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Article

Characterization of Automorphisms of (θ,ω)-Twisted Radford’s Hom-Biproduct

1
Department of Mathematics, Jining University, Qufu 273155, China
2
School of Mathematical Sciences, Qufu Normal University, Qufu 273165, China
*
Author to whom correspondence should be addressed.
Mathematics 2022, 10(3), 407; https://doi.org/10.3390/math10030407
Submission received: 18 December 2021 / Revised: 22 January 2022 / Accepted: 22 January 2022 / Published: 27 January 2022
(This article belongs to the Special Issue Rota-Baxter Algebra and Related Topics)

Abstract

:
In this paper, we study the Hom–Hopf algebra automorphism group of a ( θ , ω ) -twisted-Radford’s Hom-biproduct, which satisfies certain conditions. First, we study the endomorphism monoid and automorphism group of ( θ , ω ) -twisted Radford’s Hom-biproducts, and show that the endomorphism has a factorization closely related to the factors ( A , α ) and ( H , β ) . Then, we consider ( θ , ω ) -twisted Radford’s Hom-biproduct automorphism group Aut Hom - Hopf ( A × ω θ H , p ) as a subgroup of some semidirect product U ( C , A ) op φ G ( A ) . Finally, we characterize the automorphisms of a concrete example.

1. Introduction

The automorphism group of an algebra has a long history and extensive study, and determining the fully automorphic group of an algebra is usually a very difficult problem. For instance, the automorphism group of the polynomial ring of one or two variables has been understood. Jung [1] in 1942 first proved the case of two variables for characteristic zero and then van der Kulk [2] in 1953 for an arbitrary characteristic. However, the automorphism group of the polynomial ring of three variables is not yet fully understood, and a remarkable result is given by Shestakov and Umirbaev [3] in 2004, which shows that the Nagata automorphism is a wild automorphism. Many researchers have been successfully computing the automorphism groups of interesting infinite-dimensional noncommutative algebras, including certain quantum groups, generalized quantum Weyl algebras, skew polynomial rings, quantized algebras and many more—see [4,5,6,7,8,9,10,11,12,13,14,15], among others. The theory of algebra authomorphisms is very deep and related with mainstream, such as the Jacobian Conjecture [16] and Holonomic D -modules [17,18].
In the theory of the classical Hopf algebras, Radford’s biproducts are very important Hopf algebras, which play an important role in the theory of classification of pointed Hopf algebra [19,20], and account for many examples of semisimple Hopf algebra. Therefore, the study of the automorphism group of Radford’s biproducts is very interesting, and Radford characterized Hopf algebra automorphisms in [21]. There are many generalizations of Radford’s biproducts, such as [22] for quasi-Hopf algebras, [23] for multiplier Hopf algebras, Ref. [24], for monoidal Hom–Hopf algebras, Ref. [25], for Hom–Hopf algebras, and recently [26] for ( θ , ω ) -twisted-Radford’s Hom-biproducts.
Let A × H be the Radford’s biproduct, where A is both a left H-module algebra and a left H-comodule coalgebra. Define π : A × H H , a × h ε A ( a ) h , let Aut Hopf ( A × H , π ) be the set of Hopf algebra automorphisms F of A × H satisfying π F = π . Radford [21] characterized these Hopf algebra automorphisms, and factorized F into two suitable maps. Inspired by Radford’s idea in [21], the authors of [27] discuss the automorphisms of twisted tensor biproducts, and reveal the relationships between results and Radford’s results, and the authors of [28] study automorphisms of Radford’s Hom-biproducts introduced in [24]. Motivated by these works, we want to characterize the Hom–Hopf algebra automorphisms of ( θ , ω ) -twisted-Radford’s Hom-biproduct introduced in [26].
This paper is organized as follows. In Section 2, we recall some definitions and basic results related to Hom-(co)algebras, Hom-bialgebras, Hom–Hopf algebras, Hom-(co)module, Hom-(co)module (co)algebras, θ -twisted Hom-smash products, ω -twisted Hom-smash coproducts and ( θ , ω ) -twisted-Radford’s Hom-biproduct. In Section 3, we study the endomorphisms and automorphisms of ( θ , ω ) -twisted-Radford’s Hom-biproducts, and show that the automorphism has a factorization closely related to the factors ( A , α ) and ( H , β ) of the ( θ , ω ) -twisted-Radford’s Hom-biproduct ( A × ω θ H , α β ) in [26]. Finally, we characterize the automorphisms of a concrete example in Section 4.

2. Preliminaries

We always assume that k is a base field. All algebras, linear spaces, etc. will be over k ; the subscript of k is omitted for simplicity. Let id V denote the identity map on a k -space V. We now recall some useful definitions and terminology as follows; see [25,29,30,31,32,33,34].
Definition 1
([29]). A Hom-algebra is a quadruple ( A , μ , η , α ) (abbr. ( A , α ) ), where A is a k -linear space, μ : A × A A and η : k A are linear maps with notation μ ( a b ) = a b , η ( 1 k ) = 1 A , and α is an automorphism of A, satisfying the following conditions for all a , b , c A
(HA1)
α ( a b ) = α ( a ) α ( b ) ,
(HA2)
α ( a ) ( b c ) = ( a b ) α ( c ) ,
(HA3)
α ( 1 A ) = 1 A ,
(HA4)
a 1 A = 1 A a = α ( a ) .
Let ( A , μ A , η A , α A ) and ( B , μ B , η B , α B ) be two Hom-algebras. Then, a morphism f : A B is called a Hom-algebra morphism if α B f = f α A , f ( 1 A ) = 1 B and μ B ( f f ) = f μ A .
Definition 2
([30,32,34]). A Hom-coalgebra is a quadruple ( C , Δ , ε , α ) (abbr. ( C , α ) ), where C is a k -linear space, Δ : C C × C and η : C k are linear maps with notation Δ ( c ) = c ( 1 ) c ( 2 ) and α is an automorphism of C, satisfying the following conditions for all c C
(HC1)
α ( c ) ( 1 ) α ( c ) ( 2 ) = α ( c ( 1 ) ) α ( c ( 2 ) ) ,
(HC2)
α ( c ( 1 ) ) c ( 2 ) ( 1 ) c ( 2 ) ( 2 ) = c ( 1 ) ( 1 ) c ( 1 ) ( 2 ) α ( c ( 2 ) ) ,
(HC3)
ε α = ε ,
(HC4)
ε ( c ( 1 ) ) c ( 2 ) = c ( 1 ) ε ( c ( 2 ) ) = α ( c ) .
Let ( C , Δ C , ε C , α C ) and ( D , Δ D , ε D , α D ) be two Hom-coalgebras. Then, a morphism f : C D is called a Hom-coalgebra morphism if α D f = f α C , ε D f = ε C and Δ D f = ( f f ) Δ C .
Definition 3
([31,32]). A Hom-bialgebra is a sextuple ( H , μ , η , Δ , ε , β ) (abbr. ( H , β ) ), where ( H , μ , η , β ) is a Hom-algebra and ( H , Δ , ε , β ) is a Hom-coalgebra, such that Δ and ε are Hom-algebra morphisms, i.e.,
(HB1)
Δ ( h h ) = h ( 1 ) h ( 1 ) h ( 2 ) h ( 2 ) ,
(HB2)
Δ ( 1 H ) = 1 H 1 H ,
(HB3)
ε ( h h ) = ε ( h ) ε ( h ) ,
(HB4)
ε ( 1 H ) = 1 k .
Furthermore, if there is a linear map S : H H such that
(HS1)
S ( h ( 1 ) ) h ( 2 ) = h ( 1 ) S ( h ( 2 ) ) = ε ( h ) 1 H ,
(HS2)
S ( β ( h ) ) = β ( S ( h ) ) .
Then, we call ( H , μ , η , Δ , ε , β , S ) (abbr. ( H , β , S ) ) a Hom–Hopf algebra.
The antipode S is both a Hom-anti-algebra morphism and a Hom-anti-coalgebra morphism. A morphism between two Hom-bialgebras is called a Hom-bialgebra morphism if it is both a Hom-algebra and a Hom-coalgebra morphism. A morphism between two Hom–Hopf algebras is called a Hom–Hopf algebra morphism if it is a Hom-bialgebra morphism and compatible with the antipodes. A useful remark is that Hom-bialgebra morphisms of Hom–Hopf algebras are Hom–Hopf algebra morphisms.
Definition 4
([32]). Let ( A , β ) be a Hom-algebra. A left ( A , β ) -Hom-module is a triple ( X , θ , α ) , where X is a k -linear space, θ : A X X (write θ ( a x ) = a x , a A , x X ) is a linear map, and α is an automorphism of X, such that
(HM1)
α ( a x ) = β ( a ) α ( x ) ,
(HM2)
β ( a ) ( b x ) = ( a b ) α ( x ) ,
(HM3)
1 A x = α ( x )
are satisfied for a , b A and x X .
Let ( X , θ X , α X ) and ( Y , θ Y , α Y ) be two left ( A , β ) -Hom-modules. Then, a morphism f : X Y is called a ( A , β ) -Hom-module morphism if α Y f = f α X and θ Y ( id A f ) = f θ X .
Definition 5
([32]). Let ( C , β ) be a Hom-coalgebra. A left ( C , β ) -Hom-comodule is a triple ( X , ρ , α ) , where X is a k -linear space, ρ : X C X (write ρ ( x ) = x [ 1 ] x [ 0 ] , x X ) is a linear map, and α is an automorphism of X, such that
(HCM1)
α ( x ) [ 1 ] α ( x ) [ 0 ] = β ( x [ 1 ] ) α ( x [ 0 ] ) ,
(HCM2)
β ( x [ 1 ] ) x [ 0 ] [ 1 ] x [ 0 ] [ 0 ] = x [ 1 ] ( 1 ) x [ 1 ] ( 2 ) α ( x [ 0 ] ) ,
(HCM3)
ε C ( x [ 1 ] ) x [ 0 ] = α ( x )
are satisfied for x X .
Let ( X , ρ X , α X ) and ( Y , ρ Y , α Y ) be two left ( C , β ) -Hom-comodules. Then, a morphism f : X Y is called a ( C , β ) -Hom-comodule morphism if α Y f = f α X and ρ Y f = ( id C f ) ρ X .
Definition 6
([33]). Let ( H , β ) be a Hom-bialgebra and ( A , α ) be a Hom-algebra. If ( A , , α ) is a left ( H , β ) -Hom-module, and for all h H and a , b A ,
(HMA1)
β 2 ( h ) ( a b ) = ( h ( 1 ) a ) ( h ( 2 ) b ) ,
(HMA2)
h 1 A = ε H ( h ) 1 A ,
then, ( A , , α ) is called a left ( H , β ) -Hom-module Hom-algebra.
Definition 7
([25,34]). Let ( H , β ) be a Hom-bialgebra and ( A , α ) a Hom-algebra. If ( A , ρ , α ) is a left ( H , β ) -Hom-comodule and for all a , b A ,
(HCMA1)
ρ ( a b ) = a [ 1 ] b [ 1 ] a [ 0 ] b [ 0 ] ,
(HCMA2)
ρ ( 1 A ) = 1 H 1 A ,
then, ( A , ρ , α ) is called a left ( H , β ) -Hom-comodule Hom-algebra.
Definition 8
([25]). Let ( H , β ) be a Hom-bialgebra and ( C , α ) be a Hom-coalgebra. If ( C , , α ) is a left ( H , β ) -Hom-module and for all h H and c C ,
(HMC1)
( h c ) ( 1 ) ( h c ) ( 2 ) = ( h ( 1 ) c ( 1 ) ) ( h ( 2 ) c ( 2 ) ) ,
(HMC2)
ε C ( h c ) = ε H ( h ) ε C ( c ) ,
then ( C , , α ) is called a left ( H , β ) -Hom-module Hom-coalgebra.
Definition 9
([25]). Let ( H , β ) be a Hom-bialgebra and ( C , α ) be a Hom-coalgebra. If ( C , ρ , α ) is a left ( H , β ) -Hom-comodule and for all c C ,
(HCMC1)
β 2 ( c [ 1 ] ) c [ 0 ] [ 1 ] c [ 0 ] [ 2 ] = c ( 1 ) [ 1 ] c ( 2 ) [ 1 ] c ( 1 ) [ 0 ] c ( 2 ) [ 0 ] ,
(HCMC2)
c [ 1 ] ε C ( c [ 0 ] ) = 1 H ε C ( c ) ,
then ( C , ρ , α ) is called a left ( H , β ) -Hom-comodule Hom-coalgebra.
Now, we recall the definition of ( θ , ω ) -twisted-Radford’s Hom-biproduct from [26]. First, if ( H , β ) is a Hom-bialgebra, we denote the group of all Hom-bialgebra automorphisms of H by Aut Hom bialg ( H ) .
Definition 10
([26]). Let ( H , β ) be a Hom-bialgebra, θ Aut Hom bialg ( H ) and ( A , , α ) be a left ( H , β ) -Hom-module Hom-algebra. A θ-twisted smash Hom-product ( A θ H , α β ) of ( A , α ) and ( H , β ) is defined as follows. For all a , b A , h , k H :
(i)
as a vector space, A θ H = A H ,
(ii)
the multiplication is given by
( a θ h ) ( b θ k ) = a ( θ ( h ( 1 ) ) α 1 ( b ) ) θ β 1 ( h ( 2 ) ) k .
The θ-twisted smash Hom-product ( A θ H , α β ) is a Hom-algebra with the unit 1 A θ 1 H .
Definition 11
([26]). Let ( H , β ) be a Hom-bialgebra, ω Aut Hom bialg ( H ) and ( A , α ) be a left ( H , β ) -Hom-comodule Hom-coalgebra. A ω-twisted smash Hom-coproduct ( A ω H , α β ) of ( A , α ) and ( H , β ) is defined as follows. For all a A , h H :
(i)
as a vector space, A ω H = A H ,
(ii)
the comultiplication is given by
Δ ( a ω h ) = a ( 1 ) ω ω ( a ( 2 ) [ 1 ] ) β 1 ( h ( 1 ) ) α 1 ( a ( 2 ) [ 0 ] ) ω h ( 2 ) .
The ω-twisted smash Hom-coproduct ( A ω H , α β ) is a Hom-coalgebra with the counit ε A θ ε H .
Theorem 1
([26]). Let ( H , β ) be a Hom-bialgebra, ( A , α ) be a left ( H , β ) -Hom-module Hom-algebra with Hom-module structure ▹, and a left ( H , β ) -Hom-comodule Hom-coalgebra with Hom-comodule structure ρ, and θ , ω Aut Hom bialg ( H ) , ( A θ H , α β ) be a θ-twisted smash Hom-product, and ( A ω H , α β ) be a ω-twisted smash Hom-coproduct. Then, the following two conditions are equivalent:
(i)
( A ω θ H , α β ) is a Hom-bialgebra.
(ii)
The following conditions hold:
(T1)
ε A is a Hom-algebra map and Δ A ( 1 A ) = 1 A 1 A ,
(T2)
( A , α ) is a left ( H , β ) -Hom-module Hom-coalgebra,
(T3)
( A , α ) is a left ( H , β ) -Hom-comodule Hom-algebra,
(T4)
Δ A ( a b ) = a ( 1 ) ( θ ω β 2 ( a ( 2 ) [ 1 ] ) α 1 ( b ( 1 ) ) ) α 1 ( a ( 2 ) [ 0 ] ) b ( 2 ) ,
(T5)
( h ( 1 ) b ) [ 1 ] ω 1 θ 1 β 2 ( h ( 2 ) ) ( h ( 1 ) b ) [ 0 ] = ω 1 θ 1 β 2 ( h ( 1 ) ) β ( b [ 1 ] ) β ( h ( 2 ) ) b [ 0 ] ,
for all a , b A and h H .
Theorem 2
([26]). Suppose ( A ω θ H , α β ) is a ( θ , ω ) -twisted-Radford’s Hom-biproduct. If H is a Hom–Hopf algebra with antipode S H and S A : A A is a linear map, such that S A is a convolution inverse of id A and S A α = α S A , then ( A ω θ H , α β ) is a Hom–Hopf algebra with antipode S described by
S A ω θ H ( a h ) = θ S H ( ω ( a [ 1 ] ) β 1 ( h ) ) ( 1 ) S A ( α 2 ( a [ 0 ] ) ) β 1 S H ( ω ( a [ 1 ] ) β 1 ( h ) ) ( 2 ) .
The ( θ , ω ) -twisted-Radford’s Hom-biproduct ( A ω θ H , α β ) is a Hom–Hopf algebra and is denoted by ( A × ω θ H , α β ) , for short, and write elements a × h A × ω θ H .

3. Factorization of Certain Twisted Hom-Biproduct Endomorphisms

Let ( A × ω θ H , α β ) be a ( θ , ω ) -twisted-Radford’s Hom-biproduct. We define π : A × ω θ H H by π ( a × h ) = ε A ( a ) h for all a A , h H and j : H A × ω θ H by j ( h ) = 1 A × h for all h H , are Hom–Hopf algebra maps which satisfy π j = id H . Let End Hom - Hopf ( A × ω θ H , π ) be the set of all Hom–Hopf algebra endomorphisms F of ( A × ω θ H , α β ) such that π F = π and let Aut Hom - Hopf ( A × ω θ H , π ) be its set of units, which is the group of Hom–Hopf algebra automorphisms F of ( A × ω θ H , α β ) such that π F = π under composition. The purpose of this section is to show that F has a factorization closely related to the factors ( A , α ) and ( H , β ) of ( A × ω θ H , α β ) .
We define Π : A × ω θ H A and J : A A × ω θ H by Π ( a × h ) = a ε H ( h ) , for all a A , h H and J ( a ) = a × 1 H , for all a A , respectively. There is a fundamental relationship between these four maps given by:
J Π ( α 2 β 2 ) = id A × ω θ H ( j S H π )
The factorization of F is given in terms of F L : A A and F R : H A defined by:
F L = Π F J and F R = Π F j .
First, we shall reveal the relationships among F, F L and F R in the following lemma.
Lemma 1.
Let F End Hom - Hopf ( A × ω θ H , π ) . Then:
F L α = α F L and F R β = α F R ,
F L ( a ) × 1 H = F ( a × 1 H ) ,
F R ( β 2 ( h ) ) × 1 H = F ( 1 A × h ( 1 ) ) ( 1 A × S H ( h ( 2 ) ) ) ,
F ( a × h ) = F L ( α 1 ( a ) ) F R ( β 2 ( h ( 1 ) ) ) × β 1 ( h ( 2 ) ) .
for all a A and h H .
Proof. 
According to the definition of these four maps and F, we have
F L α = Π F J α = Π F ( α β ) J = Π ( α β ) F J = α Π F J = α F L ,
and
F R β = Π F j α = Π F ( α β ) j = Π ( α β ) F j = α Π F j = α F L
Next, we need to calculate J Π ( α 2 β 2 ) F . For a A and h H , we use (4) to compute
J Π ( α 2 β 2 ) F ( a × h ) = F ( ( a × h ) ( 1 ) ) ( j S H π ) F ( ( a × h ) ( 2 ) ) = F ( ( a × h ) ( 1 ) ) ( j S H π ) ( ( a × h ) ( 2 ) ) = F ( a ( 1 ) × ω ( a ( 2 ) [ 1 ] ) β 1 ( h ( 1 ) ) ) ( j S H π ) ( α 1 ( a ( 2 ) [ 0 ] ) × h ( 2 ) ) = F ( α ( a ) × h ( 1 ) ) ( 1 A × S H ( h ( 2 ) ) ) .
It follows that
J Π F ( α 2 ( a ) × β 2 ( h ) ) = F ( α ( a ) × h ( 1 ) ) ( 1 A × S H ( h ( 2 ) ) ) ,
for a A and h H . Equations (7) and (8) follow from the above equation via setting h = 1 H and a = 1 A respectively. As for (9), we calculate
F ( a × h ) = F ( α 1 ( a ) × 1 H ) F ( 1 A × β 1 ( h ) ) = F ( α 1 ( a ) × 1 H ) F ( 1 A × β 3 ( h ( 1 ) ) ) 1 A × S H ( β 5 ( h ( 2 ) ( 1 ) ) ) 1 A × β 5 ( h ( 2 ) ( 2 ) ) = F ( α 1 ( a ) × 1 H ) F ( 1 A × β 4 ( h ( 1 ) ) ) 1 A × S H ( β 5 ( h ( 2 ) ( 1 ) ) ) 1 A × β 4 ( h ( 2 ) ( 2 ) ) = F ( α 1 ( a ) × 1 H ) F ( 1 A × β 5 ( h ( 1 ) ( 1 ) ) ) 1 A × S H ( β 5 ( h ( 1 ) ( 2 ) ) ) 1 A × β 3 ( h ( 2 ) ) = F ( α 1 ( a ) × 1 H ) ( F R ( β 3 ( h ( 1 ) ) ) × 1 H ) ( 1 A × β 3 ( h ( 2 ) ) ) = F ( α 2 ( a ) × 1 H ) ( F R ( β 3 ( h ( 1 ) ) ) × 1 H ) ( 1 A × β 2 ( h ( 2 ) ) ) = ( F L ( α 2 ( a ) ) × 1 H ) ( F R ( β 3 ( h ( 1 ) ) ) × 1 H ) ( 1 A × β 2 ( h ( 2 ) ) ) = ( F L ( α 2 ( a ) ) F R ( β 3 ( h ( 1 ) ) ) × 1 H ) ( 1 A × β 2 ( h ( 2 ) ) ) = F L ( α 1 ( a ) ) F R ( β 2 ( h ( 1 ) ) ) × β 1 ( h ( 2 ) )
as desired. □
By (7) and (8) of the above Lemma:
( id A × ω θ H ) L = id A and ( id A × ω θ H ) R = η A ε H
Since F L ( 1 A ) = 1 A by Equation (7). By (9) of Lemma 1:
F ( 1 A × h ) = F R ( β 1 ( h ( 1 ) ) ) × β 1 ( h ( 2 ) )
for all h H . Now, we can calculate the factors of a composite.
Corollary 1.
Let F End Hom - Hopf ( A × ω θ H , π ) . Then:
(i)
( F G ) L = F L G L ,
(ii)
( F G ) R β 2 = ( F L G R ) F R .
Proof. 
(i) For any a A , by (7) of Lemma 1, we have
( F G ) L ( a ) × 1 H = ( F G ) ( a × 1 H ) = F ( G L ( a ) × 1 H ) = F L G L ( a ) × 1 H .
(ii) For all h H . Using Equation (11), the fact F is multiplicative and Equation (7), we obtain that
( F G ) R ( β 1 ( h ( 1 ) ) ) × β 1 ( h ( 2 ) ) = ( F G ) ( 1 A × h ) = F G R ( β 1 ( h ( 1 ) ) ) × β 1 ( h ( 2 ) ) = F G R ( β 2 ( h ( 1 ) ) ) × 1 H F 1 A × β 2 ( h ( 2 ) ) = F L G R ( β 2 ( h ( 1 ) ) ) × 1 H F R ( β 3 ( h ( 2 ) ( 1 ) ) ) × β 3 ( h ( 2 ) ( 2 ) ) = F L G R ( β 2 ( h ( 1 ) ) ) θ ( 1 H ) F R ( β 4 ( h ( 2 ) ( 1 ) ) ) × 1 H β 3 ( h ( 2 ) ( 2 ) ) = F L G R ( β 2 ( h ( 1 ) ) ) F R ( β 3 ( h ( 2 ) ( 1 ) ) ) × β 2 ( h ( 2 ) ( 2 ) ) = F L G R ( β 3 ( h ( 1 ) ( 1 ) ) ) F R ( β 3 ( h ( 1 ) ( 2 ) ) ) × β 1 ( h ( 2 ) )
i.e.,
( F G ) R ( β 1 ( h ( 1 ) ) ) × β 1 ( h ( 2 ) ) = F L G R ( β 3 ( h ( 1 ) ( 1 ) ) ) F R ( β 3 ( h ( 1 ) ( 2 ) ) ) × β 1 ( h ( 2 ) )
Applying id A ε H to both sides of the above equation, and replacing h with β 2 ( h ) , we can obtain part (ii). □
According to Lemma 1, in order to characterize F, we must characterize F L and F R . First, we shall characterize F L in the following lemma.
Lemma 2.
Let F End Hom - Hopf ( A × ω θ H , π ) . Then:
(i)
F L : A A is a Hom-algebra endomorphism.
(ii)
ε A F L = ε A .
(iii)
For all a A ,
Δ ( F L ( a ) ) = F L ( α 1 ( a ( 1 ) ) ) F R ( ω ( a ( 2 ) [ 1 ] ) ) F L ( α 1 ( a ( 2 ) [ 0 ] ) ) .
(iv)
For all a A ,
ρ ( F L ( a ) ) = a [ 1 ] F L ( a [ 0 ] ) .
(v)
For all a A and h H ,
F L ( θ ( h ( 1 ) ) a ) F R ( β 1 ( h ( 2 ) ) ) = F R ( β 1 ( h ( 1 ) ) ) ( θ ( h ( 2 ) ) F L ( a ) ) .
Proof. 
(i) We need to check three aspects. From the above discussion, we have known that F L ( 1 A ) = 1 A and F L α = α F L have checked in Lemma 1. Finally, we shall check that F L preserves the multiplication. In fact, for a , b A , we have
F L ( a b ) = ( Π F J ) ( a b ) = Π ( F ( a b × 1 H ) ) = Π ( F ( a × 1 H ) F ( b × 1 H ) ) = Π ( ( F L ( a ) × 1 H ) ( F L ( b ) × 1 H ) ) = F L ( a ) F L ( b ) .
It is easy to check part (ii), since ε A Π = Π ε A × ω θ H , ε A × ω θ H J = ε A and F is a Hom-coalgebra morphism.
Next, we will check that parts (iii) and (iv) hold. As a matter of fact, we compute the coproduct of F L ( a ) × 1 H = F ( a × 1 H ) in two ways. Firstly, we have
Δ ( F L ( a ) × 1 H ) = F L ( a ) ( 1 ) × β ω ( F L ( a ) ( 2 ) [ 1 ] ) α 1 ( F L ( a ) ( 2 ) [ 0 ] ) × 1 H
and secondly, since F is a Hom-coalgebra map, we have
Δ ( F ( a × 1 H ) ) = F ( ( a × 1 H ) ( 1 ) ) F ( ( a × 1 H ) ( 2 ) ) = F ( a ( 1 ) × β ω ( a ( 2 ) [ 1 ] ) ) F ( α 1 ( a ( 2 ) [ 0 ] ) × 1 H ) = ( 9 ) F L ( α 1 ( a ( 1 ) ) ) F R ( β 1 ω ( a ( 2 ) [ 1 ] ( 1 ) ) ) × ω ( a ( 2 ) [ 1 ] ( 2 ) ) F L ( α 1 ( a ( 2 ) [ 0 ] ) ) × 1 H .
It follows that
F L ( a ) ( 1 ) × β ω ( F L ( a ) ( 2 ) [ 1 ] ) α 1 ( F L ( a ) ( 2 ) [ 0 ] ) × 1 H = F L ( α 1 ( a ( 1 ) ) ) F R ( β 1 ω ( a ( 2 ) [ 1 ] ( 1 ) ) ) × ω ( a ( 2 ) [ 1 ] ( 2 ) ) F L ( α 1 ( a ( 2 ) [ 0 ] ) ) × 1 H .
Applying id A ε H id A ε H to both sides of the above equation yields (12). It follows easily that ε A F R = ε H from (11). Applying ε A id H id A ε H to both sides of the above equation again, we can gain
ω β 2 ( F L ( a ) [ 1 ] ) F L ( a ) [ 0 ] = ω β 2 ( a [ 1 ] ) F L ( a [ 0 ] ) ,
and applying β 2 ω 1 id A to both sides, we obtain (13).
(v) Finally, it is left to us to check (14). Indeed, for a A and h H , we have
F ( ( 1 A × h ) ( a × 1 H ) ) = F ( ( β θ ( h ( 1 ) ) a ) × h ( 2 ) ) = F L ( θ ( h ( 1 ) ) α 1 ( a ) ) F R ( β 2 ( h ( 2 ) ( 1 ) ) ) × β 1 ( h ( 2 ) ( 2 ) ) .
On the other hand, Since F preserves the multiplication, we compute:
F ( ( 1 A × h ) ( a × 1 H ) ) = F ( 1 A × h ) F ( a × 1 H ) = F R ( β 1 ( h ( 1 ) ) ) × β 1 ( h ( 2 ) ) ( F L ( a ) × 1 H ) = F R ( β 1 ( h ( 1 ) ) ) θ β 1 ( h ( 2 ) ( 1 ) ) α 1 F L ( a ) × β 1 ( h ( 2 ) ( 2 ) ) .
Thus,
F L ( θ ( h ( 1 ) ) α 1 ( a ) ) F R ( β 2 ( h ( 2 ) ( 1 ) ) ) × β 1 ( h ( 2 ) ( 2 ) ) = F R ( β 1 ( h ( 1 ) ) ) θ β 1 ( h ( 2 ) ( 1 ) ) α 1 F L ( a ) × β 1 ( h ( 2 ) ( 2 ) ) .
Applying id A × ε H to the above equation and replacing a with α ( a ) , we obtain (14). □
As the reader might suspect, whether or not F L is a Hom-coalgebra morphism is explained in terms of F R .
Corollary 2.
Let F End Hom - Hopf ( A × ω θ H , π ) . Then, F L is a Hom-coalgebra morphism if, and only if, F R ( ω ( c [ 1 ] ) ) c [ 0 ] = 1 A α ( c ) , for any c im ( F L ) .
Proof. 
Suppose F R ( ω ( c [ 1 ] ) ) c [ 0 ] = 1 A α ( c [ 0 ] ) , for any c im ( F L ) . Then, we have
Δ ( F L ( a ) ) = F L ( α 1 ( a ( 1 ) ) ) F R ( ω ( a ( 2 ) [ 1 ] ) ) F L ( α 1 ( a ( 2 ) [ 0 ] ) ) = ( 13 ) F L ( α 1 ( a ( 1 ) ) ) F R ( ω ( F L ( a ( 2 ) ) [ 1 ] ) ) α 1 ( F L ( a ( 2 ) ) [ 0 ] ) = F L ( a ( 1 ) ) F L ( a ( 2 ) )
Conversely, suppose that F L is a Hom-coalgebra morphism. For all a A , we compute
F R ( ω ( F L ( a ) [ 1 ] ) ) F L ( a ) [ 0 ] = ( 13 ) F R ( ω ( a [ 1 ] ) ) F L ( a [ 0 ] ) = ( HC 4 ) F L ( ε A ( α 5 ( a ( 1 ) ) ) 1 A ) F R ( ω β 2 ( a ( 2 ) [ 1 ] ) ) F L ( α 1 ( a ( 2 ) [ 0 ] ) ) = ( HS 1 ) F L ( S A ( α 5 ( a ( 1 ) ( 1 ) ) ) ) F L ( α 5 ( a ( 1 ) ( 2 ) ) ) F R ( ω β 2 ( a ( 2 ) [ 1 ] ) ) F L ( α 1 ( a ( 2 ) [ 0 ] ) ) = ( HC 2 ) F L ( S A ( α 4 ( a ( 1 ) ) ) ) F L ( α 5 ( a ( 2 ) ( 1 ) ) ) F R ( ω β 3 ( a ( 2 ) ( 2 ) [ 1 ] ) ) F L ( α 2 ( a ( 2 ) ( 2 ) [ 0 ] ) ) = ( HA 2 ) F L ( S A ( α 3 ( a ( 1 ) ) ) ) F L ( α 5 ( a ( 2 ) ( 1 ) ) ) F R ( ω β 4 ( a ( 2 ) ( 2 ) [ 1 ] ) ) α 3 ( F L ( α 5 ( a ( 2 ) ( 2 ) [ 0 ] ) ) ) = ( 12 ) F L ( S A ( α 3 ( a ( 1 ) ) ) ) F L ( α 4 ( a ( 2 ) ) ) ( 1 ) α 3 ( F L ( α 4 ( a ( 2 ) ) ) ( 2 ) ) = F L ( S A ( α 3 ( a ( 1 ) ) ) ) F L ( α 4 ( a ( 2 ) ( 1 ) ) ) F L ( α 1 ( a ( 2 ) ( 2 ) ) ) = ( HC 2 ) F L ( S A ( α 4 ( a ( 1 ) ( 1 ) ) ) ) F L ( α 4 ( a ( 1 ) ( 2 ) ) ) F L ( a ( 2 ) ) = 1 A α ( F L ( a ) ) ,
as desired. □
From Lemma 2, we have characterized the conditions that F L satisfies. Pay special attention to its part (v) describing a commutation relation between F L and F R . It is left to us to characterize F R as follows.
Lemma 3.
Let F End Hom - Hopf ( A × ω θ H , π ) . Then:
(i)
F R ( 1 H ) = 1 A .
(ii)
For all h , k H ,
F R ( h k ) = F R ( β 1 ( h ( 1 ) ) ) θ ( h ( 2 ) ) F R ( β 1 ( k ) ) .
(iii)
F R : H A is a Hom-coalgebra morphism.
(iv)
For all h H ,
ρ ( F R ( h ) ) = ω 1 β 4 ( h ( 1 ) ( 1 ) ) S H ( β 3 ( h ( 2 ) ) ) F R ( β 1 ( h ( 1 ) ( 2 ) ) ) .
Proof. 
(i) By (11), we have F ( 1 A × h ) = F R ( β 1 ( h ( 1 ) ) ) × β 1 ( h ( 2 ) ) for h H . Since F is a Hom-algebra morphism and β is an automorphism, 1 A × 1 H = F ( 1 A × 1 H ) = F R ( 1 A ) × 1 H , which implies F R ( 1 H ) = 1 A .
(ii) For h , k H , we have
F ( 1 A × h k ) = F R ( β 1 ( h ( 1 ) ) β 1 ( k ( 1 ) ) ) × β 1 ( h ( 2 ) ) β 1 ( k ( 2 ) ) .
and on the other hand,
F ( 1 A × h k ) = F ( 1 A × h ) F ( 1 A × k ) = F R ( β 1 ( h ( 1 ) ) ) × β 1 ( h ( 2 ) ) F R ( β 1 ( k ( 1 ) ) ) × β 1 ( k ( 2 ) ) = F R ( β 1 ( h ( 1 ) ) ) θ β 1 ( h ( 2 ) ( 1 ) ) F R ( β 2 ( k ( 1 ) ) ) × β 2 ( h ( 2 ) ( 2 ) ) β 1 ( k ( 2 ) )
Applying id A ε H to both expressions for F ( 1 A × h k ) , it follows that (15) holds.
To prove parts (iii) and (iv), we calculate Δ ( F ( 1 A × h ) ) , for all h H , in two ways as follows.
Δ ( F ( 1 A × h ) ) = F ( 1 A × h ( 1 ) ) F ( 1 A × h ( 2 ) ) = F R ( β 1 ( h ( 1 ) ( 1 ) ) ) × β 1 ( h ( 1 ) ( 2 ) ) F R ( β 1 ( h ( 2 ) ( 1 ) ) ) × β 1 ( h ( 2 ) ( 2 ) ) .
On the other hand,
Δ ( F ( 1 A × h ) ) = Δ ( F R ( β 1 ( h ( 1 ) ) ) × β 1 ( h ( 2 ) ) ) = F R ( β 1 ( h ( 1 ) ) ) ( 1 ) × ω ( F R ( β 1 ( h ( 1 ) ) ) ( 2 ) [ 1 ] ) β 2 ( h ( 2 ) ( 1 ) ) α 1 ( F R ( β 1 ( h ( 1 ) ) ) ( 2 ) [ 0 ] ) × β 1 ( h ( 2 ) ( 2 ) )
Applying id A ε H id A ε H to the expressions for Δ ( F ( 1 A × h ) ) gives F R ( h ) ( 1 ) F R ( h ) ( 2 ) = F R ( h ( 1 ) ) F R ( h ( 2 ) ) . since α F R = F R β and ε A F R = ε H have been discussed before, and hence, part (iii) holds.
Applying ε A id H id A ε H to the expressions for Δ ( F ( 1 A × h ) ) again yields
h ( 1 ) F R ( h ( 2 ) ) = ω β F R ( β 1 ( h ( 1 ) ) ) [ 1 ] β 1 ( h ( 2 ) ) F R ( β 1 ( h ( 1 ) ) ) [ 0 ]
Therefore,
( ω id A ) ( ρ ( F R ( h ) ) ) = β 1 ω ( F R ( β 1 ( h ( 1 ) ) ) [ 1 ] ) ε H ( β 5 ( h ( 2 ) ) ) 1 H F R ( β 1 ( h ( 1 ) ) ) [ 0 ] = ( HS 1 ) β 1 ω ( F R ( β 1 ( h ( 1 ) ) ) [ 1 ] ) β 5 ( h ( 2 ) ( 1 ) ) S H ( β 5 ( h ( 2 ) ( 2 ) ) ) F R ( β 1 ( h ( 1 ) ) ) [ 0 ] = ( HA 2 ) β 2 ω ( F R ( β 1 ( h ( 1 ) ) ) [ 1 ] ) β 5 ( h ( 2 ) ( 1 ) ) S H ( β 4 ( h ( 2 ) ( 2 ) ) ) F R ( β 1 ( h ( 1 ) ) ) [ 0 ] = ( HC 2 ) β 2 ω ( F R ( β 2 ( h ( 1 ) ( 1 ) ) ) [ 1 ] ) β 5 ( h ( 1 ) ( 2 ) ) S H ( β 3 ( h ( 2 ) ) ) F R ( β 2 ( h ( 1 ) ( 1 ) ) ) [ 0 ] = β 2 ω β ( F R ( β 3 ( h ( 1 ) ( 1 ) ) ) [ 1 ] ) β 3 ( h ( 1 ) ( 2 ) ) S H ( β 3 ( h ( 2 ) ) ) α ( F R ( β 3 ( h ( 1 ) ( 1 ) ) ) [ 0 ] ) = ( 17 ) β 4 ( h ( 1 ) ( 1 ) ) S H ( β 3 ( h ( 2 ) ) ) F R ( β 1 ( h ( 1 ) ( 2 ) ) ) .
it follows that
( ω id A ) ( ρ ( F R ( h ) ) ) = β 4 ( h ( 1 ) ( 1 ) ) S H ( β 3 ( h ( 2 ) ) ) F R ( β 1 ( h ( 1 ) ( 2 ) ) ) .
Applying ω 1 id A to both sides of the equation gives part (iv). □
Corollary 3.
Let F End Hom - Hopf ( A × ω θ H , π ) . Then, F L is a left ( H , β ) -Hom-module morphism if and only if the condition F L ( θ ( h ( 1 ) ) a ) F R ( β 1 ( h ( 2 ) ) ) = F R ( β 1 ( h ( 1 ) ) ) F L ( θ ( h ( 2 ) ) a ) holds for all a A and h H .
Proof. 
The necessary condition can be followed easily from (14) of Lemma 2. Now, we shall prove a sufficient part. Suppose that the condition holds. Note that F R is a Hom-coalgebra morphism by Lemma 3(iii). Using this fact and (14), for all a A and h H , we have
θ ( h ) F L ( a ) = S H ( F R ( β 3 ( h ( 1 ) ( 1 ) ) ) ) F R ( β 3 ( h ( 1 ) ( 2 ) ) ) θ β 2 ( h ( 2 ) ) F L ( α 1 ( a ) ) = ( HA 2 ) S H ( F R ( β 2 ( h ( 1 ) ( 1 ) ) ) ) F R ( β 3 ( h ( 1 ) ( 2 ) ) ) θ β 3 ( h ( 2 ) ) F L ( α 2 ( a ) ) = ( HC 2 ) S H ( F R ( β 1 ( h ( 1 ) ) ) ) F R ( β 3 ( h ( 2 ) ( 1 ) ) ) θ β 4 ( h ( 2 ) ( 2 ) ) F L ( α 2 ( a ) ) = ( 14 ) S H ( F R ( β 1 ( h ( 1 ) ) ) ) F L ( θ β 2 ( h ( 2 ) ( 1 ) ) α 2 ( a ) ) F R ( β 5 ( h ( 2 ) ( 2 ) ) ) = S H ( F R ( β 1 ( h ( 1 ) ) ) ) F R ( β 3 ( h ( 2 ) ( 1 ) ) ) F L ( θ β 4 ( h ( 2 ) ( 2 ) ) α 2 ( a ) ) = ( HA 2 ) S H ( F R ( β 2 ( h ( 1 ) ) ) ) F R ( β 3 ( h ( 2 ) ( 1 ) ) ) F L ( θ β 3 ( h ( 2 ) ( 2 ) ) α 1 ( a ) ) = ( HC 2 ) S H ( F R ( β 3 ( h ( 1 ) ( 1 ) ) ) ) F R ( β 3 ( h ( 1 ) ( 2 ) ) ) F L ( θ β 2 ( h ( 2 ) ) α 1 ( a ) ) = ε A ( F R ( β 3 ( h ( 1 ) ) ) ) 1 A F L ( θ β 2 ( h ( 2 ) ) α 1 ( a ) ) = F L ( θ ( h ) a ) ,
and since θ Aut Hom - Hopf ( H ) , we show that F L is a left ( H , β ) -Hom-module morphism. □
Lemma 4.
Let F End Hom - Hopf ( A × ω θ H , π ) . Then, F R is a Hom-algebra morphism if, and only if, h F R ( k ) = ε H ( h ) F R ( β ( k ) ) for all h , k H .
Proof. 
Suppose that F R is a Hom-algebra morphism. Using (ii) and (iii) of Lemma 3, for h , k H , we have
θ β 3 ( h ) F R ( k ) = ε H ( h ( 1 ) ) 1 A θ β ( h ( 2 ) ) F R ( β 1 ( k ) ) = ( HS 1 ) S A ( F R ( β 2 ( h ( 1 ) ( 1 ) ) ) ) F R ( β 2 ( h ( 1 ) ( 2 ) ) ) θ β ( h ( 2 ) ) F R ( β 1 ( k ) ) = ( HA 2 ) S A ( F R ( β 1 ( h ( 1 ) ( 1 ) ) ) ) F R ( β 2 ( h ( 1 ) ( 2 ) ) ) θ ( h ( 2 ) ) F R ( β 2 ( k ) ) = ( HC 2 ) S A ( F R ( h ( 1 ) ) ) F R ( β 2 ( h ( 2 ) ( 1 ) ) ) θ β 1 ( h ( 2 ) ( 2 ) ) F R ( β 2 ( k ) ) = ( 15 ) S A ( F R ( h ( 1 ) ) ) F R ( β 1 ( h ( 2 ) k ) ) = S A ( F R ( h ( 1 ) ) ) F R ( β 1 ( h ( 2 ) ) ) F R ( β 1 ( k ) ) = ( HA 2 ) S A ( F R ( β 1 ( h ( 1 ) ) ) ) F R ( β 1 ( h ( 2 ) ) ) F R ( k ) = ε H ( h ) F R ( β ( k ) )
i.e., θ β 3 ( h ) F R ( k ) = ε H ( h ) F R ( β ( k ) ) . Replacing h by β 3 θ 1 ( h ) yields the condition.
Conversely, if h F R ( k ) = ε H ( h ) F R ( β ( k ) ) holds, by using Equation (15), we have
F R ( h k ) = F R ( β 1 ( h ( 1 ) ) ) θ ( h ( 2 ) ) F R ( β 1 ( k ) ) = F R ( β 1 ( h ( 1 ) ) ) ε H ( h ( 2 ) ) F R ( k ) = F R ( h ) F R ( k ) .
The proof is completed. □
Corollary 4.
Let F End Hom - Hopf ( A × ω θ H , π ) . Then, F R has a convolution inverse J R defined by J R ( h ) = θ ( h ( 1 ) ) F R ( β 2 ( S H ( h ( 2 ) ) ) ) for h H .
Proof. 
Let h H . Then, by parts (i) and (ii) of Lemma 3, we have
F R J R ( h ) = F R ( h ( 1 ) ) J R ( h ( 2 ) ) = F R ( h ( 1 ) ) θ ( h ( 2 ) ( 1 ) ) F R ( β 2 ( S H ( h ( 2 ) ( 2 ) ) ) ) = F R ( β 1 ( h ( 1 ) ( 1 ) ) ) θ ( h ( 1 ) ( 2 ) ) F R ( β 1 ( S H ( h ( 2 ) ) ) ) = ( 15 ) F R ( h ( 1 ) S H ( h ( 2 ) ) ) = ε H ( h ) 1 A ,
and using the fact that ( A , α ) is a left ( H , β ) -Hom-module algebra, we have
J R F R ( h ) = θ ( h ( 1 ) ( 1 ) ) F R ( β 2 ( S H ( h ( 1 ) ( 2 ) ) ) ) F R ( h ( 2 ) ) = θ β 1 ( h ( 1 ) ( 1 ) ( 1 ) ) F R ( β 2 ( S H ( h ( 1 ) ( 2 ) ) ) ) θ β 3 ( h ( 1 ) ( 1 ) ( 2 ) ( 1 ) ) θ S H β 3 ( h ( 1 ) ( 1 ) ( 2 ) ( 2 ) ) F R ( β 1 ( h ( 2 ) ) ) = ( HM 2 ) θ β 1 ( h ( 1 ) ( 1 ) ( 1 ) ) F R ( β 2 ( S H ( h ( 1 ) ( 2 ) ) ) ) θ β 2 ( h ( 1 ) ( 1 ) ( 2 ) ( 1 ) ) θ S H β 3 ( h ( 1 ) ( 1 ) ( 2 ) ( 2 ) ) F R ( β 2 ( h ( 2 ) ) ) = ( HC 2 ) θ β 2 ( h ( 1 ) ( 1 ) ( 1 ) ( 1 ) ) F R ( β 2 ( S H ( h ( 1 ) ( 2 ) ) ) ) θ β 2 ( h ( 1 ) ( 1 ) ( 1 ) ( 2 ) ) θ S H β 2 ( h ( 1 ) ( 1 ) ( 2 ) ) F R ( β 2 ( h ( 2 ) ) ) = ( HM 1 ) θ ( h ( 1 ) ( 1 ) ( 1 ) ) F R ( β 2 ( S H ( h ( 1 ) ( 2 ) ) ) ) θ S H β 2 ( h ( 1 ) ( 1 ) ( 2 ) ) F R ( β 2 ( h ( 2 ) ) ) = ( HC 2 ) θ β ( h ( 1 ) ( 1 ) ) F R ( β 3 ( S H ( h ( 1 ) ( 2 ) ( 2 ) ) ) ) θ S H β 2 ( h ( 1 ) ( 2 ) ( 1 ) ) F R ( β 2 ( h ( 2 ) ) ) = θ β ( h ( 1 ) ( 1 ) ) F R ( β 3 ( S H ( h ( 1 ) ( 2 ) ) ( 1 ) ) ) θ β 2 ( S H ( h ( 1 ) ( 2 ) ) ( 2 ) ) F R ( β 2 ( h ( 2 ) ) ) = ( HC 2 ) θ β 2 ( h ( 1 ) ) F R ( β 3 ( S H ( h ( 2 ) ( 1 ) ) ( 1 ) ) ) θ β 2 ( S H ( h ( 2 ) ( 1 ) ) ( 2 ) ) F R ( β 3 ( h ( 2 ) ( 2 ) ) ) = ( 15 ) θ β 2 ( h ( 1 ) ) F R β 2 ( S H ( h ( 2 ) ( 1 ) ) ) β 2 ( h ( 2 ) ( 2 ) ) = θ β 3 ( h ) F R ( 1 H ) = ε H ( h ) 1 A .
The proof is completed. □
Using the above lemmas and corollaries that we have, we can gain the main result.
Theorem 3.
Let ( A × ω θ H , α β ) be the ( θ , ω ) -twisted-Radford’s Hom-biproduct, let π : A × ω θ H H be the projection from A × ω θ H onto H, and let F A , H be the set of pairs ( L , R ) , where L : A A , R : H A are morphisms which satisfy the conclusions of Lemmas 2 and 3 for F L and F R , respectively. Then,
(i)
The function Φ : F A , H End Hom - Hopf ( A × ω θ H , π ) , described by ( L , R ) F , where F ( a × h ) = L ( α 1 ( a ) ) R ( β 2 ( h ( 1 ) ) ) × β 1 ( h ( 2 ) ) , for all a H and h H , is a bijection. Furthermore, F L = L and F R = R .
(ii)
Suppose ( L , R ) F A , H , then, F Aut Hom - Hopf ( A × ω θ H , π ) if, and only if, L is a bijection.
Proof. 
(i) Assume that the function is well-defined. We define Ψ : End Hom - Hopf ( A × ω θ H , π ) F A , H by Ψ ( F ) = ( Π F J , Π F j ) . It is easily proven that Φ and Ψ are mutually inverse; see Equations (5) and (9). According to the previous results, to complete the proof of part (a), we only need to prove that elements of F A , H give rise to elements of End Hom - Hopf ( A × ω θ H , π ) , as indicated. As the reader might suspect, the proof that F End Hom - Hopf ( A × ω θ H , π ) is somewhat tedious. We will use Lemmas 1–3, which do not need special citations initially.
It is easy to see that π F = π . Obviously, ( α β ) F = F ( α β ) . Note that F ( 1 A × 1 H ) = 1 A × 1 H and
ε ( F ( a × h ) ) = ε L ( α 1 ( a ) ) R ( β 2 ( h ( 1 ) ) ) × β 1 ( h ( 2 ) ) = ε A L ( α 1 ( a ) ) R ( β 2 ( h ( 1 ) ) ) ε H ( β 1 ( h ( 2 ) ) ) = ε A L ( α 1 ( a ) ) ε A R ( β 2 ( h ( 1 ) ) ) ε H ( β 1 ( h ( 2 ) ) ) = ε A ( a ) ε H ( h ) ,
for a A and h H , which means that ε F = ε .
Let a , b A and h , k H . Then,
F ( ( a × h ) ( b × k ) ) = F ( a ( θ ( h ( 1 ) ) α 1 ( b ) ) × β 1 ( h ( 2 ) ) k ) = L ( α 1 ( a ) ) L ( θ β 1 ( h ( 1 ) ) α 2 ( b ) ) R ( β 3 ( h ( 2 ) ( 1 ) ) β 2 ( k ( 1 ) ) ) × β 2 ( h ( 2 ) ( 2 ) ) β 1 ( k ( 2 ) ) = ( 15 ) ( L ( α 1 ( a ) ) L ( θ β 1 ( h ( 1 ) ) α 2 ( b ) ) R ( β 4 ( h ( 2 ) ( 1 ) ( 1 ) ) ) θ β 3 ( h ( 2 ) ( 1 ) ( 2 ) ) R ( β 3 ( k ( 1 ) ) ) ) × β 2 ( h ( 2 ) ( 2 ) ) β 1 ( k ( 2 ) ) = ( HA 2 ) ( L ( α 1 ( a ) ) L ( θ β 2 ( h ( 1 ) ) α 3 ( b ) ) R ( β 5 ( h ( 2 ) ( 1 ) ( 1 ) ) ) θ β 2 ( h ( 2 ) ( 1 ) ( 2 ) ) R ( β 2 ( k ( 1 ) ) ) ) × β 2 ( h ( 2 ) ( 2 ) ) β 1 ( k ( 2 ) ) = ( HC 2 ) ( L ( α 1 ( a ) ) L ( θ β 3 ( h ( 1 ) ( 1 ) ) α 3 ( b ) ) R ( β 4 ( h ( 1 ) ( 2 ) ) ) θ β 1 ( h ( 2 ) ( 1 ) ) R ( β 2 ( k ( 1 ) ) ) ) × β 2 ( h ( 2 ) ( 2 ) ) β 1 ( k ( 2 ) ) = ( 14 ) ( L ( α 1 ( a ) ) R ( β 4 ( h ( 1 ) ( 1 ) ) ) ( θ β 3 ( h ( 1 ) ( 2 ) ) L ( α 3 ( b ) ) ) θ β 1 ( h ( 2 ) ( 1 ) ) R ( β 2 ( k ( 1 ) ) ) ) × β 2 ( h ( 2 ) ( 2 ) ) β 1 ( k ( 2 ) ) = ( HA 2 ) ( L ( α 1 ( a ) ) R ( β 3 ( h ( 1 ) ( 1 ) ) ) θ β 2 ( h ( 1 ) ( 2 ) ) L ( α 2 ( b ) ) θ β 2 ( h ( 2 ) ( 1 ) ) R ( β 3 ( k ( 1 ) ) ) ) × β 2 ( h ( 2 ) ( 2 ) ) β 1 ( k ( 2 ) ) = ( HC 2 ) ( L ( α 1 ( a ) ) R ( β 2 ( h ( 1 ) ) ) θ β 3 ( h ( 2 ) ( 1 ) ( 1 ) ) L ( α 2 ( b ) ) θ β 3 ( h ( 2 ) ( 1 ) ( 2 ) ) R ( β 3 ( k ( 1 ) ) ) ) × β 2 ( h ( 2 ) ( 2 ) ) β 1 ( k ( 2 ) ) = ( HMA 1 ) L ( α 1 ( a ) ) R ( β 2 ( h ( 1 ) ) ) θ β 1 ( h ( 2 ) ( 1 ) ) L ( α 2 ( b ) ) R ( β 3 ( k ( 1 ) ) ) × β 2 ( h ( 2 ) ( 2 ) ) β 1 ( k ( 2 ) ) = L ( α 1 ( a ) ) R ( β 2 ( h ( 1 ) ) ) θ β 1 ( h ( 2 ) ( 1 ) ) α 1 ( L ( α 1 ( b ) ) R ( β 2 ( k ( 1 ) ) ) ) × β 2 ( h ( 2 ) ( 2 ) ) β 1 ( k ( 2 ) ) = L ( α 1 ( a ) ) R ( β 2 ( h ( 1 ) ) ) × β 1 ( h ( 2 ) ) L ( α 1 ( b ) ) R ( β 2 ( k ( 1 ) ) ) × β 1 ( k ( 2 ) ) = F ( a × h ) F ( b × k ) .
Therefore, F is a Hom-algebra morphism.
Next, we shall check that Δ F = ( F F ) Δ . Indeed, for all a A , h H ,
Δ ( F ( a × h ) ) = Δ ( L ( α 1 ( a ) ) R ( β 2 ( h ( 1 ) ) ) × β 1 ( h ( 2 ) ) ) = ( L ( α 1 ( a ) ) R ( β 2 ( h ( 1 ) ) ) ( 1 ) × ω L ( α 1 ( a ) ) R ( β 2 ( h ( 1 ) ) ) ( 2 ) [ 1 ] β 2 ( h ( 2 ) ( 1 ) ) ) α 1 L ( α 1 ( a ) ) R ( β 2 ( h ( 1 ) ) ) ( 2 ) [ 0 ] × β 1 ( h ( 2 ) ( 2 ) ) = ( T 4 ) ( L ( α 1 ( a ) ) ( 1 ) θ ω β 2 ( L ( α 1 ( a ) ) ( 2 ) [ 1 ] ) α 1 ( R ( β 2 ( h ( 1 ) ) ) ( 1 ) ) × ω α 1 ( L ( α 1 ( a ) ) ( 2 ) [ 0 ] ) R ( β 2 ( h ( 1 ) ) ) ( 2 ) [ 1 ] β 2 ( h ( 2 ) ( 1 ) ) ) α 1 α 1 ( L ( α 1 ( a ) ) ( 2 ) [ 0 ] ) R ( β 2 ( h ( 1 ) ) ) ( 2 ) [ 0 ] × β 1 ( h ( 2 ) ( 2 ) ) = ( Lemma ( iii ) ) ( L ( α 1 ( a ) ) ( 1 ) θ ω β 2 ( L ( α 1 ( a ) ) ( 2 ) [ 1 ] ) R ( β 3 ( h ( 1 ) ( 1 ) ) ) × ω α 1 ( L ( α 1 ( a ) ) ( 2 ) [ 0 ] ) R ( β 2 ( h ( 1 ) ( 2 ) ) ) [ 1 ] β 2 ( h ( 2 ) ( 1 ) ) ) α 1 α 1 ( L ( α 1 ( a ) ) ( 2 ) [ 0 ] ) R ( β 2 ( h ( 1 ) ( 2 ) ) ) [ 0 ] × β 1 ( h ( 2 ) ( 2 ) ) = ( HCMA 1 ) ( L ( α 1 ( a ) ) ( 1 ) θ ω β 2 ( L ( α 1 ( a ) ) ( 2 ) [ 1 ] ) R ( β 3 ( h ( 1 ) ( 1 ) ) ) × ω β 1 ( L ( α 1 ( a ) ) ( 2 ) [ 0 ] [ 1 ] ) R ( β 2 ( h ( 1 ) ( 2 ) ) ) [ 1 ] β 2 ( h ( 2 ) ( 1 ) ) ) α 1 α 1 ( L ( α 1 ( a ) ) ( 2 ) [ 0 ] [ 0 ] ) R ( β 2 ( h ( 1 ) ( 2 ) ) ) [ 0 ] × β 1 ( h ( 2 ) ( 2 ) ) = ( 16 ) ( L ( α 1 ( a ) ) ( 1 ) θ ω β 2 ( L ( α 1 ( a ) ) ( 2 ) [ 1 ] ) R ( β 3 ( h ( 1 ) ( 1 ) ) ) × ω ( β 1 ( L ( α 1 ( a ) ) ( 2 ) [ 0 ] [ 1 ] ) ω 1 ( β 6 ( h ( 1 ) ( 2 ) ( 1 ) ( 1 ) ) S H ( β 5 ( h ( 1 ) ( 2 ) ( 2 ) ) ) ) ) β 2 ( h ( 2 ) ( 1 ) ) ) α 1 α 1 ( L ( α 1 ( a ) ) ( 2 ) [ 0 ] [ 0 ] ) R ( β 3 ( h ( 1 ) ( 2 ) ( 1 ) ( 2 ) ) ) × β 1 ( h ( 2 ) ( 2 ) ) = ( HA 2 ) ( L ( α 1 ( a ) ) ( 1 ) θ ω β 2 ( L ( α 1 ( a ) ) ( 2 ) [ 1 ] ) R ( β 3 ( h ( 1 ) ( 1 ) ) ) × ω β 1 ( L ( α 1 ( a ) ) ( 2 ) [ 0 ] [ 1 ] ) β 5 ( h ( 1 ) ( 2 ) ( 1 ) ( 1 ) ) S H ( β 4 ( h ( 1 ) ( 2 ) ( 2 ) ) ) β 3 ( h ( 2 ) ( 1 ) ) ) α 1 α 1 ( L ( α 1 ( a ) ) ( 2 ) [ 0 ] [ 0 ] ) R ( β 3 ( h ( 1 ) ( 2 ) ( 1 ) ( 2 ) ) ) × β 1 ( h ( 2 ) ( 2 ) ) = ( HC 2 ) ( L ( α 1 ( a ) ) ( 1 ) θ ω β 2 ( L ( α 1 ( a ) ) ( 2 ) [ 1 ] ) R ( β 2 ( h ( 1 ) ) ) × ω β 1 ( L ( α 1 ( a ) ) ( 2 ) [ 0 ] [ 1 ] ) β 4 ( h ( 2 ) ( 1 ) ( 1 ) ) S H ( β 5 ( h ( 2 ) ( 2 ) ( 1 ) ( 1 ) ) ) β 5 ( h ( 2 ) ( 2 ) ( 1 ) ( 2 ) ) ) α 1 α 1 ( L ( α 1 ( a ) ) ( 2 ) [ 0 ] [ 0 ] ) R ( β 2 ( h ( 2 ) ( 1 ) ( 2 ) ) ) × β 2 ( h ( 2 ) ( 2 ) ( 2 ) ) = ( HS 1 ) ( L ( α 1 ( a ) ) ( 1 ) θ ω β 2 ( L ( α 1 ( a ) ) ( 2 ) [ 1 ] ) R ( β 2 ( h ( 1 ) ) ) × ω ( L ( α 1 ( a ) ) ( 2 ) [ 0 ] [ 1 ] ) β 3 ( h ( 2 ) ( 1 ) ( 1 ) ) ) α 1 α 1 ( L ( α 1 ( a ) ) ( 2 ) [ 0 ] [ 0 ] ) R ( β 2 ( h ( 2 ) ( 1 ) ( 2 ) ) ) × β 1 ( h ( 2 ) ( 2 ) ) = ( 12 ) ( L ( α 2 ( a ( 1 ) ) ) R ( ω β 1 ( a ( 2 ) [ 1 ] ) ) θ ω β 2 ( L ( α 2 ( a ( 2 ) [ 0 ] ) ) [ 1 ] ) R ( β 2 ( h ( 1 ) ) ) × ω ( L ( α 2 ( a ( 2 ) [ 0 ] ) ) [ 0 ] [ 1 ] ) β 3 ( h ( 2 ) ( 1 ) ( 1 ) ) ) α 1 α 1 ( L ( α 2 ( a ( 2 ) [ 0 ] ) ) [ 0 ] [ 0 ] ) R ( β 2 ( h ( 2 ) ( 1 ) ( 2 ) ) ) × β 1 ( h ( 2 ) ( 2 ) ) = ( 13 ) ( L ( α 2 ( a ( 1 ) ) ) R ( ω β 1 ( a ( 2 ) [ 1 ] ) ) θ ω ( a ( 2 ) [ 0 ] [ 1 ] ) R ( β 2 ( h ( 1 ) ) ) × ω ( L ( α 2 ( a ( 2 ) [ 0 ] [ 0 ] ) ) [ 1 ] ) β 3 ( h ( 2 ) ( 1 ) ( 1 ) ) ) α 1 α 1 ( L ( α 2 ( a ( 2 ) [ 0 ] [ 0 ] ) ) [ 0 ] ) R ( β 2 ( h ( 2 ) ( 1 ) ( 2 ) ) ) × β 1 ( h ( 2 ) ( 2 ) ) = ( HCM 2 ) ( L ( α 2 ( a ( 1 ) ) ) R ( ω β 2 ( a ( 2 ) [ 1 ] ( 1 ) ) ) θ ω ( a ( 2 ) [ 1 ] ( 2 ) ) R ( β 2 ( h ( 1 ) ) ) × ω ( L ( α 1 ( a ( 2 ) [ 0 ] ) ) [ 1 ] ) β 3 ( h ( 2 ) ( 1 ) ( 1 ) ) ) α 1 α 1 ( L ( α 1 ( a ( 2 ) [ 0 ] ) ) [ 0 ] ) R ( β 2 ( h ( 2 ) ( 1 ) ( 2 ) ) ) × β 1 ( h ( 2 ) ( 2 ) ) = ( HA 2 ) ( L ( α 1 ( a ( 1 ) ) ) R ( ω β 2 ( a ( 2 ) [ 1 ] ( 1 ) ) ) θ ω β 1 ( a ( 2 ) [ 1 ] ( 2 ) ) R ( β 3 ( h ( 1 ) ) ) × ω ( L ( α 1 ( a ( 2 ) [ 0 ] ) ) [ 1 ] ) β 3 ( h ( 2 ) ( 1 ) ( 1 ) ) ) α 1 α 1 ( L ( α 1 ( a ( 2 ) [ 0 ] ) ) [ 0 ] ) R ( β 2 ( h ( 2 ) ( 1 ) ( 2 ) ) ) × β 1 ( h ( 2 ) ( 2 ) ) = ( 15 ) ( L ( α 1 ( a ( 1 ) ) ) R ω β 1 ( a ( 2 ) [ 1 ] ) β 2 ( h ( 1 ) ) × ω ( L ( α 1 ( a ( 2 ) [ 0 ] ) ) [ 1 ] ) β 3 ( h ( 2 ) ( 1 ) ( 1 ) ) ) α 1 α 1 ( L ( α 1 ( a ( 2 ) [ 0 ] ) ) [ 0 ] ) R ( β 2 ( h ( 2 ) ( 1 ) ( 2 ) ) ) × β 1 ( h ( 2 ) ( 2 ) ) = ( 13 ) L ( α 1 ( a ( 1 ) ) ) R ω β 1 ( a ( 2 ) [ 1 ] ) β 2 ( h ( 1 ) ) × ω β 1 ( a ( 2 ) [ 0 ] [ 1 ] ) β 3 ( h ( 2 ) ( 1 ) ( 1 ) ) L ( α 3 ( a ( 2 ) [ 0 ] [ 0 ] ) ) R ( β 3 ( h ( 2 ) ( 1 ) ( 2 ) ) ) × β 1 ( h ( 2 ) ( 2 ) ) = ( HCM 2 ) ( L ( α 1 ( a ( 1 ) ) ) R ω β 2 ( a ( 2 ) [ 1 ] ( 1 ) ) β 2 ( h ( 1 ) ) × ω β 1 ( a ( 2 ) [ 1 ] ( 2 ) ) β 3 ( h ( 2 ) ( 1 ) ( 1 ) ) ) L ( α 2 ( a ( 2 ) [ 0 ] ) ) R ( β 3 ( h ( 2 ) ( 1 ) ( 2 ) ) ) × β 1 ( h ( 2 ) ( 2 ) ) = ( HC 2 ) ( L ( α 1 ( a ( 1 ) ) ) R ω β 2 ( a ( 2 ) [ 1 ] ( 1 ) ) β 3 ( h ( 1 ) ( 1 ) ) × ω β 1 ( a ( 2 ) [ 1 ] ( 2 ) ) β 2 ( h ( 1 ) ( 2 ) ) ) L ( α 2 ( a ( 2 ) [ 0 ] ) ) R ( β 2 ( h ( 2 ) ( 1 ) ) ) × β 1 ( h ( 2 ) ( 2 ) ) = F ( ( a ( 1 ) × ω ( a ( 2 ) [ 1 ] ) β 1 ( h ( 1 ) ) ) F ( α 1 ( a ( 2 ) [ 0 ] ) × h ( 2 ) ) = F ( ( a × h ) ( 1 ) ) F ( ( a × h ) ( 1 ) ) .
We have shown that F preserves comultiplication. Therefore, F is a Hom-coalgebra morphism, and hence, is a Hom-bialgebra morphism. Since Hom-bialgebra morphisms of Hom–Hopf algebras are Hom-Hopf algebra morphisms, this completes the proof of part (i).
(ii) Suppose F Aut Hom - Hopf ( A × ω θ H , π ) . Then, F L and ( F 1 ) L are inverse by (10) and Corollary 1(i). Thus, F L is bijective and ( F L ) 1 = ( F 1 ) L .
Conversely, suppose that F End Hom - Hopf ( A × ω θ H , π ) and F L are bijective. Set G L = ( F L ) 1 . From Corollary 4, F R has a convolution inverse J R . Set G R = G L J R = ( F L ) 1 J R and define G ( a × h ) = G L ( α 1 ( a ) ) G R ( β 2 ( h ( 1 ) ) ) × β 1 ( h ( 2 ) ) for all a A and h H . Since G L is a Hom-algebra morphism, we compute
G ( F ( a × h ) ) = G F L ( α 1 ( a ) ) F R ( β 2 ( h ( 1 ) ) ) × β 1 ( h ( 2 ) ) = G L F L ( α 2 ( a ) ) F R ( β 3 ( h ( 1 ) ) ) G R ( β 3 ( h ( 2 ) ( 1 ) ) ) × β 2 ( h ( 2 ) ( 2 ) ) = α 2 ( a ) G L ( F R ( β 3 ( h ( 1 ) ) ) ) G L ( J R ( β 3 ( h ( 2 ) ( 1 ) ) ) ) × β 2 ( h ( 2 ) ( 2 ) ) = ( HA 2 ) α 1 ( a ) G L F R ( β 3 ( h ( 1 ) ) ) J R ( β 4 ( h ( 2 ) ( 1 ) ) ) × β 2 ( h ( 2 ) ( 2 ) ) = ( HC 2 ) α 1 ( a ) G L F R ( β 4 ( h ( 1 ) ( 1 ) ) ) J R ( β 4 ( h ( 1 ) ( 2 ) ) ) × β 1 ( h ( 2 ) ) = α 1 ( a ) 1 A ε H ( h ( 1 ) ) × β 1 ( h ( 2 ) ) = a × h
and
F ( G ( a × h ) ) = F G L ( α 1 ( a ) ) G R ( β 2 ( h ( 1 ) ) ) × β 1 ( h ( 2 ) ) = F L G L ( α 2 ( a ) ) G R ( β 3 ( h ( 1 ) ) ) F R ( β 3 ( h ( 2 ) ( 1 ) ) ) × β 2 ( h ( 2 ) ( 2 ) ) = α 2 ( a ) F L ( G R ( β 3 ( h ( 1 ) ) ) ) F R ( β 3 ( h ( 2 ) ( 1 ) ) ) × β 2 ( h ( 2 ) ( 2 ) ) = α 2 ( a ) J R ( β 3 ( h ( 1 ) ) ) F R ( β 3 ( h ( 2 ) ( 1 ) ) ) × β 2 ( h ( 2 ) ( 2 ) ) = ( HA 2 ) ( HC 2 ) α 1 ( a ) J R ( β 4 ( h ( 1 ) ( 1 ) ) ) F R ( β 4 ( h ( 1 ) ( 2 ) ) ) × β 1 ( h ( 2 ) ) = α 1 ( a ) 1 A ε H ( h ( 1 ) ) × β 1 ( h ( 2 ) ) = a × h .
Thus, we have shown that G F = id A × ω θ H = F G . □
Now, let F A , H denote the set of ( L , R ) F A , H , such that L is bijective. Then, the correspondence of the above theorem induces a bijection F A , H Aut Hom - Hopf ( A × ω θ H , π ) .
When F ( 1 A × H ) 1 A × H , the structure of F is particularly simple.
Proposition 1.
Let F End Hom - Hopf ( A × ω θ H , π ) . Then, the following conditions are equivalent:
(i)
F ( 1 A × H ) 1 A × H .
(ii)
F R ( h ) k 1 A for all h H .
(iii)
F R ( h ) = ε H ( h ) 1 A for all h H .
(iv)
F R = η A ε H .
(v)
F ( a × h ) = F L ( a ) × h for all a A and h H .
(vi)
F ( 1 A × h ) = 1 A × h for all h H .
Proof. 
Firstly, we prove that part (i) implies part (ii) implies part (iii). Suppose part (i) holds. Then, we have F R ( H ) k 1 A by Equation (11), and hence, F R ( h ) = ε H ( h ) 1 A for all h H , because ε A F R = ε H . Secondly, it is easy to see that parts (iii) and (iv) are equivalent. Thirdly, part (iv) implies part (v) by Equation (9), and part (v) implies part (vi) as F L ( 1 A ) = 1 A . At last, part (vi) trivially implies part (i). □
Let ( A , α ) be a Hom-algebra and ( C , β ) be a Hom-coalgebra over k . Let Hom ( C , A ) be the set of k -linear maps f : C A satisfying α f = f β . The group G ( A ) = Aut Hom - Alg ( A ) acts on the (non-associative) convolution algebra Hom ( C , A ) by f g = f g for all f G ( A ) ) and g Hom ( C , A ) . Obviously, this action satisfies:
f ( η A ε C ) = η A ε C and f ( g g ) = ( f g ) ( f g )
for all f G ( A ) and g , g Hom ( C , A ) . Then, ( Hom ( C , A ) , α 2 ) is a Hom-algebra with the unit η A ε C under the convolution *. Indeed, for f , g , g Hom ( C , A ) and c C ,
( ( α 2 f ) ( g g ) ) ( c ) = ( α 2 f ) ( c ( 1 ) ) ( g ( c ( 2 ) ( 1 ) ) g ( c ( 2 ) ( 2 ) ) ) = ( α f ) ( c ( 1 ) ( 1 ) ) ( g ( c ( 1 ) ( 2 ) ) g ( β ( c ( 2 ) ) ) ) = ( f ( c ( 1 ) ( 1 ) ) g ( c ( 1 ) ( 2 ) ) ) ( α 2 g ) ( c ( 2 ) ) = ( ( f g ) ( α 2 g ) ) ( c ) ,
and
( f ( η A ε C ) ) ( c ) = f ( c ( 1 ) ) ε H ( c ( 2 ) ) 1 A = α ( f ( β ( c ) ) ) = ( α 2 f ) ( c ) .
Thus, it follows that ( α 2 f ) ( g g ) = ( f g ) ( α 2 g ) and f ( η A ε C ) = α 2 f . Moreover, ( η A ε C ) f = α 2 f can be checked similarly.
It is easy to check that ( Hom ( C , A ) , α 2 , η A ε C ) is an ordinary associative algebra. Let U ( C , A ) be the group of units of the algebra ( Hom ( C , A ) , α 2 , η A ε C ) . Then G ( A ) U ( C , A ) U ( C , A ) ; thus, there is a group homomorphism
φ : G ( A ) Aut Group ( U ( C , A ) )
given by φ ( f ) ( g ) = f g = f g for all f G ( A ) and g U ( C , A ) . The resulting group U ( C , A ) φ G ( A ) has product given by
( g , f ) ( g , f ) = ( α 2 ( g ( f g ) ) , f f ) .
Note that the action of G ( A ) on U ( C , A ) by group homomorphisms is also one on U ( C , A ) op . For a group G, the group G op is the group whose underlying set is G and the product is given by g · g = g g for all g , g G . Therefore, we have the following result of Theorem 3 and Corollary 1.
Theorem 4.
Let ( A × ω θ H , α β ) be the ( θ , ω ) -twisted-Radford’s Hom-biproduct, let π : A × ω θ H H be the projection from A × ω θ H onto H. Then, there is an injective group homomorphism ψ : Aut Hom - Hopf ( A × ω θ H , π ) U ( C , A ) op φ G ( A ) , which is given by F ( F R , F L ) for all F Aut Hom - Hopf ( A × ω θ H , π ) .
Proof. 
For F Aut Hom - Hopf ( A × ω θ H , π ) , it is easy to see that F L and F R are bijective by Theorem 3(ii) and Equation (11). If ( F R , F L ) = ( η A ε C , id A ) , then, using (9), we have F ( a × h ) = a × h for all a A and h H . Thus, it follows that F = id A × ω θ H , i.e., ψ is injective.
For F , G Aut Hom - Hopf ( A × ω θ H , π ) , since
ψ ( F G ) = ( ( F G ) R , ( F G ) L ) = ( α 2 ( F R op ( F L G R ) ) , F L G L ) = Cor 1 ( F R , F L ) ( G R , G L ) = ψ ( F ) ψ ( G ) ,
it follows that ψ preserves the multiplication. Thus, ψ is an injective group homomorphism. □
In conclusion, by Theorem 3, we obtain the characterization of the automorphism group Aut Hom - Hopf ( A × ω θ H , π ) of a ( θ , ω ) -twisted-Radford’s Hom-biproduct A × ω θ H via the automorphisms of the factors ( A , α ) and ( H , β ) . Furthermore, by Theorem 4, the automorphism group Aut Hom - Hopf ( A × ω θ H , π ) can be regarded as a subgroup of some semidirect product U ( C , A ) op φ G ( A ) , when π is a projection. These are important for the study of automorphism groups of twisted-Radford’s Hom-biproducts.

4. Example

In this section, we consider an example in [26]. Let A = k 1 A , x and the automorphism α : A A via α ( 1 A ) = 1 A and α ( x ) = x . Then, ( A , α ) is both a Hom-algebra with product: 1 A 1 A = 1 A , 1 A x = x 1 A = x , x 2 = 0 , and a Hom-coalgebra with coproduct and counit given by
Δ A ( 1 A ) = 1 A 1 A , Δ A ( x ) = ( x ) 1 A + 1 A ( x ) , ε A ( 1 A ) = 1 A , ε A ( x ) = 0 .
We define S A : A A by S A ( 1 A ) = 1 A , S A ( x ) = x , which is the convolution inverse of id A .
Let H = k 1 H , g , g 2 be the group Hopf algebra with g 3 = 1 H , Δ H ( g ) = g g , S H ( g ) = g 2 and S H ( g 2 ) = g . Define an automorphism β : H H by β ( g ) = g 2 , β ( g 2 ) = g . Then, we can define a new product m β = β m H and a new coproduct Δ β = Δ β on H. So, we obtain a Hom–Hopf algebra H β = ( H , m β , Δ β , S H , β ) . Then, ( A , α ) is a left H β -Hom-module Hom-algebra and Hom-module Hom-coalgebra with the action : H A A given by
1 H 1 A = 1 A , 1 H x = x , g 1 A = 1 A , g x = x , g 2 1 A = 1 A , g 2 x = x .
Furthermore, ( A , α ) is a left H β -Hom-comodule Hom-algebra and Hom-comodule Hom-coalgebra with the coaction ρ : A H A given by
ρ ( 1 A ) = 1 H 1 A , ρ ( x ) = 1 H ( x ) .
Then, we can obtain many twisted Radford’s Hom-biproducts, such as ( β , β ) -twisted Radford’s Hom-biproduct, ( β , id H ) -twisted Radford’s Hom-biproduct, ( id H , β ) -twisted Radford’s Hom-biproduct, ( id H , id H ) -twisted Radford’s Hom-biproduct, and so on.
Now, we clearly write the structures of the ( β , β ) -twisted Radford’s Hom-biproduct ( A × β β H β = { 1 A × 1 H , 1 A × g , 1 A × g 2 , x × 1 H , x × g , x × g 2 } , α β ) and characterize the automorphisms of it. Its product is defined as follows:
m 1 A × 1 H 1 A × g 1 A × g 2 x × 1 H x × g x × g 2 1 A × 1 H 1 A × 1 H 1 A × g 2 1 A × g x × 1 H x × g 2 x × g 1 A × g 1 A × g 2 1 A × g 1 A × 1 H x × g 2 x × g x × 1 H 1 A × g 2 1 A × g 1 A × 1 H 1 A × g 2 x × g x × 1 H x × g 2 x × 1 H x × 1 H x × g 2 x × g 0 0 0 x × g x × g 2 x × g x × 1 H 0 0 0 x × g 2 x × g x × 1 H x × g 2 0 0 0
Its coproduct and counit are defined as follows:
Δ ( 1 A × 1 H ) = ( 1 A × 1 H ) ( 1 A × 1 H ) , ε ( 1 A × 1 H ) = 1 , Δ ( 1 A × g ) = ( 1 A × g 2 ) ( 1 A × g 2 ) , ε ( 1 A × g ) = 1 , Δ ( 1 A × g 2 ) = ( 1 A × g ) ( 1 A × g ) , ε ( 1 A × g 2 ) = 1 , Δ ( x × 1 H ) = ( x × 1 H ) ( 1 A × 1 H ) + ( 1 A × 1 H ) ( x × 1 H ) , ε ( x × 1 H ) = 0 , Δ ( x × g ) = ( x × g 2 ) ( 1 A × g 2 ) + ( 1 A × g 2 ) ( x × g 2 ) , ε ( x × g ) = 0 , Δ ( x × g 2 ) = ( x × g ) ( 1 A × g ) + ( 1 A × g ) ( x × g ) , ε ( x × g 2 ) = 0 ,
Its antipodes are as follows:
S ( 1 A × 1 H ) = 1 A × 1 H , S ( 1 A × g ) = 1 A × g 2 , S ( 1 A × g 2 ) = 1 A × g , S ( x × 1 H ) = x × 1 H , S ( x × g ) = x × g 2 , S ( x × g 2 ) = x × g .
Now, we compute the morphisms L End ( A ) satisfying the conclusions of Lemma 2. This is taking a base of End ( A ) , { L 1 , L 2 , L 3 , L 4 } , given respectively by
L 1 : 1 A 1 A , x 0 , L 2 : 1 A x , x 0 , L 3 : 1 A 0 , x 1 A , L 4 : 1 A 0 , x x .
Let L = t 1 L 1 + t 2 L 2 + t 3 L 3 + t 4 L 4 , where t 1 , t 2 , t 3 , t 4 k . If L satisfies Lemma 2(ii), it follows that t 1 = 1 and t 3 = 0 . By part (iv) of Lemma 2, we can obtain t 2 = 0 . Therefore, L = L 1 + t L 4 for some t k . So, there is a bijection between the set of the morphisms L End ( A ) satisfying the conclusions of Lemma 2 and the set { 1 0 0 t | t k } .
Now, we will discuss the morphisms of Hom ( H , A ) which satisfy Lemma 3 in a similar way as above. Taking a base of Hom ( H , A ) , { R 1 , R 2 , , R 6 } , given respectively by
R 1 : 1 H 1 A , g 0 , g 2 0 , R 2 : 1 H x , g 0 , g 2 0 , R 3 : 1 H 0 , g 1 A , g 2 0 , R 4 : 1 H 0 , g x , g 2 0 , R 5 : 1 H 0 , g 0 , g 2 1 A , R 6 : 1 H 0 , g 0 , g 2 x .
Let R = i = 1 6 k i R i , where k i k , i = 1 , 2 , 3 , 4 , 5 , 6 . Since R satisfies part (i) of Lemma 3, we have k 1 = 1 and k 2 = 0 . By Equation (16), it follows that k 3 = k 5 and k 4 = k 6 . Thus, R = R 1 + k 3 R 3 + k 4 R 4 + k 3 R 5 k 4 R 6 . Next, we shall check part (ii) of Lemma 3, and we obtain k 3 = 1 and k 4 = 0 . Therefore, R = R 1 + R 3 + R 5 , i.e., R ( 1 H ) = 1 A , R ( g ) = 1 A , R ( g 2 ) = 1 A . Hence, F A , H { 1 0 0 t | 0 t k } k × . We can obtain the concrete characterization of Aut Hom - Hopf ( A × β β H β , π ) . Let F Aut Hom - Hopf ( A × β β H β , π ) . By Theorem 3, we have
F ( 1 A × 1 H ) = 1 A × 1 H , F ( 1 A × g ) = 1 A × g , F ( 1 A × g 2 ) = 1 A × g 2 , F ( x × 1 H ) = t x × 1 H , F ( x × g ) = t x × g , F ( x × g 2 ) = t x × g 2 ,
where t k × .

Author Contributions

Conceptualization, X.W. and D.-G.W.; methodology, X.W.; software, X.W.; validation, X.W. and D.-G.W.; formal analysis, D.-G.W.; investigation, X.W.; resources, X.W.; data curation, D.-G.W.; writing—original draft preparation, X.W.; writing—review and editing, D.-G.W.; visualization, X.W.; supervision, D.-G.W.; project administration, X.W.; funding acquisition, X.W. and D.-G.W. All authors have read and agreed to the published version of the manuscript.

Funding

This work was supported by the National Natural Science Foundation of China (Grant Nos. 11871301, 11801304) and Natural Science Foundation of Shandong Province of China (Nos. ZR2019MA060, ZR2019QA015).

Institutional Review Board Statement

Not applicable.

Informed Consent Statement

Not applicable.

Data Availability Statement

Not applicable.

Acknowledgments

The authors would like to express their gratitude to the anonymous referee for their very helpful suggestions and comments, which led to the improvement of our original manuscript.

Conflicts of Interest

The authors declare no conflict of interest.

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Wang, X.; Wang, D.-G. Characterization of Automorphisms of (θ,ω)-Twisted Radford’s Hom-Biproduct. Mathematics 2022, 10, 407. https://doi.org/10.3390/math10030407

AMA Style

Wang X, Wang D-G. Characterization of Automorphisms of (θ,ω)-Twisted Radford’s Hom-Biproduct. Mathematics. 2022; 10(3):407. https://doi.org/10.3390/math10030407

Chicago/Turabian Style

Wang, Xing, and Ding-Guo Wang. 2022. "Characterization of Automorphisms of (θ,ω)-Twisted Radford’s Hom-Biproduct" Mathematics 10, no. 3: 407. https://doi.org/10.3390/math10030407

APA Style

Wang, X., & Wang, D. -G. (2022). Characterization of Automorphisms of (θ,ω)-Twisted Radford’s Hom-Biproduct. Mathematics, 10(3), 407. https://doi.org/10.3390/math10030407

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