime }}\right)^{-1}\right)\nonumber\\ \displaystyle & & \displaystyle \circ \left(r_{Y}^{R}\right)^{-1}\!,\end{eqnarray}$$

(12) $$\begin{eqnarray}\displaystyle \pmb{\unicode[STIX]{x1D713}}_{X^{\prime },X}(g) & := & \displaystyle r_{X}^{S}\circ (X\otimes _{S}m_{Y^{\prime },X^{\prime }})\circ (X\otimes _{S}g\otimes _{R}X^{\prime })\nonumber\\ \displaystyle & & \displaystyle \circ (m_{X,Y}^{-1}\otimes _{R}X^{\prime })\circ (l_{X^{\prime }}^{R})^{-1}.\end{eqnarray}$$

Therefore, we have

Similarly, one obtains the equality $i_{X}\circ \pmb{\unicode[STIX]{x1D713}}(g)=i_{X^{\prime }}$ . Now, for $X=X^{\prime }$ , $Y=Y^{\prime }$ and $f=\text{Id}_{X}$ , we get $i_{Y}\circ \pmb{\unicode[STIX]{x1D719}}(\text{Id}_{X})=i_{Y}$ which implies that $\pmb{\unicode[STIX]{x1D719}}(\text{Id}_{X})=\text{Id}_{Y}$ , since $i_{Y}$ is a monomorphism. Taking now $g=\text{Id}_{Y}$ , we obtain $\pmb{\unicode[STIX]{x1D713}}(\text{Id}_{Y})=\text{Id}_{X}$ . Both equalities $\pmb{\unicode[STIX]{x1D713}}(\text{Id}_{Y})=\text{Id}_{X}$ and $\pmb{\unicode[STIX]{x1D719}}(\text{Id}_{X})=\text{Id}_{Y}$ form precisely equation (7), and this finishes the proof of this item.

(ii) Equalities (8) and (9) are just (3) and (4) rewritten with respect to $\text{ev}$ and $\text{coev}$ .◻

Proof. The naturality of both $\unicode[STIX]{x1D719}$ and $\unicode[STIX]{x1D713}$ is clear. Now, for $f$ as above we have

If we reflect horizontally the diagrams above and we apply the substitutions $\unicode[STIX]{x1D713}\leftrightarrow \unicode[STIX]{x1D719},f\mapsto g,V\leftrightarrow W,X\mapsto Y$ , we get the diagrammatic proof for the equality $\unicode[STIX]{x1D719}(\unicode[STIX]{x1D713}(g))=g$ . The last adjunction is similarly proved.◻

Proof. We only show the last statement. The equality (13) is shown as follows:

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6FD}\circ m_{Y^{\prime },X}\circ (i_{Y,Y^{\prime }}\otimes _{R}X) & \,\overset{(4)}{=}\, & \displaystyle m_{A}^{R}\circ (i_{Y^{\prime }}\otimes _{R}i_{X})\circ (\pmb{\unicode[STIX]{x1D719}}_{Y,Y^{\prime }}(\text{Id}_{X})\otimes _{R}X)\nonumber\\ \displaystyle & = & \displaystyle m_{A}^{R}\circ \big((i_{Y^{\prime }}\circ \pmb{\unicode[STIX]{x1D719}}_{Y,Y^{\prime }}(\text{Id}_{X}))\otimes _{R}i_{X}\big)\nonumber\\ \displaystyle & = & \displaystyle m_{A}^{R}\circ (i_{Y}\otimes _{R}i_{X})\,\overset{(4)}{=}\,\unicode[STIX]{x1D6FD}\circ m_{Y,X}.\nonumber\end{eqnarray}$$

Equality (14) follows similarly.◻

Proof. Let us prove that $f_{X}$ and $\left(X\otimes _{S}m_{A}^{R}\right)\circ \left(X\otimes _{S}i_{Y}\otimes _{R}A\right)\circ \left(\text{coev}\otimes _{R}A\right)\circ \left(l_{A}^{R}\right)^{-1}$ are mutual inverses (the proof for $g_{Y}$ is analogous). One composition can be computed by means of the following tangle diagrams.

The other composition is computed as follows.

◻

Proof. First note that the required isomorphism is obtained, using the adjunctions of Proposition 1.4, as follows:

$$\begin{eqnarray}\displaystyle \text{Hom}_{R,\,S}\left(X^{\prime },X\right) & \cong & \displaystyle \text{Hom}_{R,\,S}\left(R\otimes _{R}X^{\prime },X\right)\cong \text{Hom}_{R,\,R}\left(R,X\otimes _{S}Y^{\prime }\right)\nonumber\\ \displaystyle & \cong & \displaystyle \text{Hom}_{S,\,R}\left(Y\otimes _{R}R,Y^{\prime }\right)\cong \text{Hom}_{S,\,R}\left(Y,Y^{\prime }\right)\!.\nonumber\end{eqnarray}$$

By (7), it is clear that $\pmb{\unicode[STIX]{x1D719}}_{X,X}(\text{Id}_{X})=\text{Id}_{Y}$ and $\pmb{\unicode[STIX]{x1D713}}_{Y,Y}(\text{Id}_{Y})=\text{Id}_{X}.$ The following diagrams show that

If we reflect horizontally the diagrammatic proof above we get $\pmb{\unicode[STIX]{x1D713}}(g)\circ \pmb{\unicode[STIX]{x1D713}}(g^{\prime })=\pmb{\unicode[STIX]{x1D713}}(g^{\prime }\circ g)$ .◻

Proof. First notice that both $\text{Inv}_{R,\,S}^{r}\left(A\right)$ and $\text{Inv}_{S,R}^{l}\left(A\right)$ are regarded as skeletal categories, in the sense that any two isomorphic objects are identical [Reference Lane20, page 91]. Therefore, in view of Lemma 1.5, if $X\in \mathscr{P}\left(\text{}_{R}A_{S}\right)$ has a right inverse, then this inverse is unique and denoted by $X^{r}$ . Thus to each object $X\in \text{Inv}_{R,\,S}^{r}\left(A\right)$ it corresponds a unique object $X^{r}\in \text{Inv}_{S,R}^{l}\left(A\right)$ . This establishes a bijection at the level of objects. By Lemma 1.9 we know that the maps $\pmb{\unicode[STIX]{x1D719}}$ of equation (11) induce isomorphisms $\text{Hom}_{\text{Inv}_{R,\,S}^{r}\left(A\right)}\left(X,U\right)\cong \text{Hom}_{\text{Inv}_{S,R}^{l}\left(A\right)}\left(U^{r},X^{r}\right)$ , which, by the last equations of that Lemma, give the desired contravariant category isomorphism $\boldsymbol{\boldsymbol{\unicode[STIX]{x1D719}}}.$ ◻

Proof. $(1)$ By Proposition 1.4, the functor $X\otimes _{S}(-)$ is a right adjoint and hence left exact so that $X\otimes _{S}i_{X^{\prime }}$ is a monomorphism. By Corollary 1.7, the morphism $f_{X}:X\otimes _{S}A\rightarrow A$ is an isomorphism. Thus $m_{A}^{S}\circ (i_{X}\otimes _{S}i_{X^{\prime }})=f_{X}\circ (X\otimes _{S}i_{X^{\prime }}):X\otimes _{S}X^{\prime }\rightarrow A$ is a monomorphism. Similarly, using the fact that the functor $(-)\otimes _{S}Y$ is a right adjoint (Proposition 1.4) and that $g_{Y}:=m_{A}^{S}\circ (A\otimes _{S}i_{Y})$ is an isomorphism (Corollary 1.7) one gets that $m_{A}^{S}\circ (i_{Y^{\prime }}\otimes _{S}i_{Y})$ is a monomorphism too.

$(2)$ Define $m_{X\otimes _{S}X^{\prime }}$ and $m_{Y^{\prime }\otimes _{S}Y}$ diagrammatically by setting

Now, using equation (3) for $X^{\prime }$ and $X$ , one gets the same equality for $X\otimes _{S}X^{\prime }$ as follows.

The same diagrammatic proof, once applied the substitutions $X\leftrightarrow Y^{\prime },Y\leftrightarrow X^{\prime }$ and $\unicode[STIX]{x1D6FC}\leftrightarrow \unicode[STIX]{x1D6FE}$ , yields (4) for $Y^{\prime }\otimes _{S}Y$ using the corresponding equality for $Y$ and $Y^{\prime }$ . The last statement is clear.◻

Proof. Let $X,X^{\prime }\in \text{Inv}_{S,\,S}^{r}(A)$ . By Proposition 1.12, we have $X\otimes _{S}X^{\prime }\in \text{Inv}_{S,\,S}^{r}(A)$ , with monomorphism $m_{A}^{S}\circ (i_{X}\otimes _{S}i_{X^{\prime }})$ , so that $\otimes _{S}$ is a well-defined multiplication for $\text{Inv}_{S,\,S}^{r}(A)$ . It is clearly associative. Note that $X\otimes _{S}S=X=S\otimes _{S}X$ as sub-bimodules of $A$ (these equalities make sense as the definition of subobject is given up to isomorphism, cf. [Reference Lane20, page 122]). Thus $S$ is the neutral element for this operation.

Let us check the last statement. For this consider $X\in \text{Inv}_{S,\,S}^{r}(A)$ and $Y$ its right inverse (as in Definition 1.1). Then, by equation (3), we have

$$\begin{eqnarray}i_{X\otimes _{S}Y}:=m_{A}^{S}\circ (i_{X}\otimes _{S}i_{Y})=\unicode[STIX]{x1D6FD}\circ m_{X}.\end{eqnarray}$$

This means that $X\otimes _{S}Y=S\in \text{Inv}_{S,\,S}^{r}(A)$ , as $m_{X}:X\otimes _{S}Y\rightarrow S$ is an isomorphism. Therefore, any element in $\text{Inv}_{S,\,S}^{r}(A)$ is already right invertible w.r.t. to the multiplication $\otimes _{S}$ . Thus an element $X\in \text{Inv}_{S,\,S}^{r}(A)$ belongs to the group of units if and only if it has a left $\otimes _{S}$ -inverse. Note that, in this case, the left and the right inverses should coincide as we are in a monoid. Therefore, the element $X$ is $\otimes _{S}$ -invertible if and only if $Y\otimes _{S}X=S$ . Now, $Y\otimes _{S}X$ is sub-bimodule of $A$ via $m_{A}^{S}\circ (i_{Y}\otimes _{S}i_{X})=\unicode[STIX]{x1D6FD}\circ m_{Y}$ by equation (4). Thus, the equality $Y\otimes _{S}X=S$ holds if and only if there is an isomorphism $\unicode[STIX]{x1D709}:Y\otimes _{S}X\rightarrow S$ of bimodules over $S$ such that $\unicode[STIX]{x1D6FD}\circ \unicode[STIX]{x1D709}=\unicode[STIX]{x1D6FD}\circ m_{Y}$ . Since we are assuming $\unicode[STIX]{x1D6FD}$ to be a monomorphism, this equality is equivalent to say that $m_{Y}$ is an isomorphism. This entails that $X$ is $\otimes _{S}$ - invertible if and only if $X\in \text{Inv}_{S,\,S}(A)$ .◻

Proof. $(1)$ is trivial.

$(2)$ From Lemma 1.11 we know that $i_{Y}\circ \boldsymbol{\boldsymbol{\unicode[STIX]{x1D719}}}(i)=i_{Y^{\prime }}$ . Thus $i_{Y^{\prime }}\circ j\circ \boldsymbol{\boldsymbol{\unicode[STIX]{x1D719}}}(i)=i_{Y}\circ \boldsymbol{\boldsymbol{\unicode[STIX]{x1D719}}}(i)=i_{Y^{\prime }}$ . Since $i_{Y^{\prime }}$ is a monomorphism, we get $j\circ \boldsymbol{\boldsymbol{\unicode[STIX]{x1D719}}}(i)=\text{Id}_{Y^{\prime }}.$ Similarly one proves that $\boldsymbol{\boldsymbol{\unicode[STIX]{x1D719}}}(i)\circ j=\text{Id}_{Y}$ . Thus $j$ and $\boldsymbol{\boldsymbol{\unicode[STIX]{x1D719}}}(i)$ are isomorphisms. By $(1)$ , the morphism $i$ is an isomorphism too, and this finishes the proof.◻

Proof. Straightforward. ◻

Proof. First note that the right inverse $Y$ of $X$ is unique up to isomorphism as shown in Lemma 1.5. Let us check that $\boldsymbol{\unicode[STIX]{x1D70E}}_{X}\left(g\right)$ does not depend on this isomorphism. Thus, given another right inverse $Z$ of $X$ , we want to check that the formula (18) is the same for both $Y$ and $Z$ . That is, we need to check the following equality

$$\begin{eqnarray}\displaystyle \boldsymbol{\unicode[STIX]{x1D70E}}_{X,Y}\left(g\right) & := & \displaystyle m_{A}^{S}\circ \left(i_{X}\otimes _{S}(g\rhd _{S}i_{Y})\right)\circ \left(m_{X}\right)^{-1}\nonumber\\ \displaystyle & = & \displaystyle m_{A}^{S}\circ \left(i_{X}\otimes _{S}(g\rhd _{S}i_{Z})\right)\circ \left(m_{X}\right)^{-1}:=\boldsymbol{\unicode[STIX]{x1D70E}}_{X,Z}\left(g\right)\!.\nonumber\end{eqnarray}$$

The map $\boldsymbol{\unicode[STIX]{x1D70E}}_{X,Y}(g)$ can be represented by the following diagram:

Using the isomorphisms stated in Corollary 1.5, we then have

By a similar argument, one proves that the definition of this map does not depend on the representative of the equivalence class $(X,i_{X})$ . In other words if $X=X^{\prime }$ (i.e., there is an isomorphism $f:X^{\prime }\rightarrow X$ such that $i_{X}\circ f=i_{X^{\prime }}$ ), then $\boldsymbol{\unicode[STIX]{x1D70E}}_{X}\left(g\right)=\boldsymbol{\unicode[STIX]{x1D70E}}_{X^{\prime }}\left(g\right)$ . On the other hand, the morphism $\boldsymbol{\unicode[STIX]{x1D70E}}_{X}\left(g\right)$ is a morphism in $_{R}{\mathcal{M}}_{R}$ being the composition of morphisms in $_{R}{\mathcal{M}}_{R}$ . Therefore, $\boldsymbol{\unicode[STIX]{x1D70E}}_{X}\left(g\right)$ is an element in ${\mathcal{Z}}_{R}(A)$ .

We have to check that $\boldsymbol{\unicode[STIX]{x1D70E}}_{X}$ is a ${\mathcal{Z}}(\mathbb{I})$ -algebra map. To this aim, given $g,h\in {\mathcal{Z}}_{S}(A)$ , we first show that $\boldsymbol{\unicode[STIX]{x1D70E}}_{X}\left(g\right)\ast \boldsymbol{\unicode[STIX]{x1D70E}}_{X}\left(h\right)=\boldsymbol{\unicode[STIX]{x1D70E}}_{X}\left(g\ast h\right)$ . Using the diagrammatic notation, we have

so that we get

(19) $$\begin{eqnarray}\boldsymbol{\unicode[STIX]{x1D70E}}_{X}(g)\rhd _{R}i_{X}=i_{X}\lhd _{S}g.\end{eqnarray}$$

Using this equality, we obtain

which means that $\boldsymbol{\unicode[STIX]{x1D70E}}_{X}(g)\ast \boldsymbol{\unicode[STIX]{x1D70E}}_{X}(h)=\boldsymbol{\unicode[STIX]{x1D70E}}_{X}(g\ast h)$ . Moreover, we have

$$\begin{eqnarray}\displaystyle \boldsymbol{\unicode[STIX]{x1D70E}}_{X}(1_{{\mathcal{Z}}_{S}(A)}) & = & \displaystyle \boldsymbol{\unicode[STIX]{x1D70E}}_{X}(\unicode[STIX]{x1D6FD})=m_{A}^{S}\circ (i_{X}\otimes _{S}(\unicode[STIX]{x1D6FD}\rhd _{S}i_{Y}))\circ \text{coev}\nonumber\\ \displaystyle & = & \displaystyle m_{A}^{S}\circ (i_{X}\otimes _{S}i_{Y})\circ \text{coev}\overset{(8)}{=}\unicode[STIX]{x1D6FC}=1_{{\mathcal{Z}}_{R}(A)}.\nonumber\end{eqnarray}$$

We still need to check that $\boldsymbol{\unicode[STIX]{x1D70E}}_{X}\circ \widetilde{\unicode[STIX]{x1D6FD}}\,=\,\widetilde{\unicode[STIX]{x1D6FC}}$ . To this aim, given $t\in {\mathcal{Z}}(\mathbb{I})$ , we have

(20)

and this completes the proof.◻

Proof. Consider both

$$\begin{eqnarray}\displaystyle \boldsymbol{\unicode[STIX]{x1D70E}}^{S,R} & : & \displaystyle \text{Inv}_{R,\,S}^{r}\left(A\right)\rightarrow \text{Hom}_{{\mathcal{Z}}(\mathbb{I})\text{-alg}}\left({\mathcal{Z}}_{S}(A),{\mathcal{Z}}_{R}(A)\right)\nonumber\\ \displaystyle \boldsymbol{\unicode[STIX]{x1D70E}}^{R,\,S} & : & \displaystyle \text{Inv}_{S,R}^{r}\left(A\right)\rightarrow \text{Hom}_{{\mathcal{Z}}(\mathbb{I})\text{-alg}}\left({\mathcal{Z}}_{R}(A),{\mathcal{Z}}_{S}(A)\right)\!.\nonumber\end{eqnarray}$$

Take $X\in \text{Inv}_{R,\,S}\left(A\right)$ as in (15). Since $m_{X^{r}}$ is an isomorphism, we have that $X^{r}\in \text{Inv}_{S,R}^{r}\left(A\right)$ . Hence we can consider both $\boldsymbol{\unicode[STIX]{x1D70E}}_{X}^{S,R}$ and $\boldsymbol{\unicode[STIX]{x1D70E}}_{X^{r}}^{R,\,S}$ . Let us check that these maps are mutual inverses. For $g\in {\mathcal{Z}}_{S}(A)$ , we have $\left(\boldsymbol{\unicode[STIX]{x1D70E}}_{X^{r}}^{R,\,S}\circ \boldsymbol{\unicode[STIX]{x1D70E}}_{X}^{S,R}\right)\left(g\right)$ is equal to

(22)

Thus $\boldsymbol{\unicode[STIX]{x1D70E}}_{X^{r}}^{R,\,S}\circ \boldsymbol{\unicode[STIX]{x1D70E}}_{X}^{S,R}=\text{Id}_{{\mathcal{Z}}_{S}(A)}$ . Similarly one gets $\boldsymbol{\unicode[STIX]{x1D70E}}_{X}^{S,R}\circ \boldsymbol{\unicode[STIX]{x1D70E}}_{X^{r}}^{R,\,S}=\text{Id}_{{\mathcal{Z}}_{R}(A)}$ . Therefore, $\boldsymbol{\unicode[STIX]{x1D70E}}_{X}^{S,R}$ is an isomorphism. This proves that $\unicode[STIX]{x1D6F7}^{S,R}$ is well defined.

Now assume that $R=S$ . By Corollary 1.13, we know that $(\text{Inv}_{R}(A),\otimes _{R},R)$ is a group. Let us check that $\unicode[STIX]{x1D6F7}^{R}$ is a group morphism. Take $X,X^{\prime }\in \text{Inv}_{R}\left(A\right)$ with two-sided inverses $Y$ and $Y^{\prime }$ , respectively. We have

where $g$ is as above. The second equality in diagram (22) implies that $\boldsymbol{\unicode[STIX]{x1D70E}}_{X}(\boldsymbol{\unicode[STIX]{x1D70E}}_{X^{\prime }}(g))=\boldsymbol{\unicode[STIX]{x1D70E}}_{X\otimes _{R}X^{\prime }}(g)$ . In other words $\unicode[STIX]{x1D6F7}^{R}(X\otimes _{R}X^{\prime })=\unicode[STIX]{x1D6F7}^{R}(X)\circ \unicode[STIX]{x1D6F7}^{R}(X^{\prime })$ . Moreover,

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6F7}^{R}(R)(g) & = & \displaystyle \boldsymbol{\unicode[STIX]{x1D70E}}_{R}(g)\overset{(18)}{=}m_{A}^{R}\circ (\unicode[STIX]{x1D6FC}\otimes _{R}(g\rhd _{R}\unicode[STIX]{x1D6FC}))\circ (r_{R}^{R})^{-1}\nonumber\\ \displaystyle & \overset{(\ast )}{=} & \displaystyle m_{A}^{R}\circ (\unicode[STIX]{x1D6FC}\otimes _{R}g)\circ (r_{R}^{R})^{-1}=\unicode[STIX]{x1D6FC}\rhd _{R}g=g\nonumber\end{eqnarray}$$

so that $\unicode[STIX]{x1D6F7}^{R}(R)=\text{Id}_{{\mathcal{Z}}_{R}(A)},$ where in $(\ast )$ we applied the definition of $\rhd _{R}$ .

Let us check that $\unicode[STIX]{x1D6F7}^{R}$ factors as stated. Take $p\in {\mathcal{Z}}_{R}(R)$ and set $g:={\mathcal{Z}}_{R}(\unicode[STIX]{x1D6FC})(p)=\unicode[STIX]{x1D6FC}\circ p$ . Since we already know that $\unicode[STIX]{x1D70E}_{X}$ is multiplicative, in order to conclude that $\unicode[STIX]{x1D70E}_{X}$ is ${\mathcal{Z}}_{R}(R)$ -bilinear, it suffices to check that $\unicode[STIX]{x1D70E}_{X}(g)=g$ . As $p$ is an endomorphism of the unit $R$ of the monoidal category $\text{}_{R}{\mathcal{M}}_{R}$ , we have $R\otimes _{R}p=p\otimes _{R}R$ . Hence, by using the same arguments of diagram (20), we get the desired equality.◻

Proof. (i) Fixing an element $t\in {\mathcal{Z}}(A)$ , we have

which shows that

$$\begin{eqnarray}\displaystyle (\unicode[STIX]{x1D70E}(t)\rhd A)\circ m\circ (\mathfrak{eq}_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D6FF}}\otimes _{}\mathfrak{eq}_{\unicode[STIX]{x1D6FF},\unicode[STIX]{x1D6FE}}) & = & \displaystyle (A\lhd \unicode[STIX]{x1D6FE}(t))\circ m\circ (\mathfrak{eq}_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D6FF}}\otimes _{}\mathfrak{eq}_{\unicode[STIX]{x1D6FF},\unicode[STIX]{x1D6FE}}).\nonumber\end{eqnarray}$$

Hence, by the universal property, the desired morphism is the one which turns commutative the diagrams

(30)

for every $t\in {\mathcal{Z}}(A)$ .

(ii) The equation follows from the fact that $\mathfrak{eq}_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D706}}$ is a monomorphism and the computation:

$$\begin{eqnarray}\displaystyle \mathfrak{eq}_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D706}}\circ m_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D706}}^{\unicode[STIX]{x1D6FF}}\circ (\text{}_{\unicode[STIX]{x1D70E}}J_{\unicode[STIX]{x1D6FF}}\otimes m_{\unicode[STIX]{x1D6FF},\unicode[STIX]{x1D706}}^{\unicode[STIX]{x1D6FE}}) & \overset{(27)}{=} & \displaystyle m\circ (\mathfrak{eq}_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D6FF}}\otimes _{}\mathfrak{eq}_{\unicode[STIX]{x1D6FF},\unicode[STIX]{x1D706}})\circ (\text{}_{\unicode[STIX]{x1D70E}}J_{\unicode[STIX]{x1D6FF}}\otimes m_{\unicode[STIX]{x1D6FF},\unicode[STIX]{x1D706}}^{\unicode[STIX]{x1D6FE}})\nonumber\\ \displaystyle & \overset{(27)}{=} & \displaystyle m\circ (A\otimes _{}m)\circ (\mathfrak{eq}_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D6FF}}\otimes _{}\mathfrak{eq}_{\unicode[STIX]{x1D6FF},\unicode[STIX]{x1D6FE}}\otimes _{}\mathfrak{eq}_{\unicode[STIX]{x1D6FE},\unicode[STIX]{x1D706}})\nonumber\\ \displaystyle & = & \displaystyle m\circ (m\otimes _{}A)\circ (\mathfrak{eq}_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D6FF}}\otimes _{}\mathfrak{eq}_{\unicode[STIX]{x1D6FF},\unicode[STIX]{x1D6FE}}\otimes _{}\mathfrak{eq}_{\unicode[STIX]{x1D6FE},\unicode[STIX]{x1D706}})\nonumber\\ \displaystyle & \overset{(27)}{=} & \displaystyle m\circ (\mathfrak{eq}_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D6FE}}\otimes _{}\mathfrak{eq}_{\unicode[STIX]{x1D6FE},\unicode[STIX]{x1D706}})\circ (m_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D6FE}}^{\unicode[STIX]{x1D6FF}}\otimes \text{}_{\unicode[STIX]{x1D6FE}}J_{\unicode[STIX]{x1D706}})\nonumber\\ \displaystyle & \overset{(27)}{=} & \displaystyle \mathfrak{eq}_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D706}}\circ m_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D706}}^{\unicode[STIX]{x1D6FE}}\circ (m_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D6FE}}^{\unicode[STIX]{x1D6FF}}\otimes \text{}_{\unicode[STIX]{x1D6FE}}J_{\unicode[STIX]{x1D706}}).\nonumber\end{eqnarray}$$

(iii) The associativity follows by (28). Denote $m_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D70E}}^{\unicode[STIX]{x1D70E}}:=m_{\unicode[STIX]{x1D70E}}$ and $\text{}_{\unicode[STIX]{x1D70E}}J_{\unicode[STIX]{x1D70E}}:=J$ . Now, we show that there is a morphism $u_{\unicode[STIX]{x1D70E}}$ such that the following diagrams commute

(31)

We compute

$$\begin{eqnarray}\displaystyle m\circ (A\otimes _{}\unicode[STIX]{x1D70E}(t))\circ u & = & \displaystyle m\circ (u\otimes _{}A)\circ (\mathbb{I}\otimes _{}\unicode[STIX]{x1D70E}(t))=l_{A}\circ (\mathbb{I}\otimes _{}\unicode[STIX]{x1D70E}(t))\nonumber\\ \displaystyle & = & \displaystyle \unicode[STIX]{x1D70E}(t)\circ l_{\mathbb{I}}=\unicode[STIX]{x1D70E}(t)\circ r_{\mathbb{I}}\nonumber\\ \displaystyle & = & \displaystyle r_{A}\circ (\unicode[STIX]{x1D70E}(t)\otimes _{}\mathbb{I})=m\circ (A\otimes _{}u)\circ (\unicode[STIX]{x1D70E}(t)\otimes _{}\mathbb{I})\nonumber\\ \displaystyle & = & \displaystyle m\circ (\unicode[STIX]{x1D70E}(t)\otimes _{}A)\circ u.\nonumber\end{eqnarray}$$

We now prove that the multiplication of $J$ is unitary:

$$\begin{eqnarray}\displaystyle \mathfrak{eq}_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D70E}}\circ m_{\unicode[STIX]{x1D70E}}\circ (J\otimes _{}u_{\unicode[STIX]{x1D70E}}) & \overset{(27)}{=} & \displaystyle m\circ (\mathfrak{eq}_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D70E}}\otimes _{}\mathfrak{eq}_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D70E}})\circ (J\otimes _{}u_{\unicode[STIX]{x1D70E}})\nonumber\\ \displaystyle & = & \displaystyle m\circ (A\otimes _{}u)\circ (\mathfrak{eq}_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D70E}}\otimes _{}\mathbb{I})\nonumber\\ \displaystyle & = & \displaystyle r_{A}\circ (\mathfrak{eq}_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D70E}}\otimes _{}\mathbb{I})=\mathfrak{eq}_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D70E}}\circ r_{J}.\nonumber\end{eqnarray}$$

Since $\mathfrak{eq}_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D70E}}$ is a monomorphism we obtain $m_{\unicode[STIX]{x1D70E}}\circ (J\otimes _{}u_{\unicode[STIX]{x1D70E}})=r_{J}$ . In a similar manner one gets $m_{\unicode[STIX]{x1D70E}}\circ (u_{\unicode[STIX]{x1D70E}}\otimes _{}J)=l_{J}$ . We have so proved that $(J,m_{\unicode[STIX]{x1D70E}},u_{\unicode[STIX]{x1D70E}})$ is a monoid.

(iv) It follows by (28).

(v) From (28) one gets that $m_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D6FE}}^{\unicode[STIX]{x1D6FF}}$ is balanced over $_{\unicode[STIX]{x1D6FF}}J_{\unicode[STIX]{x1D6FF}}$ .◻

Proof. By applying equation (19) to our situation we obtain

(33) $$\begin{eqnarray}\unicode[STIX]{x1D6F7}_{X}(t)\rhd i_{X}=i_{X}\lhd t,\quad \text{for every }t\in {\mathcal{Z}}(A).\end{eqnarray}$$

Equivalently $(\unicode[STIX]{x1D6F7}_{X}(t)\rhd A)\circ i_{X}\,=\,(A\lhd t)\circ i_{X}$ for every $t\in {\mathcal{Z}}(A)$ , which by the universal property of $\text{}_{\unicode[STIX]{x1D6F7}_{X}}J_{1}$ gives the desired monomorphism. Replacing $X$ by $Y$ one gets the other monomorphism.◻

Proof. Along this proof we abbreviate $\unicode[STIX]{x1D6F7}_{X}$ to $\unicode[STIX]{x1D719}$ . The morphisms $m_{\text{}_{\unicode[STIX]{x1D719}}J_{1}}=\overline{m}_{\unicode[STIX]{x1D719},\unicode[STIX]{x1D719}}^{1}$ and $m_{\text{}_{1}J_{\unicode[STIX]{x1D719}}}=\overline{m}_{1,1}^{\unicode[STIX]{x1D719}}$ defined in Proposition 1.3 for the pair $(\text{}_{\unicode[STIX]{x1D719}}J_{1},\text{}_{1}J_{\unicode[STIX]{x1D719}})$ , are given by Proposition 2.5(v), where the compatibility constrains are shown using equations (28) and (29). Consider the morphism

where $\unicode[STIX]{x1D704}_{X}$ was defined in Lemma 3.1. Now let us check that $m_{\text{}_{\unicode[STIX]{x1D719}}J_{1}}$ is invertible with inverse $\unicode[STIX]{x1D708}$ defined as the following composition of morphisms:

We have

where the last equality is given by Proposition 2.5(iii), while equality $(\star )$ is proved as follows:

$$\begin{eqnarray}\displaystyle \mathfrak{eq}_{\unicode[STIX]{x1D719},\unicode[STIX]{x1D719}}\circ m_{\unicode[STIX]{x1D719},\unicode[STIX]{x1D719}}^{1}\circ (\unicode[STIX]{x1D704}_{X}\otimes \unicode[STIX]{x1D704}_{Y}) & \overset{(27)}{=} & \displaystyle m\circ (\mathfrak{eq}_{\unicode[STIX]{x1D719},1}\otimes \mathfrak{eq}_{1,\unicode[STIX]{x1D719}})\circ (\unicode[STIX]{x1D704}_{X}\otimes \unicode[STIX]{x1D704}_{Y})=m\circ (i_{X}\otimes i_{Y})\nonumber\\ \displaystyle & \overset{(1)}{=} & \displaystyle u\circ m_{X}=\mathfrak{eq}_{\unicode[STIX]{x1D719},\unicode[STIX]{x1D719}}\circ u_{\unicode[STIX]{x1D719}}\circ \text{coev}^{-1}.\nonumber\end{eqnarray}$$

This proves that $\overline{m}_{\unicode[STIX]{x1D719},\unicode[STIX]{x1D719}}^{1}\circ \unicode[STIX]{x1D708}$ is the identity. A similar computation, which uses composition with the epimorphism $\unicode[STIX]{x1D712}_{\unicode[STIX]{x1D719},\unicode[STIX]{x1D719}}^{1}$ , shows that $\unicode[STIX]{x1D708}\circ \overline{m}_{\unicode[STIX]{x1D719},\unicode[STIX]{x1D719}}^{1}$ is the identity too so that $\overline{m}_{\unicode[STIX]{x1D719},\unicode[STIX]{x1D719}}^{1}$ is invertible. Replacing $\unicode[STIX]{x1D719}$ by $\unicode[STIX]{x1D719}^{-1}$ will show that $m_{\text{}_{\unicode[STIX]{x1D719}^{-1}}J_{1}}$ is an isomorphism which means that $m_{\text{}_{1}J_{\unicode[STIX]{x1D719}}}$ is an isomorphism, since $_{\unicode[STIX]{x1D719}^{-1}}J_{1}\,=\,\text{}_{1}J_{\unicode[STIX]{x1D719}}$ .◻

Proof. Under the assumption made on $u_{1}$ , the unit $u$ is a monomorphism by Proposition 2.5(iii). Hence, by Proposition 3.2, we know that $_{\unicode[STIX]{x1D6F7}_{X}}J_{1}$ is invertible with inverse $\text{}_{1}J_{\unicode[STIX]{x1D6F7}_{X}}$ . Therefore, we have that $\text{}_{\unicode[STIX]{x1D6F7}_{X}}J_{1}\otimes \text{}_{1}J_{\unicode[STIX]{x1D6F7}_{X}}\cong \mathbb{I}$ and $\text{}_{1}J_{\unicode[STIX]{x1D6F7}_{X}}\otimes \text{}_{\unicode[STIX]{x1D6F7}_{X}}J_{1}\cong \mathbb{I}$ . Now, by Lemma 3.1, we know that $X$ is injected in $\text{}_{\unicode[STIX]{x1D6F7}_{X}}J_{1}$ while $Y$ is injected in $\text{}_{1}J_{\unicode[STIX]{x1D6F7}_{X}}$ . Summing up, we have shown that the elements $X$ and $\text{}_{\unicode[STIX]{x1D6F7}_{X}}J_{1}$ satisfy the assumptions of Proposition 1.14(2) so that we obtain the equalities $X=\text{}_{\unicode[STIX]{x1D6F7}_{X}}J_{1}$ and $Y=\text{}_{1}J_{\unicode[STIX]{x1D6F7}_{X}}$ .◻

Proof. In view of Proposition 3.2, it is clear that $\unicode[STIX]{x1D6F7}$ corestricts to a map $\widehat{\unicode[STIX]{x1D6F7}}$ such that $\unicode[STIX]{x1D70D}\circ \widehat{\unicode[STIX]{x1D6F7}}=\unicode[STIX]{x1D6F7}$ and, by Lemma 3.3, we have that $\widehat{\unicode[STIX]{x1D6F7}}$ is injective. Let us check that it is also surjective that is that $\unicode[STIX]{x1D6F7}_{\text{}_{\unicode[STIX]{x1D703}}J_{1}}\,=\,\unicode[STIX]{x1D703}$ , for every $\unicode[STIX]{x1D703}\in {\mathcal{G}}_{{\mathcal{Z}}(\mathbb{I})}({\mathcal{Z}}(A))$ . Recall that, for a given $t\in {\mathcal{Z}}(A)$ , we have

$$\begin{eqnarray}\unicode[STIX]{x1D6F7}_{\text{}_{\unicode[STIX]{x1D703}}J_{1}}(t)\,=\,m\circ (A\otimes _{}m)\circ (\mathfrak{eq}_{\unicode[STIX]{x1D703},1}\otimes _{}A\otimes _{}\mathfrak{eq}_{1,\unicode[STIX]{x1D703}})\circ (\text{}_{\unicode[STIX]{x1D703}}J_{1}\otimes _{}t\otimes \text{}_{1}J_{\unicode[STIX]{x1D703}})\circ (m_{\unicode[STIX]{x1D703},\unicode[STIX]{x1D703}}^{1})^{-1}\circ u_{\unicode[STIX]{x1D703}}.\end{eqnarray}$$

The desired equality can be checked by diagrams as follows

(36)

We have so proved that the map $\widehat{\unicode[STIX]{x1D6F7}}:\text{Inv}_{\mathbb{I}}(A)\rightarrow {\mathcal{G}}_{{\mathcal{Z}}(\mathbb{I})}({\mathcal{Z}}(A))$ is bijective. As a consequence, since $\unicode[STIX]{x1D6F7}$ is a group morphism, we have for free that ${\mathcal{G}}_{{\mathcal{Z}}(\mathbb{I})}({\mathcal{Z}}(A))$ is a subgroup of the automorphisms group $\text{Aut}_{{\mathcal{Z}}(\mathbb{I})\text{-alg}}({\mathcal{Z}}(A))$ and that both $\widehat{\unicode[STIX]{x1D6F7}}$ and $\unicode[STIX]{x1D70D}$ are group morphisms.◻

Proof. Let $g_{i}$ denote the resulting maps from the assumption $A_{\unicode[STIX]{x1D6FE}}|A_{\unicode[STIX]{x1D6FF}}$ . Since each $g_{i}$ is left $A$ -linear, we have

(41) $$\begin{eqnarray}\mathop{\sum }_{i}f_{i}\lhd (g_{i}\circ u)=\mathop{\sum }_{i}g_{i}\circ (f_{i}\lhd u)=\mathop{\sum }_{i}g_{i}\circ f_{i}=A.\end{eqnarray}$$

We set $\unicode[STIX]{x1D709}_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D6FE}}^{\unicode[STIX]{x1D6FF}}:=({\sum _{i}}_{\unicode[STIX]{x1D70E}}f_{i}\,\otimes \,_{\unicode[STIX]{x1D6FF}}g_{i})\circ (\text{}_{\unicode[STIX]{x1D70E}}J_{\unicode[STIX]{x1D6FE}}\otimes u_{\unicode[STIX]{x1D6FF}})\circ r_{_{\unicode[STIX]{x1D70E}}J_{\unicode[STIX]{x1D6FE}}}^{-1}$ , where $_{\unicode[STIX]{x1D70E}}f_{i}$ and $_{\unicode[STIX]{x1D6FF}}g_{j}$ are defined as in equation (39). By its own definition the morphism $\unicode[STIX]{x1D709}_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D6FE}}^{\unicode[STIX]{x1D6FF}}$ satisfies

(42)

where the summation is understood. So we compute

Since $\mathfrak{eq}_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D6FE}}$ is a monomorphism, we conclude that $m_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D6FE}}^{\unicode[STIX]{x1D6FF}}\circ \unicode[STIX]{x1D709}_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D6FE}}^{\unicode[STIX]{x1D6FF}}=\text{}_{\unicode[STIX]{x1D70E}}J_{\unicode[STIX]{x1D6FE}}$ and hence $m_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D6FE}}^{\unicode[STIX]{x1D6FF}}$ is a split epimorphism. Now let us check, under the assumptions $u_{\unicode[STIX]{x1D6FF}}:\mathbb{I}\rightarrow \text{}_{\unicode[STIX]{x1D6FF}}J_{\unicode[STIX]{x1D6FF}}$ is an isomorphism and $A$ is right flat, that $m_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D6FE}}^{\unicode[STIX]{x1D6FF}}$ is indeed an isomorphism. Using diagrams, with summation understood again, we have

where in the first equality we have used simultaneously equations (42), (39) and the fact that the unit $u_{\unicode[STIX]{x1D6FF}}$ of $\text{}_{\unicode[STIX]{x1D6FF}}J_{\unicode[STIX]{x1D6FF}}$ satisfies $\mathfrak{eq}_{\unicode[STIX]{x1D6FF},\unicode[STIX]{x1D6FF}}\circ u_{\unicode[STIX]{x1D6FF}}\,=\,u$ the unit of $A$ . We continue then our computation

If $\mathfrak{eq}_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D6FF}}\otimes _{}\mathfrak{eq}_{\unicode[STIX]{x1D6FF},\unicode[STIX]{x1D6FE}}$ is a monomorphism, then we get $\unicode[STIX]{x1D709}_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D6FE}}^{\unicode[STIX]{x1D6FF}}\circ m_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D6FE}}^{\unicode[STIX]{x1D6FF}}=\text{}_{\unicode[STIX]{x1D70E}}J_{\unicode[STIX]{x1D6FF}}\otimes \text{}_{\unicode[STIX]{x1D6FF}}J_{\unicode[STIX]{x1D6FE}}$ , and so $m_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D6FE}}^{\unicode[STIX]{x1D6FF}}$ is an isomorphism.

The fact that $\mathfrak{eq}_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D6FF}}\otimes _{}\mathfrak{eq}_{\unicode[STIX]{x1D6FF},\unicode[STIX]{x1D6FE}}$ is a monomorphism is proved as follows. First, using Proposition 2.5 and that $A_{\unicode[STIX]{x1D6FE}}|A_{\unicode[STIX]{x1D6FF}}$ , one shows that $(\text{}_{\unicode[STIX]{x1D6FE}}J_{\unicode[STIX]{x1D6FF}},\text{}_{\unicode[STIX]{x1D6FF}}J_{\unicode[STIX]{x1D6FE}},\overline{m}_{\unicode[STIX]{x1D6FF},\,\unicode[STIX]{x1D6FF}}^{\unicode[STIX]{x1D6FE}},\overline{\unicode[STIX]{x1D709}}_{\unicode[STIX]{x1D6FE},\unicode[STIX]{x1D6FE}}^{\unicode[STIX]{x1D6FF}})$ is a right dualizable datum between the monoids $\text{}_{\unicode[STIX]{x1D6FE}}J_{\unicode[STIX]{x1D6FE}}$ and $\text{}_{\unicode[STIX]{x1D6FF}}J_{\unicode[STIX]{x1D6FF}}$ , where $\overline{\unicode[STIX]{x1D709}}_{\unicode[STIX]{x1D6FE},\unicode[STIX]{x1D6FE}}^{\unicode[STIX]{x1D6FF}}=\unicode[STIX]{x1D712}_{\unicode[STIX]{x1D6FE},\unicode[STIX]{x1D6FE}}^{\unicode[STIX]{x1D6FF}}\circ \unicode[STIX]{x1D709}_{\unicode[STIX]{x1D6FE},\unicode[STIX]{x1D6FE}}^{\unicode[STIX]{x1D6FF}}$ is a splitting of $\overline{m}_{\unicode[STIX]{x1D6FE},\unicode[STIX]{x1D6FE}}^{\unicode[STIX]{x1D6FF}}$ . Secondly, by Proposition 1.4, we obtain that $\text{}_{\unicode[STIX]{x1D6FF}}J_{\unicode[STIX]{x1D6FE}}$ is left flat as $\mathbb{I}\cong \text{}_{\unicode[STIX]{x1D6FF}}J_{\unicode[STIX]{x1D6FF}}$ . In this way, we deduce that $\mathfrak{eq}_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D6FF}}\otimes _{}\mathfrak{eq}_{\unicode[STIX]{x1D6FF},\unicode[STIX]{x1D6FE}}=(A\otimes _{}\mathfrak{eq}_{\unicode[STIX]{x1D6FF},\unicode[STIX]{x1D6FE}})\circ (\mathfrak{eq}_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D6FF}}\otimes \text{}_{\unicode[STIX]{x1D6FF}}J_{\unicode[STIX]{x1D6FE}})$ is a monomorphism since $A$ is right flat.◻

Proof. $(1)$ Let $f\in \text{Hom}_{A,A}\left(M,\,N\right)$ . Then $f\in \text{Hom}_{A\text{-}}(M,N)$ and, for every $z\in {\mathcal{Z}}(A)$ , we have

$$\begin{eqnarray}f\circ (M\lhd \unicode[STIX]{x1D70E}(z))=f\circ \unicode[STIX]{x1D71A}_{N}\circ (M\otimes \unicode[STIX]{x1D70E}(z))=\unicode[STIX]{x1D71A}_{M}\circ (f\otimes A)\circ (M\otimes \unicode[STIX]{x1D70E}(z))=f\lhd \unicode[STIX]{x1D70E}(z).\end{eqnarray}$$

$(2)$ It follows by $(1)$ .

$(3)$ Let $f:M\rightarrow N$ be an isomorphism of $A$ -bimodules. Set $k:=1,f_{1}:=f$ and $g_{1}:=f^{-1}$ . Clearly $\sum _{i=1}^{k}g_{i}\circ f_{i}=M$ so that, by $(2)$ , we get $M_{\unicode[STIX]{x1D70E}}|N_{\unicode[STIX]{x1D70E}}$ . If we interchange the roles of $M$ and $N$ we get also $N_{\unicode[STIX]{x1D70E}}|M_{\unicode[STIX]{x1D70E}}$ .

$(4)$ Let $k\geqslant 1$ and let $f_{i}\in \text{Hom}_{A,A}\left(M,\,N\right)$ , $g_{i}\in \text{Hom}_{A,A}\left(N,\,M\right)$ , for $i=1,\ldots ,k$ , such that $\sum _{i}g_{i}\circ f_{i}=M$ . Then $f_{i}^{\prime }:=f_{i}\otimes _{A}L\in \text{Hom}_{A,A}\left(M\otimes _{A}L,\,N\otimes _{A}L\right)$ , $g_{i}^{\prime }:=g_{i}\otimes _{A}L\in \text{Hom}_{A,A}\left(N\otimes _{A}L,\,M\otimes _{A}L\right)$ . Moreover, $\sum _{i}g_{i}^{\prime }\circ f_{i}^{\prime }=M\otimes _{A}L$ .

$(5)$ Let $f\in {\mathcal{M}}_{A,{\mathcal{Z}}(A)}(M_{\unicode[STIX]{x1D70E}},N_{\unicode[STIX]{x1D70F}})$ and $g\in {\mathcal{M}}_{A,{\mathcal{Z}}(A)}(L_{\unicode[STIX]{x1D70C}},M_{\unicode[STIX]{x1D70E}})$ . Then $f\circ g\in \text{Hom}_{A\text{-}}(L,N)$ . Moreover,

$$\begin{eqnarray}\displaystyle f\circ g\circ (L\lhd \unicode[STIX]{x1D70C}(z)) & = & \displaystyle f\circ (g\lhd \unicode[STIX]{x1D70E}(z))=f\circ (M\lhd \unicode[STIX]{x1D70E}(z))\circ g\nonumber\\ \displaystyle & = & \displaystyle (f\lhd \unicode[STIX]{x1D70F}(z))\circ g=(f\circ g)\lhd \unicode[STIX]{x1D70F}(z).\nonumber\end{eqnarray}$$

$(6)$ By assumption there is $k\geqslant 1$ and morphisms $f_{i}\in {\mathcal{M}}_{A,{\mathcal{Z}}(A)}(M_{\unicode[STIX]{x1D70E}},N_{\unicode[STIX]{x1D70F}})$ , $g_{i}\in {\mathcal{M}}_{A,{\mathcal{Z}}(A)}(N_{\unicode[STIX]{x1D70F}},M_{\unicode[STIX]{x1D70E}})$ , for $i=1,\ldots ,k$ , such that $\sum _{i}g_{i}\circ f_{i}=M$ . Moreover, there is $k^{\prime }\geqslant 1$ and morphisms $f_{i}^{\prime }\in {\mathcal{M}}_{A,{\mathcal{Z}}(A)}(L_{\unicode[STIX]{x1D70C}},M_{\unicode[STIX]{x1D70E}})$ , $g_{i}^{\prime }\in {\mathcal{M}}_{A,{\mathcal{Z}}(A)}(M_{\unicode[STIX]{x1D70E}},L_{\unicode[STIX]{x1D70C}})$ , for $i=1,\ldots ,k^{\prime }$ , such that $\sum _{i}g_{i}^{\prime }\circ f_{i}^{\prime }=L$ . Set $f_{i,j}:=f_{i}\circ f_{j}^{\prime }$ and $g_{i,j}:=g_{i}^{\prime }\circ g_{j}$ . Then $\sum _{i,j}g_{i,j}\circ f_{j,i}=\sum _{i,j}g_{i}^{\prime }\circ g_{j}\circ f_{j}\circ f_{i}^{\prime }=L$ and, by $(5)$ , we have that $f_{i,j}\in {\mathcal{M}}_{A,{\mathcal{Z}}(A)}(L_{\unicode[STIX]{x1D70C}},N_{\unicode[STIX]{x1D70F}})$ and $g_{i,j}\in {\mathcal{M}}_{A,{\mathcal{Z}}(A)}(N_{\unicode[STIX]{x1D70F}},L_{\unicode[STIX]{x1D70C}})$ .

$(7)$ It follows from the obvious inclusion ${\mathcal{M}}_{A,{\mathcal{Z}}(A)}(M_{\unicode[STIX]{x1D70E}},N_{\unicode[STIX]{x1D70F}})\subseteq {\mathcal{M}}_{A,{\mathcal{Z}}(A)}(M_{\unicode[STIX]{x1D70E}\unicode[STIX]{x1D70C}},N_{\unicode[STIX]{x1D70F}\unicode[STIX]{x1D70C}})$ .◻

Proof. Set $k:=1,f_{1}:=M\otimes \unicode[STIX]{x1D6FD}\unicode[STIX]{x1D6FC}^{-1}$ and $g_{1}:=f_{1}^{-1}$ . Clearly $f_{1}\in {\mathcal{M}}_{A,-}(M\otimes A,M\otimes A)$ . Moreover,

$$\begin{eqnarray}\displaystyle f_{1}\circ ((M\otimes A)\lhd \unicode[STIX]{x1D70E}(z)) & = & \displaystyle (M\otimes \unicode[STIX]{x1D6FD})\circ (M\otimes \unicode[STIX]{x1D6FC}^{-1})\circ (M\otimes (A\lhd \unicode[STIX]{x1D70E}(z)))\nonumber\\ \displaystyle & = & \displaystyle M\otimes [\unicode[STIX]{x1D6FD}\circ \unicode[STIX]{x1D6FC}^{-1}\circ (A\lhd \unicode[STIX]{x1D70E}(z))]\nonumber\\ \displaystyle & = & \displaystyle M\otimes [(\unicode[STIX]{x1D6FD}\circ \unicode[STIX]{x1D6FC}^{-1})\lhd (\unicode[STIX]{x1D6FD}\circ \unicode[STIX]{x1D6FC}^{-1}\circ \unicode[STIX]{x1D70E}(z))]\nonumber\\ \displaystyle & = & \displaystyle M\otimes [(\unicode[STIX]{x1D6FD}\circ \unicode[STIX]{x1D6FC}^{-1})\lhd (\unicode[STIX]{x1D6FD}\circ \unicode[STIX]{x1D6FC}^{-1}\circ \unicode[STIX]{x1D6FC}\circ z)]\nonumber\\ \displaystyle & = & \displaystyle M\otimes [(\unicode[STIX]{x1D6FD}\circ \unicode[STIX]{x1D6FC}^{-1})\lhd (\unicode[STIX]{x1D6FD}\circ z)]\nonumber\\ \displaystyle & = & \displaystyle M\otimes [(\unicode[STIX]{x1D6FD}\circ \unicode[STIX]{x1D6FC}^{-1})\lhd \unicode[STIX]{x1D70F}(z)]\nonumber\\ \displaystyle & = & \displaystyle f_{1}\lhd \unicode[STIX]{x1D70F}(z)\nonumber\end{eqnarray}$$

so that $f_{1}\in {\mathcal{M}}_{A,{\mathcal{Z}}(A)}(M\otimes A_{\unicode[STIX]{x1D70E}},M\otimes A_{\unicode[STIX]{x1D70F}})$ . A similar argument shows that $g_{1}\in {\mathcal{M}}_{A,{\mathcal{Z}}(A)}(M\otimes A_{\unicode[STIX]{x1D70F}},M\otimes A_{\unicode[STIX]{x1D70E}})$ . Since $\sum _{i}g_{i}\circ f_{i}=M\otimes A$ , we conclude that $M\otimes A_{\unicode[STIX]{x1D70E}}|M\otimes A_{\unicode[STIX]{x1D70F}}$ . If we interchange the roles of $M$ and $N$ and the ones of $\unicode[STIX]{x1D70E}$ and $\unicode[STIX]{x1D6FC}$ we get also $M\otimes A_{\unicode[STIX]{x1D70F}}|M\otimes A_{\unicode[STIX]{x1D70E}}$ .◻

Proof. We split the proof in three steps.

(1) Since $A\cong A\otimes _{A}A$ , by Proposition 3.8(3), we get $A_{\unicode[STIX]{x1D70E}}\sim (A\otimes _{A}A)_{\unicode[STIX]{x1D70E}}=A\otimes _{A}A_{\unicode[STIX]{x1D70E}}$ .

(2) Since $A\sim A\otimes A$ , in view of Proposition 3.8(4), we have $A\otimes _{A}A\sim A\otimes A\otimes _{A}A$ and hence $A\otimes _{A}A_{\unicode[STIX]{x1D70E}}\sim A\otimes A\otimes _{A}A_{\unicode[STIX]{x1D70E}}$ , by Proposition 3.8(2).

(3) Since $A\otimes A\otimes _{A}A\cong A\otimes A$ , by Proposition 3.8(3), we get $A\otimes A\otimes _{A}A_{\unicode[STIX]{x1D70E}}\sim A\otimes A_{\unicode[STIX]{x1D70E}}$ .

Now, if we glue together (1), (2) and (3) using Proposition 3.8(6), we obtain $A_{\unicode[STIX]{x1D70E}}\sim A\otimes A_{\unicode[STIX]{x1D70E}}$ . Similarly one gets $A_{\unicode[STIX]{x1D70F}}\sim A\otimes A_{\unicode[STIX]{x1D70F}}.$ The conclusion follows by Propositions 3.9 and 3.8(6).◻

Proof. By Theorem 3.10, we have that $A_{\unicode[STIX]{x1D70E}}\sim A_{1}$ . Henceforth, we apply Corollary 3.7 to obtain that $\unicode[STIX]{x1D70E}\in {\mathcal{G}}_{{\mathcal{Z}}(\mathbb{I})}({\mathcal{Z}}(A))$ , which means that $\text{}_{\unicode[STIX]{x1D70E}}J_{1}\in \text{Inv}_{\mathbb{I}}(A)$ with inverse $\text{}_{1}J_{\unicode[STIX]{x1D70E}}$ . By applying now Theorem 3.4, it is clear that $\unicode[STIX]{x1D70E}=\widehat{\unicode[STIX]{x1D6F7}}\widehat{\unicode[STIX]{x1D6F7}}^{-1}(\unicode[STIX]{x1D70E})=\unicode[STIX]{x1D6F7}_{\text{}_{\unicode[STIX]{x1D70E}}J_{1}}$ . The particular statement is immediate.◻

Proof. (1) Since $\unicode[STIX]{x1D6FA}=\boldsymbol{\unicode[STIX]{x1D70D}}\circ \unicode[STIX]{x1D714}$ is surjective, it is clear that $\boldsymbol{\unicode[STIX]{x1D70D}}$ is surjective, and so bijective as it is by construction injective. Therefore, by Theorem 3.4, this implies that $\unicode[STIX]{x1D6F7}=\boldsymbol{\unicode[STIX]{x1D70D}}\circ \widehat{\unicode[STIX]{x1D6F7}}$ is bijective as well. (2) If ${\mathcal{Z}}$ is faithful, then clearly $\unicode[STIX]{x1D6FA}$ is injective, hence it is also bijective. Thus $\unicode[STIX]{x1D714}=\boldsymbol{\unicode[STIX]{x1D70D}}^{-1}\circ \unicode[STIX]{x1D6FA}$ is also bijective. The stated chain of isomorphisms of groups is now immediate.◻

Proof. In view of Corollary 1.13 (here the assumption $u$ is monomorphism is used), it suffices to prove that we have a morphism of monoids

$$\begin{eqnarray}\boldsymbol{\unicode[STIX]{x1D6E4}}:(\text{Inv}_{\mathbb{I}}^{r}(A),\otimes ,\mathbb{I})\longrightarrow (\text{End}_{alg}(A),\circ ,\text{Id}_{A}),\qquad \Big(X\longmapsto \unicode[STIX]{x1D6E4}_{X}\Big).\end{eqnarray}$$

First we have to check that this map is well defined, that is $\boldsymbol{\unicode[STIX]{x1D6E4}}_{X}$ is a monoid endomorphism of $A$ . Let us check that $\boldsymbol{\unicode[STIX]{x1D6E4}}_{X}$ is unitary

Let us check that $\boldsymbol{\unicode[STIX]{x1D6E4}}_{X}$ is multiplicative. We start by computing $m\circ (\boldsymbol{\unicode[STIX]{x1D6E4}}_{X}\otimes _{}\boldsymbol{\unicode[STIX]{x1D6E4}}_{X})\,=\,$

It is clear that $\unicode[STIX]{x1D6E4}_{\mathbb{I}}=\text{Id}_{A}$ . Let now $X,X^{\prime }\in \text{Inv}_{\mathbb{I}}^{r}(A)$ have right inverses $Y,Y^{\prime }$ respectively. Then, we compute $\boldsymbol{\unicode[STIX]{x1D6E4}}_{X\otimes _{}X^{\prime }}\,=\,$

It is now clear that, for $X\in \text{Inv}_{\mathbb{I}}(A)$ with inverse $Y$ , one has $\boldsymbol{\unicode[STIX]{x1D6E4}}_{X}^{-1}=\boldsymbol{\unicode[STIX]{x1D6E4}}_{Y}$ . The particular statement is clear, and this finishes the proof.◻

Proof. (i). By [Reference Femić10, Proposition 3.3.], we can identify $\text{}_{A^{e}}[A,A]$ with the equalizer

(43)

where $\boldsymbol{\unicode[STIX]{x1D701}}^{A}$ is the unit of the adjunction $A\otimes _{}-\dashv [A,-]$ . Now, using the universal property of $\text{}_{1}J_{1}$ given by equation (25), one shows the existence of the triangle. We leave to the reader to check that the stated diagram is in fact a triangle of monoids.

(ii) It follows from the fact that the diagram in (i) commutes and that the morphisms involved are unitary.

(iii) The definition of the functor $\text{}_{A^{e}}[A,-]$ obviously implies the isomorphism ${\mathcal{Z}}(\text{}_{A^{e}}[A,A])\cong \text{Hom}_{A^{e}\text{-}}(A,A)$ of ${\mathcal{Z}}(\mathbb{I})$ -algebras. On the other hand, under the assumption made on the functor ${\mathcal{Z}}$ a tedious computation gives a chase to an isomorphism ${\mathcal{Z}}(\text{}_{1}J_{1})\cong \text{Hom}_{A^{e}\text{-}}(A,A)$ of ${\mathcal{Z}}(\mathbb{I})$ -algebras. Combining these two isomorphisms shows that ${\mathcal{Z}}(\unicode[STIX]{x1D706})$ is an isomorphism. Therefore, $\text{}_{1}J_{1}\cong \text{}_{A^{e}}[A,A]\cong \mathbb{I}$ , since $A$ is central.

(iv) We only need to check the direct implication since the converse follows by the second item. This implication clearly follows from part (iii), since a faithful functor, whose domain is an abelian category, reflects isomorphisms. ◻

Remarks 4.4. Consider an Azumaya monoid $(A,m,u)$ in ${\mathcal{M}}$ as in Definition 4.3.

1. (i) It is noteworthy to mention that one cannot expect to have for free that $A$ is a dualizable object in ${\mathcal{M}}$ , that is dualizable as in Definition 1.2 with the trivial structure of $(\mathbb{I},\mathbb{I})$ -bimodule. This perhaps happens when $\mathbb{I}$ is a small generator in ${\mathcal{M}}$ . However, as was shown in [Reference Vitale30, Proposition 1.2], $A$ is in fact a dualizable object in a certain monoidal category of monoids whose dual object is exactly the opposite monoid $A^{o}$ . On the other hand, one can show as follows that $A$ forms part of a two-sided dualizable datum in the sense of Definition 1.2 by taking $\mathbb{I}$ and $A^{e}$ as base monoids. Following the proof of [Reference Vitale30, Proposition 1.2] since $A$ can be considered either as an $(\mathbb{I},A^{e})$ -bimodule or as an $(A^{e},\mathbb{I})$ -bimodule, $A\otimes _{}-:{\mathcal{M}}\rightarrow \text{}_{A^{e}}{\mathcal{M}}$ is an equivalence of categories if and only if $A\otimes _{A^{e}}-:\text{}_{A^{e}}{\mathcal{M}}\rightarrow {\mathcal{M}}$ is so. In this way, by applying twice [Reference Pareigis25, Proposition 5.1] or [Reference Vitale30, Proposition 1.3], we obtain the existence of two objects: $[A,\mathbb{I}]$ and $[A,A]$ together with the following natural isomorphisms $\text{Hom}_{{\mathcal{M}}}\left(A\otimes _{}-,\,\mathbb{I}\right)\cong \text{Hom}_{{\mathcal{M}}}\left(-,\,[A,\mathbb{I}]\right)$ and $\text{Hom}_{{\mathcal{M}}}\left(A\otimes _{}-,\,A\right)\cong \text{Hom}_{{\mathcal{M}}}\left(-,\,[A,A]\right)$ . Furthermore, there are isomorphisms: $[A,\mathbb{I}]\otimes _{}A\cong A^{e}$ of $A^{e}$ -bimodules, $\mathbb{I}\cong A\otimes _{A^{e}}[A,\mathbb{I}]$ of objects in ${\mathcal{M}}$ and $A^{e}=A\otimes _{}A^{o}\cong [A,A]$ of monoids. Thus, the pair $(A,[A,\mathbb{I}])$ , within these two first isomorphisms, is a two-sided dualizable datum relating the monoids $\mathbb{I}$ and $A^{e}$ in the sense of Definition 1.2, as claimed above.

2. (ii) Since $A\otimes _{}-:{\mathcal{M}}\rightarrow \text{}_{A^{e}}{\mathcal{M}}$ is an equivalence of categories, it has a right adjoint functor which, as in Appendix B, is denoted by $\text{}_{A^{e}}[A,-]:\text{}_{A^{e}}{\mathcal{M}}\rightarrow {\mathcal{M}}$ . In contrast with Proposition B.1, here we only know that $\text{}_{A^{e}}[A,-]$ exists and there are no indications about its construction. Indeed, at this level of generality, it is not clear whether $A$ has a left internal hom functor in ${\mathcal{M}}$ . Thus a very interesting class of Azumaya monoids consists of those for which the underlying object has this property. This is, of course, the case of the usual class of Azumaya algebras over commutative rings.

Proof. It is clear that ${\mathcal{Z}}(A)$ is a finitely generated and projective ${\mathcal{Z}}(\mathbb{I})$ -module. Using the characterization of Azumaya algebras (see for example [Reference DeMeyer and Ingraham7, Reference Kadison16]), we need to check that $\text{End}_{{\mathcal{Z}}(\mathbb{I})}({\mathcal{Z}}(A))$ is isomorphic as a ${\mathcal{Z}}(\mathbb{I})$ -algebra to ${\mathcal{Z}}(A)\otimes _{{\mathcal{Z}}(\mathbb{I})}{\mathcal{Z}}(A)^{o}$ , which is proved as follows. Following the ideas of [Reference Fausk, Lewis and May.9, Proposition 1.2], for any object $X$ in ${\mathcal{M}}$ , we know that ${\mathcal{Z}}(X)\otimes _{{\mathcal{Z}}(\mathbb{I})}{\mathcal{Z}}(\mathbb{I}^{n})\,\cong \,{\mathcal{Z}}(X\otimes _{}\mathbb{I}^{n})$ . Thus, the same isomorphism is inherited by any direct summands of some $\mathbb{I}^{n}$ . Therefore, we have an isomorphism ${\mathcal{Z}}(X)\otimes _{{\mathcal{Z}}(\mathbb{I})}{\mathcal{Z}}(A)\,\cong \,{\mathcal{Z}}(X\otimes _{}A)$ and by symmetry ${\mathcal{Z}}(A)\otimes _{{\mathcal{Z}}(\mathbb{I})}{\mathcal{Z}}(X)\,\cong \,{\mathcal{Z}}(A\otimes _{}X)$ , for any object $X$ in ${\mathcal{M}}$ . In particular, we have

$$\begin{eqnarray}{\mathcal{Z}}(A)\otimes _{{\mathcal{Z}}(\mathbb{I})}{\mathcal{Z}}(A)\cong {\mathcal{Z}}(A\otimes _{}A),\,\text{ and }\,{\mathcal{Z}}(A)\otimes _{{\mathcal{Z}}(\mathbb{I})}{\mathcal{Z}}([A,\mathbb{I}])\cong {\mathcal{Z}}(A\otimes _{}[A,\mathbb{I}]).\end{eqnarray}$$

Therefore, ${\mathcal{Z}}([A,\mathbb{I}])\cong {\mathcal{Z}}(A)^{\ast }$ as ${\mathcal{Z}}(\mathbb{I})$ -modules, where ${\mathcal{Z}}(A)^{\ast }$ is the ${\mathcal{Z}}(\mathbb{I})$ -linear dual of ${\mathcal{Z}}(A)$ , since $[A,\mathbb{I}]$ is a dual object of $A$ in ${\mathcal{M}}$ . By Remark 4.4(i), we have a chain of isomorphisms $A^{e}\cong [A,A]\cong A\otimes _{}[A,\mathbb{I}]$ , from which we deduce the following isomorphisms

$$\begin{eqnarray}\displaystyle {\mathcal{Z}}(A)\otimes _{{\mathcal{Z}}(\mathbb{I})}{\mathcal{Z}}(A)^{o} & \cong & \displaystyle {\mathcal{Z}}(A\otimes _{}A^{o})\,\cong \,{\mathcal{Z}}([A,A])\,\cong \,{\mathcal{Z}}(A)\otimes _{{\mathcal{Z}}(\mathbb{I})}{\mathcal{Z}}(A)^{\ast }\nonumber\\ \displaystyle & \cong & \displaystyle \text{End}_{{\mathcal{Z}}(\mathbb{I})}({\mathcal{Z}}(A)),\nonumber\end{eqnarray}$$

whose composition leads to the desired ${\mathcal{Z}}(\mathbb{I})$ -algebra isomorphism.◻

Proof. The functor $A\otimes _{}-:{\mathcal{M}}\rightarrow {\mathcal{M}}$ is clearly the composition of the functor $A\otimes _{}-:{\mathcal{M}}\rightarrow \text{}_{A^{e}}{\mathcal{M}}$ with the forgetful functor $\mathscr{O}:\text{}_{A^{e}}{\mathcal{M}}\rightarrow {\mathcal{M}}$ , so that it is left exact. Thus $A$ is a flat object.

We know that the unit and the counit

$$\begin{eqnarray}\boldsymbol{\unicode[STIX]{x1D700}}_{M}:A\otimes _{\,A^{e}}[A,M]\longrightarrow M,\qquad \boldsymbol{\unicode[STIX]{x1D701}}_{X}:X\longrightarrow _{A^{e}}[A,A\otimes _{}X],\end{eqnarray}$$

of the adjunction are natural isomorphisms, for every pair ofobjects $(X,M)$ in ${\mathcal{M}}\times _{A^{e}}{\mathcal{M}}$ . Thus, $\boldsymbol{\unicode[STIX]{x1D701}}_{\mathbb{I}}:\mathbb{I}\rightarrow _{A^{e}}[A,A]$ is an isomorphism, and so $A$ is central. The retraction of $u$ is given by the following dashed arrow

◻