Hopf Quasigroup Galois Extensions and a Morita Equivalence

Abstract

For H, a Hopf coquasigroup, and A, a left quasi-H-module algebra, we show that the smash product $A\#H$ is linked to the algebra of H invariants $A^H$ by a Morita context. We use the Morita setting to prove that for finite dimensional H, there are equivalent conditions for $A/A^H$ to be Galois parallel in the case of H finite dimensional Hopf algebra.

1. Introduction

The notion of Hopf algebra was given by Borel in 1953 (see [1]), honoring the foundational work of Heinz Hopf in algebraic topology. Nowadays, progress obtained in understanding the structure of Hopf algebras and their representations have been entwined with the development of different areas of mathematics such as knot theory [2], topology, conformal field theory, algebraic geometry, ring theory [3], category theory (see, [4]), combinatorics, etc. More generally, a Hopf quasigroup (non-associative algebra) was introduced by Klim and Majid [5], whose dual notion is a Hopf coquasigroup, a particular case of the notion of unital coassociative bialgebra introduced in [6]. For more research on these aspects, see the papers [7,8,9]. For some basic and recent papers related to non-associative BCC-algebras and B-filters in this field, we refer to [10,11,12].

In the paper [13], the authors constructed the relevance between an algebra A and its subalgebra of invariants $A^H$ under the action of a finite-dimensional Hopf algebra H by using the smash product $A\#H$ of A by H. Furthermore, they studied the conception of Hopf Galois extension $A/A^H$ and gave some equivalent conditions.

It is now natural to ask whether the results in [13] hold in the framework of Hopf coquasigroups. This becomes our motivation of writing this paper. We will overcome non-associativity in Hopf quasigroups and non-coassociativity in Hopf coquasigroups by introducing some new notions.

Actually, in this paper, we introduce some several actions of Hopf coquasigroup H on an algebra A and consider the existence of a Morita context relating $A^H$ and $A♯H$. Then, we introduce the definition of a Hopf quasigroup Galois extension and apply the Morita setting to prove that for finite dimensional H, there are equivalent conditions for $A/A^H$ to be Galois. Finally, we consider a special case of a quasigroup.

This paper is organized as follows.

In Section 2 of this paper, the notions of a Hopf (co)quasigroups, a smash product and an integral are recalled. Meanwhile, we introduce and study the notion of quasigroup-graded algebras.

In Section 3 of this paper, we mainly construct a Morita context linking $A^H$ and $A\#H$ by the connecting bimodules being both A (see Theorem 1).

In Section 4, we introduce the notion of a Hopf quasigroup Galois extension and study the surjectivity of the Morita context map (see Theorems 5 and 6), and in the final Section 5, we treat a special case of a quasigroup (see Theorem 7).

Throughout this paper, unless stated otherwise, k is a field, and all vector spaces are over k. By linear maps, we means k-linear maps. Unadorned ⊗ means ${\otimes }_k$. We make use of the Sweedler’s notation for the comultiplication, $△\left(h\right)=h_{\left(1\right)}\otimes h_{\left(2\right)}$, where the summation sign is suppressed (see [1]).

2. Preliminaries

We will investigate the notion of a quasigroup graded algebra and mainly recall some basic definitions and properties of Hopf (co)quasigroups and smash products.

2.1. Quasigroup Graded Algebras

Recall from [14] that a quasigroup is a non-empty set G with a multiplication and an identity $1_G$ such that for any $g\in G$ there is $g^{-1}\in G$ satisfying

$$\left(hg\right)g^{-1}=h,\mathrm{and}{g}^{-1}\left(gh\right)=h$$

for all $h\in G$. We note that in any quasigroup G, we have unique inverses and

$${\left(g^{-1}\right)}^{-1}=g,\mathrm{and}{\left(gh\right)}^{-1}=h^{-1}{g}^{-1}$$

for all $g,h\in G$. An associative quasigroup is a group.

By an algebra A, we always assume it to be an associative algebra with identity $1=1_A$. We fix a multiplicative quasigroup G with identity $1=1_G$.

Definition 1.

A G-graded algebra A is an algebra, also denoted by A, together with a direct sum decomposition:

$$A=\underset{\sigma \in G}{⨁}{A}_{\sigma }\left(asadditivegroups\right)$$

(1)

into additive subgroups $A_{\sigma },\sigma \in G$, such that:

$$A_{\sigma }{A}_{\tau }\subseteq A_{\sigma \tau },forall\sigma ,\tau \in G.$$

(2)

We call Equation (1) the G-grading of A and $A_{\sigma }$ the σ-component of A for any $\sigma \in G$.

We say that A is a strongly G-graded algebra if

$$A_{\sigma }{A}_{\tau }=A_{\sigma \tau },forall\sigma ,\tau \in G.$$

(3)

Example 1.

(1) 

Every algebra A is a trivially graded algebra by letting $A_1$ = A and $A_{\sigma }$ = 0 for all $\sigma \ne 1_G$.

(2) 

Consider the  quasigroup algebra  $k\left(Q\right)$ of some quasigroup Q with coefficients in k, which is a k-vector space with basis Q and with multiplication defined distributively using the given multiplication of Q. In other words, for the latter, one has

$$(\sum _{x\in Q}{a}_{x}x)(\sum _{y\in Q}{b}_{y}y)=\sum _{x,y\in Q}\left(a_x{b}_y\right)\left(xy\right)=\sum _{z\in Q}{c}_{z}z$$

where $c_z={\sum }_{xy=z}{a}_x{b}_y={\sum }_{x\in Q}{a}_x{b}_{x^{-1}z}$. Certainly, the non-associative law in Q guarantees the non-associativity of multiplication in $k\left(Q\right)$, so $k\left(Q\right)$ is a non-associative algebra.

Any epimorphism $\pi$ of Q onto a quasigroup G turns $k\left(Q\right)$ into a strongly G-graded algebra with the $\sigma$-components:

$$k{\left(Q\right)}_{\sigma }=\underset{\rho \in {\pi }^{-1}\left(\sigma \right)}{⨁}k\rho ,\mathrm{for}\mathrm{all}\sigma \in G.$$

Proposition 1.

Let A be a G-graded algebra. Then

(1) 

The $1_G$-component $A_1$ is a subalgebra of A.

(2) 

Each σ-component $A_{\sigma }$, $\sigma \in G$, is a two-sided $A_1$-submodule of A.

(3) 

The subalgebra $A_1$ contains the identity $1_A$.

Proof. 

(1) and (2) follow immediately from Equation (2).

(3) By Equation (1), write $1_A={\displaystyle \sum _{\sigma \in G}}{a}_{\sigma }$, where each $a_{\sigma }\in A_{\sigma }$ and all but a finite number of $a_{\sigma }$’s are zero. Then, fixing some $\tau \in G$, we have $a_{\tau }=1_A·a_{\tau }={\displaystyle \sum _{\sigma \in G}}{a}_{\sigma }{a}_{\tau }$.

By comparing degrees, we see that $a_{\tau }=a_1{a}_{\tau }\in A_{\tau }$, for every $\tau$, for $\sigma \ne 1_G$ must be zero. Thus,

$$a_1=a_1·1_A=\sum _{\sigma \in G}{a}_1{a}_{\sigma }=\sum _{\sigma \in G}{a}_{\sigma }=1_A\in A_1.$$

This finishes the proof. □

Proposition 2.

A G-graded algebra A is strongly G-graded if and only if $1_A\in A_{\sigma }{A}_{{\sigma }^{-1}},\forall \sigma \in G$.

Proof. 

If A is strongly G-graded, then $A_{\sigma }{A}_{{\sigma }^{-1}}=A_1$, and Proposition 1(3) implies that $1_A\in A_{\sigma }{A}_{{\sigma }^{-1}}$.

Conversely, if $1_A\in A_{\sigma }{A}_{{\sigma }^{-1}}$, then $A_{\sigma \tau }=1_A{A}_{\sigma \tau }\subseteq \left(A_{\sigma }{A}_{{\sigma }^{-1}}\right)A_{\sigma \tau }\subseteq A_{\sigma }\left(A_{{\sigma }^{-1}}{A}_{\sigma \tau }\right)\subseteq A_{\sigma }{A}_{{\sigma }^{-1}\left(\sigma \tau \right)}=A_{\sigma }{A}_{\tau }\subseteq A_{\sigma \tau },\forall \sigma ,\tau \in G$, which shows that $A_{\sigma }{A}_{\tau }=A_{\sigma \tau }$.

This completes the proof. □

Proposition 3.

A is a G-graded algebra if and only if there exists a map $\beta :G⟶End\left(A\right)$ such that the following conditions hold:

(1) ${\beta }_g\circ {\beta }_h=0$ if $g\ne h$; ${\beta }_g\circ {\beta }_h={\beta }_g$, if $g=h$,

(2) ${\displaystyle \sum _{g\in G}}{\beta }_g=Id$,

(3) For each $g\in G,a,b\in A$, ${\beta }_g\left(ab\right)={\displaystyle \sum _{uv=g}}{\beta }_u\left(a\right){\beta }_v\left(b\right)$.

Proof. 

$⟹)$ Assume that A is graded by G, and an element $a\in A$ has a unique decomposition as $a={\displaystyle \sum _{\sigma \in G}}{a}_{\sigma }$ with $a_{\sigma }\in A_{\sigma }$. Then, let ${\beta }_g\in$ End(A) defined by ${\beta }_g({\displaystyle \sum _{\sigma \in G}}{a}_{\sigma })=a_g$ for all $g\in G$. It is easy to see that ${\beta }_g$ is well-defined.

$({\beta }_g\circ {\beta }_h)({\displaystyle \sum _{\sigma \in G}}{a}_{\sigma })={\beta }_g\left(a_h\right)$, if $g\ne h$, then ${\beta }_g\left(a_h\right)$ = 0, if $g=h$, then ${\beta }_g\left(a_h\right)=a_g$, (1) is satisfied; $({\displaystyle \sum _{g\in G}}{\beta }_g)({\displaystyle \sum _{\sigma \in G}}{a}_{\sigma })={\displaystyle \sum _{g\in G}}{a}_g=a$, (2) is satisfied; $\forall g\in G,a={\displaystyle \sum _{\sigma \in G}}{a}_{\sigma }\in A,b={\displaystyle \sum _{\tau \in G}}{b}_{\tau }\in A$, we have ${\beta }_g\left(ab\right)={\beta }_g({\displaystyle \sum _{\sigma \in G}}{a}_{\sigma }{\displaystyle \sum _{\tau \in G}}{b}_{\tau })={\beta }_g({\displaystyle \sum _{\sigma \in G}}{\displaystyle \sum _{\tau \in G}}{a}_{\sigma }{b}_{\tau })={\displaystyle \sum _{uv=g}}{a}_u{b}_v={\displaystyle \sum _{uv=g}}{\beta }_u\left(a\right){\beta }_v\left(b\right)$, (3) is satisfied.

$⟸)$ Assume that there exists a map $\beta :G⟶End\left(A\right)$ satisfying (1), (2) and (3); we will show that A is a G-graded algebra. We point out that $A_g={\beta }_g\left(A\right)$; in fact, $A={\displaystyle \sum _{g\in G}}{A}_g$ by (2), and the sum is direct by (1), so formula Equation (1) is established. As for $A_u{A}_v\subseteq A_{uv}$, for all $u,v\in G$, (3) is obvious.

The proof is complete. □

2.2. Hopf (co)quasigroups

Recall from [5] that a Hopf coquasigroup H consists of the following datum:

(1)

H is an associative algebra with an identity 1;

(2)

There are two counital algebra homomorphisms $\Delta :H⟶H\otimes H$ and $\epsilon :H⟶k$;

(3)

There is a linear map $S:H⟶H$ such that

$$\begin{array}{ccc}& & \sum S\left(a_{\left(1\right)}\right)a_{\left(2\right)\left(1\right)}\otimes a_{\left(2\right)\left(2\right)}=1\otimes a=\sum a_{\left(1\right)}S\left(a_{\left(2\right)\left(1\right)}\right)\otimes a_{\left(2\right)\left(2\right)},\hfill \end{array}$$

(4)

$$\begin{array}{ccc}& & \sum a_{\left(1\right)\left(1\right)}\otimes S\left(a_{\left(1\right)\left(2\right)}\right)a_{\left(2\right)}=a\otimes 1=\sum a_{\left(1\right)\left(1\right)}\otimes a_{\left(1\right)\left(2\right)}S\left(a_{\left(2\right)}\right)\hfill \end{array}$$

(5)

for $a\in H$.

Dually, a unital counital coassociative bialgebra $(H,\Delta ,\epsilon )$ is called a Hopf quasigroup if there is a linear map $S:H⟶H$ such that

$$\sum S\left(a_1\right)\left(a_{2}b\right)=\epsilon \left(a\right)b=\sum a_1\left(S\left(a_2\right)b\right),\sum \left(a{b}_1\right)S\left(b_2\right)=a\epsilon \left(b\right)=\sum \left(aS\left(b_1\right)\right)b_2$$

(6)

for any $a,b\in H$.

Some properties of S are used in later proofs as S is an anti (co)algebra map. For them, we refer to [5]. A Hopf (co)quasigroup is the usual Hopf algebra if and only if its (co)product is (co)associative.

Let G be a quasigroup. Then, quasigroup algebra $H=kG$ is a Hopf quasigroup with $\Delta \left(h\right)=h\otimes h$, $\epsilon \left(h\right)=1$ and $S\left(h\right)=h^{-1}$ for $h\in G$ (see [5]). If G is finite, then the linear dual ${\left(kG\right)}^{*}$ of $kG$ is a Hopf coquasigroup.

If H is a finite dimensional Hopf coquasigroup, its linear dual $H^{*}$ is not a Hopf coquasigroup but a Hopf quasigroup, and one has a non-degenerate bilinear form $⟨,⟩:H^{*}\times H⟶k$ given by $⟨h^{*},h⟩=h^{*}\left(h\right)$ for all $h^{*}\in H^{*}$ and $h\in H$. Then, the left action of $H^{*}$ on H is given by $h^{*}\rightharpoonup h=\sum ⟨h^{*},h_{\left(2\right)}⟩h_{\left(1\right)}$ for any $h^{*}\in H^{*}$ and $h\in H$. Similarly, the right action of $H^{*}$ on H is given by $h\leftharpoonup h^{*}=\sum ⟨h^{*},h_{\left(1\right)}⟩h_{\left(2\right)}$ for any $h^{*}\in H^{*}$ and $h\in H$.

2.3. Quasi-Module Algebras and Comodule Algebras

Let H be a Hopf coquasigroup. Recall from [15] that a (associative and unital) algebra A is called a left quasi-H-module algebra if A is a left H-module such that for all $a,b\in A$,

$$\begin{array}{ccc}& & h_{\left(1\right)}·\left(ab\right)\otimes h_{\left(2\right)}=(h_{\left(1\right)}·a)(h_{\left(2\right)\left(1\right)}·b)\otimes h_{\left(2\right)\left(2\right)}\hfill \end{array}$$

(7)

$$\begin{array}{ccc}& & h·1_A=\epsilon \left(h\right)1_A.\hfill \end{array}$$

(8)

Let H be a Hopf coquasigroup and A a left quasi-H-module algebra. The smash product of A by H, denoted by $A♯H$, is defined as the tensor product $A\otimes H$, with the multiplication given by

$$\begin{array}{c}\hfill (a♯h)(b♯k)=a(h_{\left(1\right)}·b)♯h_{\left(2\right)}k\end{array}$$

(9)

where $a,b\in A$ and $h,k\in H$.

Remark 1.

(1) 

It can be observed by direct calculations that $A♯H$ is an associative algebra with $1_A♯1_H$, and $A♯H=(A♯1_H)(1_A♯H)$; therefore, we sometimes identify $A♯1_H$ with A and identify $1_A♯H$ with H.

(2) 

If we apply $I_A\otimes \epsilon$ to Equation (7) on both sides, we have

$$\begin{array}{c}\hfill h·\left(ab\right)=(h_{\left(1\right)}·a)(h_{\left(2\right)}·b)\end{array}$$

(10)

This is in line with the usual definition of left H-module algebra when H is Hopf algebra. However, if we replace Equation (7) with Equation (10), $A♯H$ is not always associative.

Assume H is a Hopf coquasigroup and M is a left H-module; then, we denote $M^H$ by the set of H invariants under the modular action, that is,

$$M^H=\{m\in M\mid h·m=\epsilon \left(h\right)m,\forall h\in H\}.$$

It is easy to see that $M^H$ is the submodule of M. When A is a left quasi-H-module algebra, $A^H$ is a subalgebra of A, which is analogous to the classical situation.

Recall from [16] that the notion of comodule algebra is given by the following.

Given a Hopf quasigroup H, an algebra A is a right H-comodule algebra if there is a map $\rho :A⟶A\otimes H,\rho \left(a\right)=a_{\left[0\right]}\otimes a_{\left[1\right]}$, such that for all $a,b\in A$, the following conditions hold

$$\begin{array}{ccc}& & a_{\left[0\right]\left[0\right]}\otimes a_{\left[0\right]\left[1\right]}\otimes a_{\left[1\right]}=a_{\left[0\right]}\otimes a_{\left[1\right]\left(1\right)}\otimes a_{\left[1\right]\left(2\right)},\hfill \end{array}$$

(11)

$$\begin{array}{ccc}& & a_{\left[0\right]}\epsilon \left(a_{\left[1\right]}\right)=a,\hfill \end{array}$$

(12)

$$\begin{array}{ccc}& & \rho \left(ab\right)=a_{\left[0\right]}{b}_{\left[0\right]}\otimes a_{\left[1\right]}{b}_{\left[1\right]},\hfill \end{array}$$

(13)

$$\begin{array}{ccc}& & \rho \left(1_A\right)=1_A\otimes 1_H.\hfill \end{array}$$

(14)

We denote $A^{coH}$ by the set of H-coinvariants under the comodular action, that is,

$$A^{coH}=\{a\in A\mid \rho \left(a\right)=a\otimes 1,\forall a\in A\}.$$

If A is an H-comodule algebra, then $A^{coH}$ is a subalgebra of A.

Proposition 4.

Let A be a left quasi-H-module algebra, and consider A as a left or right $A♯H$-module. Then:

(1) 

${\left(A^H\right)}^{op}\cong End{(}_{A♯H}A)$,

(2) 

If S is bijective, then also $A^H\cong End\left(A_{A♯H}\right)$.

Proof. 

Let $L_a$ (resp. $R_a):A⟶A$ denote left (resp. right) multiplication by a.

(1)

Define $\varphi :{\left(A^H\right)}^{op}⟶End{(}_{A♯H}A)$ by $\varphi \left(a\right)=R_a$, for all $a\in A^H$. Clearly $\varphi$ is injective. Now, given any $\sigma \in End{(}_{A♯H}A)$ and $b\in A$, we have $\sigma \left(b\right)=\sigma ((b♯1_H)⊳1_A)=(b♯1_H)⊳\sigma \left(1_A\right)=b\sigma \left(1_A\right)$, so $\sigma =R_{\sigma \left(1_A\right)}$. Moreover, $a=\sigma \left(1_A\right)\in A^H$. For, $\forall h\in H$, $h·\sigma \left(1_A\right)=(1_A♯h)⊳\sigma \left(1_A\right)=\sigma ((1_A♯h)⊳1_A)=\sigma (h·1_A)=\epsilon \left(h\right)\sigma \left(1_A\right)$.

For all $a,a^{\prime }\in A^H,b\in A$, $\varphi \left(a{a}^{\prime }\right)\left(b\right)=R_{a{a}^{\prime }}\left(b\right)=b\left(a{a}^{\prime }\right)=\varphi \left(a^{\prime }\right)\left(ba\right)=(\varphi \left(a^{\prime }\right)\circ \varphi \left(a\right))\left(b\right)$, that is, $\varphi$ is an algebra homomorphism.

(2)

The process is similar although slightly more complicated. We define $\psi :A^H⟶End\left(A_{A♯H}\right)$, by $a⟼L_a,\forall a\in A^H$. It is easy to see that $L_a$ is a right $A♯H$-homomorphism, for $a\in A^H$: if $b\in A,c♯h\in A♯H$, then $L_a\left(b\right)⊲(c♯h)=\left(ab\right)⊲(c♯h)=S^{-1}\left(h\right)·\left(abc\right)=(S^{-1}{\left(h\right)}_{\left(1\right)}·a)(S^{-1}{\left(h\right)}_{\left(2\right)}·\left(bc\right))=(S^{-1}\left(h_{\left(2\right)}\right)·a)(S^{-1}\left(h_{\left(1\right)}\right)·\left(bc\right))=\left(\epsilon \left(h_{\left(2\right)}\right)a\right)(S^{-1}\left(h_{\left(1\right)}\right)·\left(bc\right))=a(S^{-1}\left(h\right)·\left(bc\right))=L_a(b⊲(c♯h))$.

Analogous to (1), $\psi$ is bijective and is an algebra homomorphism. □

It is straightforward to obtain the following:

Proposition 5.

Let H be a finite dimensional Hopf coquasigroup. If A is a right $H^{*}$-comodule algebra, then A is a left quasi-H-module algebra via $h·a=⟨a_{\left[1\right]},h⟩a_{\left[0\right]}$ for any $a\in A$ and $h\in H$. Conversely, if A is a left quasi-H-module algebra and $\{h_i,h_i^{*}\}$ are dual bases for H and $H^{*}$, then A becomes a right $H^{*}$-comodule algebra via $\rho \left(a\right)={\sum }_i(h_i·a)\otimes h_i^{*}\in A\otimes H^{*}$ for any $a\in A$.

Proposition 6.

Let H be a finite dimensional Hopf coquasigroup and A a left quasi-H-module algebra. Then $A^H=A^{co{H}^{*}}$.

2.4. Integrals

The notion of an integral as a very important research content in classical Hopf algebra has also been extended to Hopf (co)quasigroup (see, [17]). Here, we give the following slight different definition from the one given in [17] (see below Remark 2).

Definition 2.

Let H be a Hopf coquasigroup. A left integral in H is an element $t\in H$ such that,

$$ht=\epsilon \left(h\right)t,\forall h\in H.$$

We denote the space of left integrals in H by ${\int }_H^{\ell }$.

Remark 2.

(1) 

Let H be a finite dimensional vector space. If H is a Hopf quasigroup, then $H^{*}$ is a Hopf coquasigroup with natural structure induced by H. Conversely, if $H^{*}$ is a Hopf coquasigroup, then $H\cong {\left(H^{*}\right)}^{*}$ is a Hopf quasigroup.

(2) 

Although the definition given by Klim does not seem to be quite the same as Definition 2 formally, they both are consistent by ([17], Lemma 3.2); our definition here is closer to the classical form for Hopf algebras.

(3) 

If H is a finite-dimensional Hopf coquasigroup, then a left integral exists and is unique up to a scalar, i.e., $dim\left({\int }_H^{\ell }\right)=1$.

(4) 

If H is a Hopf coquasigroup, $T\in {\int }_{H^{*}}^r$, then for all $h,g\in H$,

$$\begin{array}{c}\hfill T\left(S\left(h\right)g_{\left(1\right)}\right)g_{\left(2\right)}=h_{\left(1\right)}T\left(S\left(h_{\left(2\right)}\right)g\right).\end{array}$$

(15)

In fact, for this, we have

$$\begin{array}{cc}\hfill T\left(S\left(h\right)g_{\left(1\right)}\right)g_{\left(2\right)}& =T\left(S\left(h_{\left(2\right)\left(2\right)}\right)g_{\left(1\right)}\right)h_{\left(1\right)}S\left(h_{\left(2\right)\left(1\right)}\right)g_{\left(2\right)}\hfill \\ \hfill & =T\left(S{\left(h_{\left(2\right)}g\right)}_{\left(1\right)}\right)h_{\left(1\right)}S{\left(h_{\left(2\right)}g\right)}_{\left(2\right)}\hfill \\ & =T\left(S\left(h_{\left(2\right)}\right)g\right)h_{\left(1\right)}.\hfill \end{array}$$

3. The Morita Context

We will establish a Morita context linking $A♯H$ and $A^H$ for A our left quasi-H-module algebra.

Lemma 1.

Let H be a Hopf coquasigroup and A a left quasi-H-module algebra. Then, A is a left or right $A^H$-module via left or right multiplication.

Proposition 7.

Let H be a Hopf coquasigroup and A a left quasi-H-module algebra. Then,

(i) 

A is a left $A♯H$-module, via $(a♯h)⊳b=a(h·b)$, for any $a,b\in A$ and $h\in H$.

(ii) 

If H has a bijective antipode S, then A is a right $A♯H$-module, via $b⊲(a♯h)=S^{-1}\left(h\right)·\left(ba\right)$, for any $a,b\in A$ and $h\in H$.

Proof. 

(i)

By calculations, we have $(1_A♯1_H)⊳b=1_A(1_H·b)=b$ and

$$\begin{array}{cc}\hfill (a^{\prime }♯h^{\prime })⊳((a♯h)⊳b)& =(a^{\prime }♯h^{\prime })⊳\left(a(h·b)\right)\hfill \\ & =a^{\prime }(h^{\prime }·\left(a(h·b)\right))\hfill \\ \hfill & =a^{\prime }\left((h_{\left(1\right)}^{\prime }·a)(h_{\left(2\right)}^{\prime }·(h·b))\right)\hfill \\ & =a^{\prime }\left((h_{\left(1\right)}^{\prime }·a)(\left(h_{\left(2\right)}^{\prime }h\right)·b)\right)\hfill \\ & =(a^{\prime }(h_{\left(1\right)}^{\prime }·a)♯\left(h_{\left(2\right)}^{\prime }h\right))⊳b\hfill \\ & =\left((a^{\prime }♯h^{\prime })(a♯h)\right)⊳b\hfill \end{array}$$

for all $a,a^{\prime },b\in A,h,h^{\prime }\in H$.

(ii)

The process for the part (ii) is generally similar, but it is slightly more complicated because of the non-coassociativity. When the antipode S of H is bijective, we have for all $a,a^{\prime },b\in A$ and $h,h^{\prime }\in H$,

$$b⊲(1_A♯1_H)=S^{-1}\left(1_H\right)·\left(b{1}_A\right)=b.$$

In addition,

$$\begin{array}{cc}\hfill (b⊲(a♯h))⊲(a^{\prime }♯h^{\prime })& =(S^{-1}\left(h\right)·\left(ba\right))⊲(a^{\prime }♯h^{\prime })\hfill \\ & =S^{-1}\left(h^{\prime }\right)·\left((S^{-1}\left(h\right)·\left(ba\right))a^{\prime }\right)\hfill \\ \hfill & =(S^{-1}{\left(h^{\prime }\right)}_{\left(1\right)}·(S^{-1}\left(h\right)·\left(ba\right)))(S^{-1}{\left(h^{\prime }\right)}_{\left(2\right)}·a^{\prime })\hfill \\ & =(S^{-1}\left(h_{\left(2\right)}^{\prime }\right)·(S^{-1}\left(h\right)·\left(ba\right)))(S^{-1}\left(h_{\left(1\right)}^{\prime }\right)·a^{\prime })\hfill \\ & =(S^{-1}\left(h{h}_{\left(2\right)}^{\prime }\right)·\left(ba\right)))(S^{-1}\left(h_{\left(1\right)}^{\prime }\right)·a^{\prime }),\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill b⊲\left((a♯h)(a^{\prime }♯h^{\prime })\right)& =b⊲(a(h_{\left(1\right)}·a^{\prime })♯h_{\left(2\right)}{h}^{\prime })\hfill \\ & =S^{-1}\left(h_{\left(2\right)}{h}^{\prime }\right)·\left(ba(h_{\left(1\right)}·a^{\prime })\right)\hfill \\ \hfill & =(S^{-1}{\left(h_{\left(2\right)}{h}^{\prime }\right)}_{\left(1\right)}·\left(ba\right))(S^{-1}{\left(h_{\left(2\right)}{h}^{\prime }\right)}_{\left(2\right)}·(h_{\left(1\right)}·a^{\prime }))\hfill \\ & =(S^{-1}\left(h_{\left(2\right)\left(2\right)}{h}_{\left(2\right)}^{\prime }\right)·\left(ba\right))(S^{-1}\left(h_{\left(2\right)\left(1\right)}{h}_{\left(1\right)}^{\prime }\right)·(h_{\left(1\right)}·a^{\prime }))\hfill \\ & =(S^{-1}\left(h_{\left(2\right)\left(2\right)}{h}_{\left(2\right)}^{\prime }\right)·\left(ba\right))(\left(S^{-1}\left(h_{\left(1\right)}^{\prime }\right)S^{-1}\left(h_{\left(2\right)\left(1\right)}\right)h_{\left(1\right)}\right)·a^{\prime })\hfill \\ & =(S^{-1}\left(h{h}_{\left(2\right)}^{\prime }\right)·\left(ba\right)))(S^{-1}\left(h_{\left(1\right)}^{\prime }\right)·a^{\prime })\hfill \end{array}$$

where Equation (4) was used to obtain the last equality. □

From now until the end of this section, we will assume that H is a finite dimensional Hopf coquasigroup. Let us here make some comments on integrals for the Hopf coquasigroup, which will be helpful for our later discussion. By a duality to ([17], Corollary 3.6), we conclude that if H is a finite dimensional Hopf coquasigroup, then the antipode S is always bijective.

We will use the following lemmas later.

Lemma 2.

Let V be a n-dimensional vector space with a base of $\{v_1,v_2,\cdots ,v_n\}$, and $k_1,k_2,\cdots ,k_n$ be any n scalars. Then, there exists a unique linear function λ on V such that $\lambda \left(v_i\right)=k_i$ for all $i=1,\cdots ,n$.

Lemma 3

([18]). Let A be an algebra. Then,

(i) 

$G\left(A^{\circ }\right)=Alg(A,k)$, where $Alg(A,k)$ denotes the set of algebra homomorphisms from A to k.

(ii) 

$Alg(A,k)$ is a linearly independent subset of $A^{*}$.

Now, choose a non-zero $t\in {\int }_H^{\ell }$, also $th\in {\int }_H^{\ell }$ for any $h\in H$. Since $dim({\int }_H^{\ell })=1$, it follows from Lemma 3 that $th=\lambda \left(h\right)t$, for some $\lambda \in H^{*}$. Moreover, clearly, $\lambda$ is an algebra homomorphism from A to k, and so by Lemma 4, it is a group-like element of $H^{*}$. We note that $\lambda$ is determined by the above equation and that $\lambda$ does not depend on the choice of t. Similar to the classical one, the element $\lambda$ of $G\left(H^{*}\right)$ constructed above is called an H-distinguished grouplike element of $H^{*}$.

By means of the above $\lambda$, we can define a map from H to itself as follows:

$$\begin{array}{c}\hfill f:H⟶H,h\mapsto h^{\lambda }=\lambda \rightharpoonup h=h_{\left(1\right)}\lambda \left(h_{\left(2\right)}\right),\end{array}$$

(16)

for all $h\in H$.

Proposition 8.

With the above f.

(i) 

We have that f is an algebraic automorphism on H.

(ii) 

We can extend f to an automorphism of $A♯H$ by the following formula:

$$\begin{array}{c}\hfill {(a♯h)}^{\lambda }=a♯h^{\lambda }.\end{array}$$

(17)

Proof. 

(i) The algebraic endomorphism is easy. Define a map

$$g:H⟶H,h\mapsto h_{\left(1\right)}\lambda \left(S\left(h_{\left(2\right)}\right)\right),$$

for all $h\in H$. By computing, one has

$$(f\circ g)\left(h\right)=f\left(h_{\left(1\right)}\lambda \left(S\left(h_{\left(2\right)}\right)\right)\right)=h_{\left(1\right)\left(1\right)}\lambda \left(h_{\left(1\right)\left(2\right)}\right)\lambda \left(S\left(h_{\left(2\right)}\right)\right)=h_{\left(1\right)\left(1\right)}\lambda \left(h_{\left(1\right)\left(2\right)}S\left(h_{\left(2\right)}\right)\right)=h,$$

where Equation (5) was used in the last equation. Similarly, $g\circ f=i{d}_H$.

(ii) Straightforward. □

In order to construct a Morita context better, we now define a new right action of $A♯H$ on A using Equation (17). For any $b\in A$ and $a♯h\in A♯H$, set

$$\begin{array}{c}\hfill b◂(a♯h):=b⊲(a♯h^{\lambda })=S^{-1}\left(h^{\lambda }\right)·\left(ba\right).\end{array}$$

(18)

Notice that if $0\ne t\in {\int }_H^{\ell }$, we have the following linear isomorphism

$$\phi :H^{*}⟶H,h^{*}⟼t\leftharpoonup h^{*}=h^{*}\left(t_{\left(1\right)}\right)t_{\left(2\right)}.$$

Then, if $0\ne T\in {\int }_{H^{*}}^r$, it follows from ([17], Lemma 4.2) that $T\left(t\right)1_H=T\left(t_{\left(1\right)}\right)t_{\left(2\right)}=\phi \left(T\right)\ne 0$, and consequently, $T\left(t\right)$ is distinct of zero. Thus, it always possible to choose some $t\in {\int }_H^{\ell }$ and $T\in {\int }_{H^{*}}^r$ satisfying $T\left(t\right)=1$.

Consider the right action of H on $H^{*}$ defined by

$$(h^{*}\prec h)\left(g\right)=h^{*}\left(hg\right),\forall h,g\in H,h^{*}\in H^{*}.$$

It is easy to verify that $H^{*}$ is a right H-module under ≺. It should be emphasized that in the previous discussion, H is not a right $H^{*}$-module, because when H is a Hopf coquasigroup, $H^{*}$ is a Hopf quasigroup, and it is non-associative as algebra. Dually, if $0\ne T\in {\int }_{H^{*}}^r$, we also have the following linear isomorphism

$$\psi :H⟶H^{*},h⟼T\prec S\left(h\right)=T_{\left(1\right)}\left(S\left(h\right)\right)T_{\left(2\right)},\forall h\in H,$$

which is called a Fourier transformation in H defined by Klim in [17].

Combining the above two situations, we have

Proposition 9.

For $t\in {\int }_H^{\ell }$ with $\epsilon \left(t\right)\ne 0$ and $T\in {\int }_{H^{*}}^r$, we have $T\left(t\right)=T\left(S\right(t\left)\right)$.

Proof. 

On the one hand, since t is a left integral in H, we have

$$T\left(S\right(t\left)t\right)=T\left(\epsilon \right(S\left(t\right)\left)t\right)=\epsilon \left(t\right)T\left(t\right).$$

Notice that $S\left(t\right)$ is a right integral in H; on the other hand, we have

$$T\left(S\right(t\left)t\right)=T\left(\epsilon \right(t\left)S\right(t\left)\right)=\epsilon \left(t\right)T\left(S\right(t\left)\right).$$

This completes the proof. □

Proposition 10.

Let φ, ψ, t and T be as above. If $T\left(t\right)=1$, then

$$\begin{array}{c}\hfill t\leftharpoonup (T\prec S(h\left)\right)=h,\end{array}$$

(19)

$$\begin{array}{c}\hfill T\prec S(t\leftharpoonup h^{*})=h^{*}.\end{array}$$

(20)

In particular, $t\leftharpoonup T=1$ and $T\prec S\left(t\right)=\epsilon$.

Proof. 

For all $h\in H,h^{*}\in H^{*}$, we have

$$\begin{array}{cc}\hfill (\phi \circ \psi )\left(h\right)& =t\leftharpoonup (T\prec S(h\left)\right)\hfill \\ \hfill & =t\leftharpoonup \left(T_{\left(1\right)}\left(S\left(h\right)\right)T_{\left(2\right)}\right)\hfill \\ \hfill & =T_{\left(1\right)}\left(S\left(h\right)\right)T_{\left(2\right)}\left(t_{\left(1\right)}\right)t_{\left(2\right)}\hfill \\ \hfill & =T\left(S\left(h\right)t_{\left(1\right)}\right)t_{\left(2\right)}\hfill \\ \hfill & \stackrel{\left(1.15\right)}{=}{h}_{\left(1\right)}T\left(S\left(h_{\left(2\right)}\right)t\right)=T\left(t\right)h=h,\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill (\psi \circ \phi )\left(h^{*}\right)\left(h\right)& =(T\prec S(t\leftharpoonup h^{*}))\left(h\right)\hfill \\ \hfill & =(T\prec S\left(h^{*}\left(t_{\left(1\right)}\right)t_{\left(2\right)}\right))\left(h\right)\hfill \\ \hfill & =h^{*}\left(t_{\left(1\right)}\right)T_{\left(1\right)}\left(S\left(t_{\left(2\right)}\right)\right)T_{\left(2\right)}\left(h\right)\hfill \\ \hfill & =h^{*}\left(t_{\left(1\right)}T\left(S\left(t_{\left(2\right)}\right)h\right)\right)\hfill \\ \hfill & =h^{*}\left(T\left(S\left(t\right)h_{\left(1\right)}\right)h_{\left(2\right)}\right)\hfill \\ \hfill & =h^{*}\left(T\left(S\left(t\right)\right)h\right)=h^{*}\left(h\right).\hfill \end{array}$$

This finishes the proof. □

Proposition 11.

Let t, λ and f be as above. Then, $t^{\lambda }=S\left(t\right)$.

Proof. 

First of all, we remind that $t^{\lambda }$ and $S\left(t\right)$ are right integrals in H. In fact, when $\forall h\in H$ there exists a unique $h^{\prime }\in H$ such that $f\left(h^{\prime }\right)=h$ because of the bijectivity of f; then, $t^{\lambda }h=f\left(t\right)f\left(h^{\prime }\right)=f\left(t{h}^{\prime }\right)=f\left(\lambda \left(h^{\prime }\right)t\right)=\lambda \left(h^{\prime }\right)f\left(t\right)=\lambda \left(f^{-1}\left(h\right)\right)f\left(t\right)=\epsilon \left(h\right)t^{\lambda }$, and it follows that $t^{\lambda }$ is a right integral in H.

It follows from ([17], Lemma 3.3(1)) that $S\left(t\right)$ is a right integral in H as well. Therefore, $S\left(t\right)$ and $t^{\lambda }$ are linearly dependent.

Without loss of generality, we may assume that $T\left(t\right)=1$ for some $T\in {\int }_{H^{*}}^r$, and we have the following calculations:

$$T\left(t^{\lambda }\right)=T(\lambda \rightharpoonup t)=T\left(t_{\left(1\right)}\right)\lambda \left(t_{\left(2\right)}\right)=(T*\lambda )\left(t\right)={\epsilon }_{H^{*}}\left(\lambda \right)T\left(t\right)=\lambda \left(1_H\right)T\left(t\right)=1$$

which shows that $T\left(t^{\lambda }\right)=T\left(S\left(t\right)\right)$. Therefore, $t^{\lambda }=S\left(t\right)$. □

Proposition 12.

If $t\in {\int }_H^{\ell }$, then $t_{\left(1\right)}\otimes h{t}_{\left(2\right)}=S\left(h\right)t_{\left(1\right)}\otimes t_{\left(2\right)}$ for all $h\in H$.

Proof. 

Suppose that $t\in {\int }_H^{\ell }$ and $h\in H$; then, we have

$$\begin{array}{cc}\hfill t_{\left(1\right)}\otimes h{t}_{\left(2\right)}& =S\left(h_{\left(1\right)}\right)h_{\left(2\right)\left(1\right)}{t}_{\left(1\right)}\otimes h_{\left(2\right)\left(2\right)}{t}_{\left(2\right)}\hfill \\ \hfill & =S\left(h_{\left(1\right)}\right){\left(h_{\left(2\right)}t\right)}_{\left(1\right)}\otimes {\left(h_{\left(2\right)}t\right)}_{\left(2\right)}\hfill \\ \hfill & =S\left(h_{\left(1\right)}\right){\left(\epsilon \left(h_{\left(2\right)}\right)t\right)}_{\left(1\right)}\otimes {\left(\epsilon \left(h_{\left(2\right)}\right)t\right)}_{\left(2\right)}=S\left(h\right)t_{\left(1\right)}\otimes t_{\left(2\right)}\hfill \end{array}$$

from which the proposition follows. □

In the remainder of this section, we will show the existence of a Morita context relating $A♯H$ and $A^H$ when H is a finite dimensional Hopf coquasigroup. Before we do that, let us give a few results that we will use later.

Proposition 13.

Let A be a left quasi-H-module algebra. Then, $A♯H=(1_A♯H)(A♯1_H)$, where the element form is given by

$$\begin{array}{c}\hfill a♯h=(1_A♯h_{\left(2\right)})((S^{-1}\left(h_{\left(1\right)}\right)·a)♯1_H)\end{array}$$

(21)

for all $a♯h\in A♯H$.

Proof. 

For any $h\in H$ and $a\in A$, we have

$$\begin{array}{cc}\hfill (1_A♯h_{\left(2\right)})((S^{-1}\left(h_{\left(1\right)}\right)·a)♯1_H)& =1_A(h_{\left(2\right)\left(1\right)}·(S^{-1}\left(h_{\left(1\right)}\right)·a))♯h_{\left(2\right)\left(2\right)}{1}_H\hfill \\ \hfill & =(h_{\left(2\right)\left(1\right)}·(S^{-1}\left(h_{\left(1\right)}\right)·a))♯h_{\left(2\right)\left(2\right)}\hfill \\ \hfill & =(\left(h_{\left(2\right)\left(1\right)}{S}^{-1}\left(h_{\left(1\right)}\right)\right)·a)♯h_{\left(2\right)\left(2\right)}=a♯h\hfill \end{array}$$

where Equation (4) was used to obtain the last equality. □

Proposition 14.

Let A be a left quasi-H-module algebra. Suppose that t is a left integral in H and λ is an H-distinguished grouplike element in $H^{*}$. Then, for all $a\in A$, consider A as a right $A♯H$-module as in Lemma 2; then, for any $a\in A$ and $h\in H$,

(i) 

$(1_A♯t)(a♯h)=(1_A♯t)((S^{-1}\left(h^{\lambda }\right)·a)♯1_H)=(1_A♯t)((a⊲(1_A♯h^{\lambda }))♯1_H)$.

(ii) 

$t·a=a⊲(1_A♯t^{\lambda })$.

Proof. 

(i) For any $a\in A$ and $h\in H$, we compute.

$$\begin{array}{cc}\hfill (1_A♯t)(a♯h)& =(1_A♯t)(1_A♯h_{\left(2\right)})((S^{-1}\left(h_{\left(1\right)}\right)·a)♯1_H)\mathrm{by}\mathrm{Equation}\left(21\right)\hfill \\ \hfill & =(1_A(t_{\left(1\right)}·1_A)♯t_{\left(2\right)}{h}_{\left(2\right)})((S^{-1}\left(h_{\left(1\right)}\right)·a)♯1_H)\hfill \\ \hfill & =(1_A♯t{h}_{\left(2\right)})((S^{-1}\left(h_{\left(1\right)}\right)·a)♯1_H)\hfill \\ \hfill & =(1_A♯t)((S^{-1}\left(h_{\left(1\right)}\lambda \left(h_{\left(2\right)}\right)\right)·a)♯1_H)\hfill \\ \hfill & =(1_A♯t)((S^{-1}\left(h^{\lambda }\right)·a)♯1_H)\hfill \\ \hfill & =(1_A♯t)((S^{-1}\left(h^{\lambda }\right)·\left(a{1}_A\right))♯1_H)\hfill \\ \hfill & =(1_A♯t)((a⊲(1_A♯h^{\lambda }))♯1_H).\hfill \end{array}$$

(ii) For any $a\in A$, one also has

$$a⊲(1_A♯t^{\lambda })=S^{-1}\left(t^{\lambda }\right)·a=S^{-1}\left(S\left(t\right)\right)·a=t·a.$$

This completes the proof. □

At last, after some preparations, we give the main theorem of this section, which generalizes the well-known results in [13] to the case of finite dimensional Hopf coquasigroups.

Theorem 1.

Let A be a left quasi-H-module algebra. Then, we can form a Morita context $(A^H,A\#H,{}_{A\#H}{A}_{A^H},{}_{A^H}{A}_{A\#H},\tau ,\mu )$ with the following module actions and maps:

(1) 

Consider A as a left (respectively, right) $A^H$ module via left (respectively, right) multiplication;

(2) 

Consider A as a left (respectively, right) $A\#H$-module via Proposition 7(i) (respectively, Equation (18));

(3) 

$\tau =(,):A{\otimes }_{A\#H}A⟶A^H,(a,b)=t·\left(ab\right);$

(4) 

$\mu =[,]:A{\otimes }_{A^H}A⟶A\#H,[a,b]=(a♯t)(b♯1_H).$

Proof. 

First, we show that A is an $A^H$-$A♯H$-bimodule. Based on Lemma 1 and Proposition 7, the associativity conditions need to be supplemented. In fact, for all $a\in A,a^{\prime }\in A^H$ and $b♯h\in A♯H$, we have

$$\begin{array}{cc}\hfill \left(a^{\prime }a\right)◂(b♯h)& =S^{-1}\left(h^{\lambda }\right)·\left(a^{\prime }ab\right)=(S^{-1}{\left(h^{\lambda }\right)}_{\left(1\right)}·a^{\prime })(S^{-1}{\left(h^{\lambda }\right)}_{\left(2\right)}·\left(ab\right))\hfill \\ \hfill & =\epsilon \left(S^{-1}{\left(h^{\lambda }\right)}_{\left(1\right)}\right)a^{\prime }(S^{-1}{\left(h^{\lambda }\right)}_{\left(2\right)}·\left(ab\right))\hfill \\ \hfill & =a^{\prime }(S^{-1}\left(h^{\lambda }\right)·\left(ab\right))=a^{\prime }(a◂(b♯h)).\hfill \end{array}$$

Similarly, A is an $A♯H$-$A^H$-bimodule.

Next, we will show the linearity of these two maps $\tau$ and $\mu$, respectively.

For the Morita map $\mu$, it is expected to be left and right $A♯H$-linear and to be middle $A^H$-linear at the same time. Explicitly, for all $a,b\in A,a^{\prime }\in A^H$ and $c♯h\in A♯H$, we have

$$\begin{array}{cc}\hfill \left[\right(c♯h)⊳a,b]& =[c(h·a),b]=(c(h·a)♯t)(b♯1_H)\hfill \\ \hfill & =(c(h_{\left(1\right)}·a)♯\epsilon \left(h_{\left(2\right)}\right)t)(b♯1_H)\hfill \\ \hfill & =(c(h_{\left(1\right)}·a)♯h_{\left(2\right)}t)(b♯1_H)\hfill \\ \hfill & =(c♯h)(a♯t)(b♯1_H)\hfill \\ \hfill & =(c♯h)[a,b].\hfill \end{array}$$

This complete the proof of left $A♯H$-linear.

As for right $A♯H$-linear, we still need detailed verification, since the definition of left and right $A♯H$-action on A is not symmetrical. For all $a,b\in A,a^{\prime }\in A^H$, and $c♯h\in A♯H$, we have

$$\begin{array}{cc}\hfill [a,b◂(c♯h\left)\right]& =[a,S^{-1}\left(h^{\lambda }\right)·\left(bc\right)]=(a♯t)((S^{-1}\left(h^{\lambda }\right)·\left(bc\right))♯1_H)\hfill \\ \hfill & =a(t_{\left(1\right)}·(S^{-1}\left(h^{\lambda }\right)·\left(bc\right)))♯t_{\left(2\right)}{1}_H\hfill \\ \hfill & =a(\left(t_{\left(1\right)}{S}^{-1}\left(h^{\lambda }\right)\right)·\left(bc\right))♯t_{\left(2\right)}\hfill \\ \hfill & =a(t_{\left(1\right)}·\left(bc\right))S^{-1}\left(h_{\left(1\right)}\right)\lambda \left(h_{\left(2\right)}\right)♯t_{\left(2\right)}\hfill \\ \hfill & =(a♯t)((b◂(c♯h))♯1_H)\hfill \\ \hfill & =(a♯t)((b◂\left((c♯1_H)(1_A♯h)\right))♯1_H)\hfill \\ \hfill & =(a♯t)((\left(bc\right)◂(1_A♯h))♯1_H).\hfill \end{array}$$

On the other hand, we have

$$\begin{array}{cc}\hfill [a,b](c♯h)& =(a♯t)(b♯1_H)(c♯h)=(a♯1_H)(1_A♯t)(bc♯h)\hfill \\ \hfill & =(a♯1_H)(1_A♯t)((\left(bc\right)⊲(1_A♯h^{\lambda }))♯1_H)\hfill \\ \hfill & =(a♯t)((\left(bc\right)⊲(1_A♯h^{\lambda }))♯1_H)\hfill \\ \hfill & =a(t_{\left(1\right)}·(\left(bc\right)⊲(1_A♯h^{\lambda })))♯t_{\left(2\right)}\hfill \\ \hfill & =a((\left(bc\right)⊲(1_A♯h^{\lambda }))⊲(1_A♯t_{\left(1\right)}^{\lambda }))♯t_{\left(2\right)}\hfill \\ \hfill & =(a♯t)((\left(bc\right)⊲{(1_A♯h)}^{\lambda })♯1_H)\hfill \\ \hfill & =(a♯t)((\left(bc\right)◂(1_A♯h))♯1_H)\hfill \end{array}$$

where Proposition 14(i) was used for the third equality. Hence, right $A♯H$-linear.

As the middle $A^H$-linear, we have for all $a,b\in A$ and $a^{\prime }\in A^H$,

$$\begin{array}{c}\hfill [a{a}^{\prime },b]=(a{a}^{\prime }♯t)(b♯1_H)=a{a}^{\prime }(t_{\left(1\right)}·b)♯t_{\left(2\right)}\end{array}$$

and

$$\begin{array}{cc}\hfill [a,a^{\prime }b]& =(a♯t)(a^{\prime }b♯1_H)=a(t_{\left(1\right)}·\left(a^{\prime }b\right))♯t_{\left(2\right)}\hfill \\ \hfill & =a\left((t_{\left(1\right)\left(1\right)}·a^{\prime })(t_{\left(1\right)\left(2\right)}·b)\right)♯t_{\left(2\right)}=a{a}^{\prime }(t_{\left(1\right)}·b)♯t_{\left(2\right)}.\hfill \end{array}$$

Analogously, for the Morita map $\tau$, we need prove three linearities as well, respectively, left $A^H$-linear, right $A^H$-linear, and middle $A♯H$-linear. $\forall a,b\in A,a^{\prime }\in A^H$ and $c♯h\in A♯H$, we have

$$\begin{array}{cc}\hfill (a^{\prime }a,b)=& t·\left(a^{\prime }ab\right)=(t_{\left(1\right)}·a^{\prime })(t_{\left(2\right)}·\left(ab\right))\hfill \\ \hfill & =\epsilon \left(t_{\left(1\right)}\right)a^{\prime }(t_{\left(2\right)}·\left(ab\right))=a^{\prime }(t·\left(ab\right))=a^{\prime }(a,b).\hfill \end{array}$$

This shows that $(,)$ is a left $A^H$-linear. Likewise, it is a right $A^H$-linear.

Furthermore, for all $a,b\in A$ and $c♯h\in A♯H$, we have

$$\begin{array}{cc}\hfill (a◂(c♯h),b)& =(S^{-1}\left(h^{\lambda }\right)·\left(ac\right),b)=t·\left((S^{-1}\left(h^{\lambda }\right)·\left(ac\right))b\right)\hfill \\ \hfill & =t·\left((\left(S^{-1}\left(h_{\left(1\right)}\right)\lambda \left(h_{\left(2\right)}\right)\right)·\left(ac\right))b\right)\hfill \\ \hfill & =\left(t{h}_{\left(2\right)}\right)·\left((S^{-1}\left(h_{\left(1\right)}\right)·\left(ac\right))b\right)\hfill \\ \hfill & =t·(h_{\left(2\right)}·\left((S^{-1}\left(h_{\left(1\right)}\right)·\left(ac\right))b\right))\hfill \\ \hfill & =t·((h_{\left(2\right)\left(1\right)}·(S^{-1}\left(h_{\left(1\right)}\right)·\left(ac\right))(h_{\left(2\right)\left(2\right)}·b))\hfill \\ \hfill & =t·\left((\left(h_{\left(2\right)\left(1\right)}{S}^{-1}\left(h_{\left(1\right)}\right)\right)·\left(ac\right))(h_{\left(2\right)\left(2\right)}·b)\right)\hfill \\ \hfill & =t·\left(ac\right(h·b\left)\right)=t·\left(a\right((c♯h)⊳b\left)\right)\hfill \\ \hfill & =(a,(c♯h)⊳b)\hfill \end{array}$$

and so $\tau$ is middle $A♯H$-linearity.

At last, we are left with only the proof of compatibility conditions of $\mu$ with $\tau$. For all $a,b,c\in A$, we have

$$\begin{array}{cc}\hfill [a,b]⊳c& \doteq \left((a♯t)(b♯1_H)\right)⊳c=(a♯t)⊳((b♯1_H)⊳c)=(a♯t)⊳\left(bc\right)\hfill \\ & =a(t·(bc\left)\right)=a(b,c),\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill a◂[b,c]& =a◂\left((b♯t)(c♯1_H)\right)=(a◂(b♯t))◂(c♯1_H)\hfill \\ \hfill & =(a◂(b♯t))c=(S^{-1}\left(t^{\lambda }\right)·\left(ab\right))c=(t·\left(ab\right))c=(a,b)c.\hfill \end{array}$$

So far, we have constructed the Morita context. □

Then, it follows immediately that:

Corollary 1.

Let $[,]$ and $(,)$ be as above. Then, we have

(i) 

$[A,A]$ is an ideal of $A♯H$.

(ii) 

$(A,A)$ is an ideal of $A^H$.

For a right ideal I of $A^H$ and a left ideal J of $A^H$, we put

$${\eta }_1\left(I\right)=IA\mathrm{and}{\eta }_2\left(J\right)=AJ$$

respectively.

For a right $A\#H$-submodule N of $A_{A\#H}$ and a left $A\#H$-submodule P of ${}_{A\#H}A$, one puts

$${\xi }_1\left(N\right)=(N,A)\mathrm{and}{\xi }_2\left(P\right)=(A,P),$$

respectively.

We now obtain the following results (see [19]).

Theorem 2.

With the above notation as in Theorem 1. Let $\tau =(,)$ be surjective. Then,

(1) 

${\eta }_1$ and ${\xi }_1$ are inverses and are lattice isomorphisms between the lattice of right ideals of $A^H$ and the lattice of submodules of $A_{A\#H}$. Moreover, these induce lattice isomorphisms between the lattice of ideals of $A^H$ and the lattice of submodules of ${}_{A^H}{A}_{A\#H}$.

(2) 

${\eta }_2$ and ${\xi }_2$ are inverses and are lattice isomorphisms between the lattice of left ideals of $A^H$ and the lattice of submodules of ${}_{A\#H}A$. Moreover, these induce lattice isomorphisms between the lattice of ideals of $A^H$ and the lattice of submodules of ${}_{A\#H}{A}_{A^H}$.

Similar treats for $A\#H$: For a right ideal X of $A\#H$ and a left ideal Y of $A\#H$, we put

$${\eta }_3\left(X\right)=X(A\#H)and{\eta }_4\left(Y\right)=(A\#H)Y$$

respectively.

For a right $A^H$-submodule U of $A_{A^H}$ and a left $A^H$-submodule V of ${}_{A^H}A$, one puts

$${\xi }_3\left(U\right)=[U,A]and{\xi }_4\left(V\right)=[A,V]$$

respectively.

Theorem 3.

With the above notation as in Theorem 1. Let $\mu =[,]$ be surjective. Then,

(1) 

${\eta }_3$ and ${\xi }_3$ are inverses and are lattice isomorphisms between the lattice of right ideals of $A\#H$ and the lattice of submodules of $A_{A^H}$. Moreover, these induce lattice isomorphisms between the lattice of ideals of $A\#H$ and the lattice of submodules of ${}_{A^H}{A}_{A\#H}$.

(2) 

${\eta }_4$ and ${\xi }_4$ are inverses and are lattice isomorphisms between the lattice of left ideals of $A\#H$ and the lattice of submodules of $A_{A^H}$. Moreover, these induce lattice isomorphisms between the lattice of ideals of $A\#H$ and the lattice of submodules of ${}_{A\#H}{A}_{A^H}$.

Corollary 2.

With the above notation as in Theorem 1. Let the Morita maps $\tau =(,)$ and $\mu =[,]$ be surjective. Then, $A^H$ and $A\#H$ have isomorphic lattices of ideals.

4. Surjectivity of the Morita Maps $\mathbf{\tau }$ and $\mathbf{\mu }$

Let ${\mathbf{Hom}}_{A^H}(A,A^H)$ be the set of all right $A^H$-module homomorphisms from A to $A^H$. Similarly, let ${}_{A^H}\mathbf{Hom}(A,A^H)$ be the set of all left $A^H$-module homomorphisms from A to $A^H$. We denote by ${\mathbf{Hom}}_{A\#H}(A,A\#H)$ the set of all right $A\#H$-module homomorphisms from A to $A\#H$ and similarly for ${}_{A\#H}\mathbf{Hom}(A,A\#H)$. Then, we have that ${\mathbf{Hom}}_{A^H}(A,A^H)$ and ${}_{A\#H}\mathbf{Hom}(A,A\#H)$ are $A^H$-$A\#H$-bimodules with the structures given respectively, by:

$$[x·f·(a\#h)]\left(b\right)=f(xb◂(a\#h)),\forall x\in A^H,f\in {\mathbf{Hom}}_{A^H}(A,A^H),a,b\in A,h\in H.$$

and

$$[x·g·(a\#h)]\left(b\right)=g(xb◂(a\#h)),\forall x\in A^H,g\in {}_{A\#H}\mathbf{Hom}(A,A\#H),a,b\in A,h\in H.$$

Similarly, ${}_{A^H}\mathbf{Hom}(A,A^H)$ and ${\mathbf{Hom}}_{A\#H}(A,A\#H)$ are $A\#H$-$A^H$-bimodule with the respective structures given by:

$$[(a\#h)·f·x]\left(b\right)=f((a\#h)⊳bx),\forall x\in A^H,f\in {}_{A^H}\mathbf{Hom}(A,A^H),a,b\in A,h\in H.$$

and

$$[(a\#h)·g·x]\left(b\right)=g((a\#h)⊳bx),\forall x\in A^H,g\in {\mathbf{Hom}}_{A\#H}(A,A\#H),a,b\in A,h\in H.$$

Define the following maps:

$$\begin{array}{ccc}& & {\Phi }_1:A⟶{\mathbf{Hom}}_{A^H}(A,A^H)by{\Phi }_1\left(a\right)\left(b\right)=(a,b),\forall a,b\in A.\hfill \\ & & {\Phi }_2:A⟶{}_{A\#H}\mathbf{Hom}(A,A\#H)by{\Phi }_2\left(a\right)\left(b\right)=[a,b],\forall a,b\in A.\hfill \\ & & {\Phi }_3:A⟶{}_{A^H}\mathbf{Hom}(A,A^H)by{\Phi }_3\left(a\right)\left(b\right)=(a,b),\forall a,b\in A.\hfill \\ & & {\Phi }_4:A⟶{\mathbf{Hom}}_{A\#H}(A,A\#H)by{\Phi }_4\left(a\right)\left(b\right)=[a,b],\forall a,b\in A.\hfill \end{array}$$

We also consider some rings: ${\mathbf{End}}_{A^H}\left(A\right)$, ${}_{A\#H}\mathbf{End}\left(A\right)$, ${\mathbf{End}}_{A\#H}\left(A\right)$ and ${}_{A^H}\mathbf{End}\left(A\right)$. We have the following maps:

$$\begin{array}{ccc}& & {\Psi }_1:A\#H⟶{\mathbf{End}}_{A^H}\left(A\right)by{\Psi }_1(a\#h)\left(b\right)=(a\#h)⊳b,\forall a,b\in A,h\in H.\hfill \\ & & {\Psi }_2:{\left(A^H\right)}^{op}⟶{}_{A\#H}\mathbf{End}\left(A\right)by{\Psi }_2\left(x\right)\left(a\right)=xa\forall x\in A^H,a\in A.\hfill \\ & & {\Psi }_3:{\left(A^H\right)}^{op}⟶{\mathbf{End}}_{A\#H}\left(A\right)by{\Psi }_3\left(x\right)\left(a\right)=xa\forall x\in A^H,a\in A.\hfill \\ & & {\Psi }_4:A\#H⟶{}_{A^H}\mathbf{End}\left(A\right)by{\Psi }_4(a\#h)\left(b\right)=(a\#h)⊳b,\forall a,b\in A,h\in H.\hfill \end{array}$$

We can obtain the following result (see [19]).

Theorem 4.

With the above notation as in Theorem 1. Let τ and μ be surjective. Then,

(1) 

τ and μ are isomorphisms.

(2) 

$A_{A^H}$, ${}_{A\#H}A$, $A_{A\#H}$ and ${}_{A^H}A$ are progenerators.

(3) 

The pair of functors $-{\otimes }_{A^H}A$ and $-{\otimes }_{A\#H}A$ define an equivalence of categories ${\mathbf{Mod}}_{A^H}$ and ${\mathbf{Mod}}_{A\#H}$.

(4) 

The pair of functors $A{\otimes }_{A^H}$ and $A{\otimes }_{A\#H}$ define an equivalence of categories ${}_{A^H}\mathbf{Mod}$ and ${}_{A\#H}\mathbf{Mod}$.

(5) 

The centers of A and $A\#H$ are isomorphic.

(6) 

Every ${\Phi }_i$ with $i=1,2,3,4$ is a bimodule isomorphism.

(7) 

Every ${\Psi }_i$ with $i=1,2,3,4$ is a ring isomorphism.

Definition 3.

Let H be a Hopf quasigroup and A a right H-comodule algebra. Then, $A/A^{coH}$ is said to be a right H-Galois extension if the map

$$\gamma :A{\otimes }_{A^{coH}}A⟶A\otimes H,a\otimes b\mapsto a{b}_{\left[0\right]}\otimes b_{\left[1\right]}$$

is surjective.

Theorem 5.

Let H be a finite dimensional Hopf coquasigroup and A a left quasi-H-module algebra. Then, the following statements are equivalent:

(1) 

$A/A^H$ is a right $H^{*}$-Galois extension.

(2) 

The map τ in Theorem 1 is surjective.

(3) 

For any $M\in {}_{A\#H}\mathbf{Mod}$, the map $\Phi :A{\otimes }_{A^H}{M}^H⟶M$ given by $a\otimes x\mapsto ax$ is a left $A\#H$-module isomorphism, where the $A\#H$-module structure on $A\otimes M^H$ is given via

$$(a\#h)·(b\otimes x)=a(h·b)\otimes x$$

for all $a,b\in A$, $h\in H$ and $x\in A^H$.

(4) 

A is a left $A\#H$-generator.

(5) 

A is a finitely generated projective right $A^H$-module and the map $\pi :A\#H⟶End\left(A_{A^H}\right)$ via $\pi (a\#h)\left(b\right)=a(h·b)$, for all $a,b\in A$ and $h\in H$.

Proof. 

After considering Propositions 5 and 6, the proof of this theorem essentially follows ([13], Proof of Theorem 1.2), (also see [20]). □

We have the other version of our theorem as follows.

Theorem 6.

Let H be a finite dimensional Hopf coquasigroup and A a left quasi-H-module algebra. Then, the following (1)–(5) are equivalent:

(1) 

$A/A^H$ is a right $H^{*}$-Galois extension.

(2) 

The map τ in Theorem 1 is surjective.

(3) 

For any $M\in {\mathbf{Mod}}_{A\#H}$, the map $\Psi :A{\otimes }_{A^H}{M}^H⟶M$ given by $x\otimes a\mapsto xa$ is a right $A\#H$-module isomorphism, where the $A\#H$-module structure on $M^H\otimes A$ is given via

$$(x\otimes b)·(a\#h)=x\otimes S^{-1}\left(h\right)·\left(ba\right)$$

for all $a,b\in A$, $h\in H$ and $x\in A^H$.

(4) 

A is a right $A\#H$-generator.

(5) 

A is a finitely generated projective left $A^H$-module and the map $\omega :A\#H⟶End\left({}_{A^H}A\right)$ via $\omega (a\#h)\left(b\right)=S^{-1}\left(h\right)·\left(ab\right)$, for all $a,b\in A$ and $h\in H$.

5. A Special Case

We will treat a special case in order to illustrate results in Section 3 about our Morita theory.

It was known that a G-grading of an algebra A is the same as an ${\left(kG\right)}^{*}$-module algebra action of ${\left(kG\right)}^{*}$, the Hopf algebra dual to $kG$, on A. Similarly, we have

Proposition 15.

Let G be a quasigroup. Then, a G-grading of an algebra A is the same as a left quasi-${\left(kG\right)}^{*}$-module algebra action of ${\left(kG\right)}^{*}$ on A.

Proof. 

Suppose that A is graded by G; then, we have that $A={\displaystyle \underset{g\in G}{⨁}}{A}_g$. Any $a\in A$ may be written uniquely as $a={\displaystyle \sum _{g\in G}}{a}_g$, where $a_g\in A_g$. We define the action of $k{\left[G\right]}^{*}$ on A by $p_g·a=a_g$, where $\left\{p_g\right\}$ forms the dual basis for $k{\left[G\right]}^{*}$. That is, $p_g$ is the projection onto the gth part of any element of A. Using $p_g{p}_h={\delta }_{g,h}{p}_g$, it is clear that A is a $k{\left[G\right]}^{*}$-module. In addition, for any $p_g\in k{\left[G\right]}^{*},a,b\in A$, we have $p_g·\left(ab\right)={\left(ab\right)}_g={\displaystyle \sum _{uv=g}}{p}_u\left(a\right)p_v\left(b\right)={\displaystyle \sum _{uv=g}}(p_u·a)(p_v·b)=({p_g}_{\left(1\right)}·a)({p_g}_{\left(2\right)}·b)$, and $p_g·1_A=\epsilon \left(p_g\right)1_A$ is trivial. Therefore, A is a quasi-$k{\left[G\right]}^{*}$-module algebra.

Conversely, we say that A is a quasi-$k{\left[G\right]}^{*}$-module algebra, and we denote the action of $k{\left[G\right]}^{*}$ on A by $p_g·a$ for any $a\in A$. Let $A_g=\{p_g·a|a\in A\}$; since $p_g{p}_h={\delta }_{g,h}{p}_g$, and ${\displaystyle \sum _{g\in G}}{p}_g=1$, it is clear that $A={\displaystyle \underset{g\in G}{⨁}}{A}_g$. Since the given action satisfies $p_g·\left(ab\right)=({p_g}_{\left(1\right)}·a)({p_g}_{\left(2\right)}·b)$, and $△\left(p_g\right)={\displaystyle \sum _{uv=g}}{p}_u\otimes p_v$, it follows that $p_g·\left(ab\right)={\displaystyle \sum _{uv=g}}(p_u·a)(p_v·b)$. Thus, $A_u{A}_v\subseteq A_g$, for any $g\in G$ so that A is graded by G. □

Proposition 16.

A is a G-graded algebra if and only if A is a (right) $k\left(G\right)$-comodule algebra.

Proof. 

$⟸)$ We already know from Proposition 15 that $A={\displaystyle \underset{g\in G}{⨁}}{A}_g$, a G-graded vector space, and that for $a_g\in A_g,\rho \left(a_g\right)=a_g\otimes g$. Thus, $\rho \left(a_g{b}_h\right)=a_g{b}_h\otimes gh$; that is, $a_g{b}_h\in A_{gh}$, for all $g,h\in G$, and also $1_A\in A_1$. Thus, A is a G-graded algebra.

$⟹)$ If A is a G-graded algebra, perform the same process: write $\rho \left(a_g\right)=a_g\otimes g,a_g\in A_g$, and it is not difficult to check that Equations (13) and (14) hold, and so A is a (right) $k\left(G\right)$-comodule algebra.

This completes the proof. □

If A is graded, then we have the smash product $A♯k{\left[G\right]}^{*}$. For $a,b\in A$, and basis elements $p_g,p_h\in k{\left[G\right]}^{*}$, the product is given by

$$(a♯p_g)(b♯p_h)=a({p_g}_{\left(1\right)}·b)♯{p_g}_{\left(2\right)}{p}_h={\displaystyle \sum _{uv=g}}a(p_u·b)♯p_v{p}_h={\displaystyle \sum _{uv=g}}a{b}_u♯p_v{p}_h=a{b}_{g{h}^{-1}}♯p_h$$

using $\left(uh\right)h^{-1}=u$ and given the fact that $\left\{p_g\right\}$ represents orthogonal idempotents and that $p_u·b=b_u$.

In addition, A may be identified with $A♯1_{k{\left[G\right]}^{*}}$, and $k{\left[G\right]}^{*}$ with $1_A♯k{\left[G\right]}^{*}$ in $A♯k{\left[G\right]}^{*}$. We have some results about $A♯k{\left[G\right]}^{*}$ below.

Proposition 17.

Let A be graded by the finite quasigroup G. Then, $A♯k{\left[G\right]}^{*}$ is the free right and left A-module with basis $\{1♯p_g\mid g\in G\}$, a set of orthogonal idempotents whose sum is 1, and with the product given as above. Furthermore, the following statements hold:

(1) 

$(1♯p_h)(a♯1)=\sum _g{a}_{h{g}^{-1}}♯p_g$, for all $a\in A,h\in G$.

(2) 

$(1♯p_h)(a_g♯1)=a_g♯p_{g^{-1}h}$, for all $a_g\in A_g$.

(3) 

Each $p_h$ centralizes $A_1$.

(4) 

$p_h(I♯k{\left[G\right]}^{*})p_g=I_{h{g}^{-1}}♯p_g=p_{h}I{p}_g$, for all $g,h\in G$ and for all graded ideal I.

(5) 

$p_1(I♯k{\left[G\right]}^{*})p_1=I_1{p}_1$, which is isomorphic as a ring to $I_1$.

Proof. 

The $\left\{p_g\right\}$ values are a free k-basis for $k{\left[G\right]}^{*}$, and so the $\{1♯p_g\}$ values are a free left A-basis for $A♯k{\left[G\right]}^{*}$. Using (1), A may be identified with $A♯1_{k{\left[G\right]}^{*}}$, and it is clear they are also a free right A-basis. Since $\left\{p_g\right\}$ are orthogonal idempotents in $k{\left[G\right]}^{*}$, $(1♯p_h)(1♯p_g)=1_{g{h}^{-1}}♯p_g={\delta }_{g,h}♯p_g$, and the $\{1♯p_g\}$ values are also orthogonal idempotents whose sum is 1 in $A♯k{\left[G\right]}^{*}$.

(1) $(1♯p_h)(a♯1)={p_h}_{\left(1\right)}·a♯{p_h}_{\left(2\right)}=\sum _{uv=h}{a}_u♯p_v=\sum _g{a}_{h{g}^{-1}}♯p_g$, for all $a\in A,h\in G$.

(2) By (1), $(1♯p_h)(a_g♯1)=\sum _t{\left(a_g\right)}_{h{t}^{-1}}♯p_t$, for all $a_g\in A_g,h\in G$. However, ${\left(a_g\right)}_{h{t}^{-1}}$ = 0 unless $h{t}^{-1}=g$, in which case $t=g^{-1}h$ and ${\left(a_g\right)}_g=a_g$. Hence, $(1♯p_h)(a_g♯1)=a_g♯p_{g^{-1}h}$.

(3) By (2), take g=1, $(1♯p_h)(a_1♯1)=a_1♯p_h=(a_1♯1)(1♯p_h)$.

(4) and (5) are straightforward. □

As a corollary of Theorem 1, we have

Theorem 7.

Let G be a finite quasigroup and A be a G-graded algebra. Then, we can form a Morita context $(A_1,A\#k{\left[G\right]}^{*},{}_{A\#k{\left[G\right]}^{*}}{A}_{A_1},{}_{A_1}{A}_{A\#k{\left[G\right]}^{*}},\tau ,\mu )$ with the following module actions and maps:

(1) 

Consider A as a left (respectively, right) $A_1$-module via left (respectively, right) multiplication;

(2) 

Consider A as a left (respectively, right)$A\#k{\left[G\right]}^{*}$-module via Proposition 7(i) (respectively, Equation (18));

(3) 

$\tau =(,):A{\otimes }_{A\#k{\left[G\right]}^{*}}A⟶A_1,(a,b)={\left(ab\right)}_1$;

(4) 

$\mu =[,]:A{\otimes }_{A_1}A⟶A\#k{\left[G\right]}^{*},[a,b]=(a♯t)(b♯1_H)$.

Proof. 

We here give a sketch of the proof: given a G-graded algebra A with finite quasigroup G, then A is also a quasi-$k{\left[G\right]}^{*}$-module algebra by Proposition 15.

(1) The invariants subalgebra of A is $A^{k{\left[G\right]}^{*}}=\{a\in A\mid h·a\doteq \epsilon \left(h\right)a,\forall h\in k{\left[G\right]}^{*}\}$. Notice that $h={\displaystyle \sum _{g\in G}}{\alpha }_g{p}_g$ and $\epsilon \left(p_g\right)={\delta }_{g,1},\forall g\in G$; this means for all $a\in A^{k{\left[G\right]}^{*}}$, $h·a=({\displaystyle \sum _{g\in G}}{\alpha }_g{p}_g)·a=\left({\alpha }_1{p}_1\right)·a={\alpha }_1{a}_1=a$, where ${\alpha }_1\in k$, that is, $A^{k{\left[G\right]}^{*}}=A_1$. It is obvious that A is a left (respectively, right) $A_1$-module via left (respectively, right) multiplication.

(2) By Proposition 7(i), A is a left $A\#k{\left[G\right]}^{*}$-module via $(b♯p_g)⊳a=b(p_g·a)=b{a}_g$, for all $a\in A,b♯p_g\in A\#k{\left[G\right]}^{*}$.

The right modular action is slightly more complicated. First, we compute the left integral space of $k{\left[G\right]}^{*}$, ${\int }_{k{\left[G\right]}^{*}}^{\ell }=\{t\in k{[G]}^{*}\mid ({\displaystyle \sum _{g\in G}}{\alpha }_g{p}_g)t=\epsilon ({\displaystyle \sum _{g\in G}}{\alpha }_g{p}_g)t\}$, and we assume that $t={\displaystyle \sum _{g\in G}}{\beta }_g{p}_g$, so we have ${\displaystyle \sum _{g\in G}}{\alpha }_g{\beta }_g{p}_g={\alpha }_1{\beta }_1{p}_1+{\displaystyle \sum _{g\ne 1\in G}}{\beta }_g{p}_g$, for which we have used the fact that $\left\{p_g\right\}$ are orthogonal idempotents in $k{\left[G\right]}^{*}$ and $\epsilon \left(p_g\right)={\delta }_{g,1},\forall g\in G$. So, by comparing degrees, it can be concluded that the coefficients ${\beta }_g$ of $p_g$ are all 0 unless ${\beta }_1$. Therefore, t=${\beta }_1{p}_1$, that is, ${\int }_{k{\left[G\right]}^{*}}^{\ell }=k{p}_1$. Second, we compute the distinguished group-like element $\lambda$ (as mentioned in Section 2), and choosing a left integral $p_1\in k{\left[G\right]}^{*}$, we have $p_1({\displaystyle \sum _{g\in G}}{\alpha }_g{p}_g)={\alpha }_1{p}_1=\lambda ({\displaystyle \sum _{g\in G}}{\alpha }_g{p}_g)p_1$, which means $\lambda ({\displaystyle \sum _{g\in G}}{\alpha }_g{p}_g)={\alpha }_1$, the coefficient of the $p_1$ term. The fact that $p_g\left(h\right)={\delta }_{g,h},\forall g,h\in G$ shows that $\lambda =1$, the identity element in G, where we identified ${\left(k{\left[G\right]}^{*}\right)}^{*}$ with k[G]. Finally, ${p_g}^{\lambda }={\displaystyle \sum _{uv=g}}{p}_u{p}_v\left(1\right)=p_g$.

By Equation (18), we have that A is a right $A\#k{\left[G\right]}^{*}$ module via $a◂(b♯p_g)=S^{-1}\left({p_g}^{\lambda }\right)·\left(ab\right)=p_{g^{-1}}·\left(ab\right)={\left(ab\right)}_{g^{-1}}$, for all $a\in A,b♯p_g\in A\#k{\left[G\right]}^{*}$.

(3) Applying Theorem 1(3) here, we have $\tau =(,):A{\otimes }_{A\#k{\left[G\right]}^{*}}A⟶A_1,(a,b)=p_1·\left(ab\right)={\left(ab\right)}_1$.

(4) Applying Theorem 1(4) here, we have $\mu =[,]:A{\otimes }_{A_1}A⟶A\#k{\left[G\right]}^{*},[a,b]=(a♯p_1)(b♯1_{k{\left[G\right]}^{*}})=a({p_1}_{\left(1\right)}·b)♯{p_1}_{\left(2\right)}={\displaystyle \sum _{g\in G}}a(p_g·b)♯p_{g^{-1}}={\displaystyle \sum _{g\in G}}a{b}_g♯p_{g^{-1}}$.

This finishes the proof. □