# Double crossed biproducts and related structures

Tianshui Ma* School of Mathematics and Information Science, Henan Normal
University, Xinxiang 453007, China<EMAIL_ADDRESS>,  Jie Li  School of
Mathematics and Information Science, Henan Normal University, Xinxiang 453007,
China<EMAIL_ADDRESS>,  Haiyan Yang  School of Mathematics and
Information Science, Henan Normal University, Xinxiang 453007, China
<EMAIL_ADDRESS>and  Shuanhong Wang  School of Mathematics, Southeast
University, Nanjing 210096, China<EMAIL_ADDRESS>
###### Abstract.

Let $H$ be a bialgebra. Let $\sigma:H\otimes H\rightarrow A$ be a linear map,
where $A$ is a left $H$-comodule coalgebra, and an algebra with a left
$H$-weak action $\triangleright$. Let $\tau:H\otimes H\rightarrow B$ be a
linear map, where $B$ is a right $H$-comodule coalgebra, and an algebra with a
right $H$-weak action $\triangleleft$. In this paper, we improve the necessary
conditions for the two-sided crossed product algebra
$A\\#^{\sigma}H~{}{{}^{\tau}\\#}B$ and the two-sided smash coproduct coalgebra
$A\times H\times B$ to form a bialgebra (called double crossed biproduct) such
that the condition $b_{[1]}\triangleright a_{0}\otimes b_{[0]}\triangleleft
a_{-1}=a\otimes b$ in Majid’s double biproduct (or double-bosonization) is one
of the necessary conditions. On the other hand, we provide a more general two-
sided crossed product algebra structure via Brzezński’s crossed product and
give some applications.

###### Key words and phrases:

Majid’s double biproduct;Brzeziński’s crossed product; Radford’s biproduct

###### 2010 Mathematics Subject Classification:

16T05,16T10,16T25

*Corresponding author

###### Contents

  1. 1 Introduction and preliminaries
  2. 2 Brzeziński’s two-sided crossed product
    1. 2.1 Right Brzeziński’s crossed product and its special cases
    2. 2.2 Brzeziński’s two-sided crossed product algebra and its special cases
    3. 2.3 Brzeziński’s two-sided crossed product bialgebra
  3. 3 An extended version of Majid’s double biproduct
  4. 4 Conclusion

##  1\. Introduction and preliminaries

Radford’s biproduct ([21]) is one of the celebrated objects in the theory of
Hopf algebras, which also can provide examples for Rota-Baxter bialgebras
introduced by Ma and Liu in ([13]). This plays a fundamental role in the
classification of finite-dimensional pointed Hopf algebras ([3]). Majid
realized a categorical interpretation of Radford’s biproduct ([15, 16]):
$A\star H$ is a Radford’s biproduct if and only if $A$ is a bialgebra in the
Yetter-Drinfel’d category ${}^{H}_{H}{\mathbb{YD}}$.

Let $H$ be a bialgebra, $A$ a bialgebra in ${}^{H}_{H}{\mathbb{YD}}$, and $B$
a bialgebra in ${\mathbb{YD}}{}^{H}_{H}$. In [17, 18], Majid gave a sufficient
condition

$\displaystyle b_{[1]}\triangleright a_{0}\otimes b_{[0]}\triangleleft
a_{-1}=a\otimes b$ (1)

for a two-sided smash product algebra $A\\#H\\#B$ and a two-sided smash
coproduct coalgebra $A\times H\times B$ to form a bialgebra, named the Majid’s
double biproduct (or double-bosonization) and denoted by $A\star H\star B$.
Majid’s double biproduct provided a powerful tool to construct quantum groups
(see [17, 18]), some related studies can be found in the literature [7, 8, 13,
12, 14, 10, 20, 23, 25, 26]. As an application of our results in this paper,
we find that the condition $(\ref{eq:DB})$ is also necessary in $A\star H\star
B$. This is a motivation of writing this paper.

In [24], as a generalization of Radford’s results in [21], Wang, Jiao and Zhao
presented the necessary and sufficient conditions for the (left) crossed
product algebra (introduced in [4] and extended the smash product algebra) and
the (left) smash coproduct coalgebra to be a bialgebra (denoted by
$A\star^{\sigma}H$) and also gave a mapping description. Similarly, in order
to extend Majid’s double biproduct, Ma and Liu in [14] gave the necessary and
sufficient conditions for the two-sided crossed product algebra
$A\\#^{\sigma}H~{}{{}^{\tau}\\#}B$ and the two-sided smash coproduct coalgebra
$A\times H\times B$ to form a bialgebra (denoted by
$A\star^{\sigma}H~{}{{}^{\tau}\star}B$). As can be seen from the notations of
these structures, if we substitute $B$ in $A\star H\star B$ (or
$A\star^{\sigma}H~{}{{}^{\tau}\star}B$) for the ground field $K$, then $A\star
H$ (or $A\star^{\sigma}H$) is obtained. Comparing the necessary conditions in
$A\star H\star B$ and $A\star^{\sigma}H~{}{{}^{\tau}\star}B$, we find that the
latter is a little more complicated, especially the condition $(C13)$ in [14,
Theorem 2.1] and that the former just contains the necessary and sufficient
conditions for the smash product algebra $A\\#H$ and the smash coproduct
coalgebra $A\times H$ to form a bialgebra $A\star H$. It is worth mentioning
that these conditions in the former case are just right for the use of lifting
method in the classification of finite-dimensional pointed Hopf algebras. For
these reasons, we will improve the necessary conditions in [14, Theorem 2.1]
such that the necessary and sufficient conditions for the crossed product
algebra $A\\#^{\sigma}H$ and the smash coproduct coalgebra $A\times H$ to form
a bialgebra $A\star^{\sigma}H$ can be included and also $(C13)$ above can be
decomposed into several simpler ones (of course containing $(\ref{eq:DB})$).
This is another motivation of writing this paper.

In [5], Brzeziński gave the necessary and sufficient conditions for a very
extensive crossed product to be an algebra (called Brzeziński’s crossed
product) which includes the crossed product ([4]) and the twisted tensor
product ([6]). We will provide a more general two-sided crossed product
algebra structure via Brzezński’s crossed product and list some special cases.

Throughout this paper, we follow the definitions and terminologies in [22],
with all algebraic systems supposed to be over the field $K$. Now, let $C$ be
a coalgebra. We use the simple Sweedler’s notation for the comultiplication:
$\Delta(c)=c_{1}\otimes c_{2}$ for any $c\in C$. Denote the category of left
$H$-comodules by ${}^{H}\hbox{Mod}$. For $(M,\rho)\in$ ${}^{H}\hbox{Mod}$,
write: $\rho(x)=x_{-1}\otimes x_{0}\in H\otimes M$, for all $x\in M$. Denote
the category of right $H$-comodules by $\hbox{Mod}^{H}$. For $(M,\psi)\in$
$\hbox{Mod}^{H}$, write: $\psi(x)=x_{[0]}\otimes x_{[1]}\in M\otimes H$, for
all $x\in M$. We denote the left-left Yetter-Drinfel’d category by
${}^{H}_{H}{\mathbb{YD}}$, and the right-right Yetter-Drinfel’d category by
${\mathbb{YD}}{}^{H}_{H}$. Given a $K$-space $M$, we write $id_{M}$ for the
identity map on $M$.

First of all, we rewrite the Brzeziński’s crossed product as follows.

######  Proposition 1.1.

(Left Brzeziński’s crossed product) ([5]): Let $A$ be an algebra, $H$ a vector
space and $1_{H}\in H$. Let $G:H\otimes H\longrightarrow A\otimes H$ (write
$G(x\otimes x^{\prime})=x^{G}\otimes x^{\prime}_{G}$ for all $x,x^{\prime}\in
H$) and $R:H\otimes A\longrightarrow A\otimes H$ (write $R(x\otimes
a)=a_{R}\otimes x_{R}$ for all $x\in H$ and $a\in A$) be linear maps such that

$(LB1)~{}~{}a_{R}\otimes 1_{HR}=a\otimes{1_{H}},~{}~{}~{}{1_{A}}_{R}\otimes
x_{R}={1_{A}}\otimes x$

holds. Then $A\\#^{G}_{R}H$ ($=A\otimes H$ as vector space) is an algebra with
unit $1_{A}\otimes 1_{H}$ and a product

$\displaystyle(a\otimes x)(a^{\prime}\otimes
x^{\prime})=aa^{\prime}_{R}{x_{R}}^{G}\otimes x^{\prime}_{G}$ (2)

if and only if the following conditions hold:

$\displaystyle(LB2)$ $\displaystyle(aa^{\prime})_{R}\otimes
x_{R}=a_{R}a^{\prime}_{r}\otimes x_{Rr};$ $\displaystyle(LB3)$ $\displaystyle
x^{G}\otimes{1_{H}}_{G}={1_{H}}^{G}\otimes x_{G}=1_{A}\otimes x;$
$\displaystyle(LB4)$ $\displaystyle{x^{\prime G}}_{R}{x_{R}}^{g}\otimes
x^{\prime\prime}_{Gg}=x^{G}{x^{\prime}_{G}}^{g}\otimes x^{\prime\prime}_{g};$
$\displaystyle(LB5)$ $\displaystyle a_{Rr}{x_{r}}^{G}\otimes
x^{\prime}_{RG}=x^{G}a_{R}\otimes x^{\prime}_{GR}$

for all $a,a^{\prime}\in A$, $x,x^{\prime},x^{\prime\prime}\in H$, where
$G=g$, $R=r$.

###### Proof.

The sufficiency can be gotten by [5]. So we only prove the necessity. By
$1_{A}\otimes 1_{H}$ is unit and Eqs.$(LB1)$ and (2), we can get Eq.$(LB3)$.
By the associativity and Eq.(2), we have

$\displaystyle
a{a^{\prime}_{R}}{x_{R}}^{G}{a^{\prime\prime}_{r}}{x^{\prime}_{Gr}}^{g}\otimes
x^{\prime\prime}_{g}=a(a^{\prime}a^{\prime\prime}_{R}{x^{\prime}_{R}}^{G})_{r}{x_{r}}^{g}\otimes
x^{\prime\prime}_{Gg}.$ (3)

Set $x^{\prime}=x^{\prime\prime}=1_{H}$ in Eq.(3), one can obtain Eq.$(LB2)$.
Let $a^{\prime}=a^{\prime\prime}=1_{A}$ in Eq.(3), then Eq.$(LB4)$ holds.
Likewise, setting $a^{\prime}=1_{A},x^{\prime\prime}=1_{H}$ in Eq.(3), we can
get Eq.$(LB5)$. These finish the proof. ∎

######  Remark 1.2.

When Eqs.$(LB1)$ and $(LB3)$ are interchanged, Proposition 1.1 still holds.

In what follows, we list some important special cases of left Brzeziński’s
crossed product.

######  Example 1.3.

(Left twisted crossed product) Let $A$ be an algebra, $H$ a bialgebra,
$\sigma:H\otimes H\longrightarrow A$ and $R:H\otimes A\longrightarrow A\otimes
H$ two linear maps such that Eq.(LB1) holds. Then $A\\#^{\sigma}_{R}H$
(=$A\otimes H$ as a vector space) is an associative algebra with unit
$1_{A}\otimes{1_{H}}$ and multiplication

$(a\otimes x)(a^{\prime}\otimes
x^{\prime})=aa^{\prime}_{R}\sigma(x_{R1},x^{\prime}_{1})\otimes
x_{R2}x^{\prime}_{2}$

for all $a,a^{\prime}\in A$ and $x,x^{\prime}\in H$ if and only if Eq.(LB2)
and the following conditions hold:

$\displaystyle(LTC3)$
$\displaystyle\sigma(x,1_{H})=\sigma(1_{H},x)=\varepsilon(x)1_{A};$
$\displaystyle(LTC4)$
$\displaystyle\sigma(x^{\prime}_{1},x^{\prime\prime}_{1})_{R}\sigma(x_{R1},x^{\prime}_{2}x^{\prime\prime}_{2})\otimes
x_{R2}x^{\prime}_{3}x^{\prime\prime}_{3}=\sigma(x_{1},x^{\prime}_{1})\sigma(x_{2}x^{\prime}_{2},x^{\prime\prime}_{1})\otimes
x_{3}x^{\prime}_{3}x^{\prime\prime}_{2};$ $\displaystyle(LTC5)$ $\displaystyle
a_{Rr}\sigma(x_{r1},x^{\prime}_{R1})\otimes
x_{r2}x^{\prime}_{R2}=\sigma(x_{1},x^{\prime}_{1})a_{R}\otimes(x_{2}x^{\prime}_{2})_{R},$

where $a,a^{\prime}\in A$, $x,x^{\prime},x^{\prime\prime}\in H$ and $r=R$.

###### Proof.

Let $G(x\otimes x^{\prime})=\sigma(x_{1},x^{\prime}_{1})\otimes
x_{2}x^{\prime}_{2}$ in $A\\#^{G}_{R}H$. ∎

######  Example 1.4.

(Left crossed product) ([4]): Let $H$ be a Hopf algebra with a left weak
action $\triangleright$ on the algebra $A$. Let $\sigma:H\otimes
H\longrightarrow A$ be a $K$-linear map. Let $A\\#^{\sigma}_{\triangleright}H$
be the (usually nonassociative) algebra whose underlying space is $A\otimes H$
and whose multiplication is given by

$(a\otimes x)(a^{\prime}\otimes x^{\prime})=a(x_{1}\triangleright
a^{\prime})\sigma(x_{2},x^{\prime}_{1})\otimes x_{3}x^{\prime}_{2}$

for all $a,a^{\prime}\in A$ and $x,x^{\prime}\in H$. This
$A\\#^{\sigma}_{\triangleright}H$ is an associative algebra with unit
$1_{A}\otimes{1_{H}}$ if and only if $\sigma$ satisfies Eq.(LTC3) and the
following conditions:

$\displaystyle(LC2)$
$\displaystyle(x_{1}\triangleright\sigma(x^{\prime}_{1},x^{\prime\prime}_{1}))\sigma(x_{2},x^{\prime}_{2}x^{\prime\prime}_{2})=\sigma(x_{1},x^{\prime}_{1})\sigma(x_{2}x^{\prime}_{2},x^{\prime\prime});$
$\displaystyle(LC3)$
$\displaystyle(x_{1}\triangleright(x^{\prime}_{1}\triangleright
a))\sigma(x_{2},x^{\prime}_{2})=\sigma(x_{1},x^{\prime}_{1})(x_{2}x^{\prime}_{2}\triangleright
a)$

for all $a\in A$ and $x,x^{\prime},x^{\prime\prime}\in H$.

###### Proof.

Let $R(x\otimes a)=x_{1}\triangleright a\otimes x_{2}$ in
$A\\#^{\sigma}_{R}H$. ∎

######  Example 1.5.

(Left twisted product) ([4]): Let $H$ be a bialgebra and $A$ an algebra. Let
$\sigma:H\otimes H\longrightarrow A$ be a $K$-linear map. Let $A\\#^{\sigma}H$
(=$A\otimes H$ as a vector space) is an associative algebra with unit
$1_{A}\otimes{1_{H}}$ and multiplication

$(a\otimes x)(a^{\prime}\otimes
x^{\prime})=aa^{\prime}\sigma(x_{1},x^{\prime}_{1})\otimes
x_{2}x^{\prime}_{2}$

for all $a,a^{\prime}\in A$ and $x,x^{\prime}\in H$ if and only if $\sigma$
satisfies Eq.(LTC3) and the following conditions:

$\displaystyle(L2)$
$\displaystyle\sigma(x^{\prime\prime}_{1},x^{\prime}_{1})\sigma(x,x^{\prime\prime}_{2}x^{\prime}_{2})=\sigma(x_{1},x^{\prime\prime}_{1})\sigma(x_{2}x^{\prime\prime}_{2},x^{\prime});$
$\displaystyle(L3)$ $\displaystyle
a\sigma(x,x^{\prime})=\sigma(x,x^{\prime})a$

for all $a\in A$ and $x,x^{\prime},x^{\prime\prime}\in H$.

###### Proof.

Let $x\triangleright a=\varepsilon(x)a$ in $A\\#^{\sigma}_{\triangleright}H$.
∎

######  Example 1.6.

(Left unified product) ([2]): Let $A$, $H$ be two bialgebras and
$\triangleleft:H\otimes A\longrightarrow H$, $\triangleright:H\otimes
A\longrightarrow A$, $\sigma:H\otimes H\longrightarrow A$ linear maps such
that

$(LU1)~{}~{}~{}~{}{1_{H}}\triangleright a_{1}\otimes{1_{H}}\triangleleft
a_{2}=a\otimes{1_{H}};~{}~{}x_{1}\triangleright 1_{A}\otimes
x_{2}\triangleleft 1_{A}=1_{A}\otimes x$

holds. Then $A\\#^{\sigma}_{\triangleright\triangleleft}H$ (=$A\otimes H$ as a
vector space) is an associative algebra with unit $1_{A}\otimes{1_{H}}$ and
multiplication

$(a\otimes x)(a^{\prime}\otimes x^{\prime})=a(x_{1}\triangleright
a^{\prime}_{1})\sigma((x_{2}\triangleleft
a^{\prime}_{2})_{1},x^{\prime}_{1})\otimes(x_{2}\triangleleft
a^{\prime}_{2})_{2}x^{\prime}_{2}$

for all $a,a^{\prime}\in A$ and $x,x^{\prime}\in H$ if and only if the
following conditions hold:

$\displaystyle(LU2)$ $\displaystyle
x_{1}\triangleright(aa^{\prime})_{1}\otimes
x_{2}\triangleleft(aa^{\prime})_{2}=(x_{1}\triangleright
a_{1})((x_{2}\triangleleft a_{2})_{1}\triangleright
a^{\prime}_{1})\otimes((x_{2}\triangleleft a_{2})_{2}\triangleright
a^{\prime}_{2});$ $\displaystyle(LU3)$
$\displaystyle\sigma(x_{1},{1_{H}})\otimes x_{2}=\sigma({1_{H}},x_{1})\otimes
x_{2}=1_{A}\otimes x;$ $\displaystyle(LU4)$
$\displaystyle~{}~{}(x_{1}\triangleright\sigma(x^{\prime}_{1},x^{\prime\prime}_{1})_{1})\sigma((x_{2}\triangleleft\sigma(x^{\prime}_{1},x^{\prime\prime}_{1})_{2})_{1},(x^{\prime}_{2}x^{\prime\prime}_{2})_{1})\otimes(x_{2}\triangleleft\sigma(x^{\prime}_{1},x^{\prime\prime}_{1})_{2})_{2}(x^{\prime}_{2}x^{\prime\prime}_{2})_{2}$
$\displaystyle=\sigma(x_{1},x^{\prime}_{1})\sigma((x_{2}x^{\prime}_{2})_{1},x^{\prime\prime}_{1})\otimes(x_{2}x^{\prime}_{2})_{2}x^{\prime\prime}_{2};$
$\displaystyle(LU5)$
$\displaystyle(x_{1}\triangleright(x^{\prime}_{1}\triangleright
a_{1})_{1})\sigma((x_{2}\triangleleft(x^{\prime}_{1}\triangleright
a_{1})_{2})_{1},(x^{\prime}_{2}\triangleleft
a_{2})_{1})\otimes(x_{2}\triangleleft(x^{\prime}_{1}\triangleright
a_{1})_{2})_{2}$ $\displaystyle\times(x^{\prime}_{2}\triangleleft
a_{2})_{2}=\sigma(x_{1},x^{\prime}_{1})((x_{2}x^{\prime}_{2})_{1}\triangleright
a_{1})\otimes((x_{2}x^{\prime}_{2})_{2}\triangleleft a_{2}),$

where $a,a^{\prime}\in A$, $x,x^{\prime},x^{\prime\prime}\in H$.

###### Proof.

Let $R(x\otimes a)=x_{1}\triangleright a_{1}\otimes x_{2}\triangleleft a_{2}$
in $A\\#^{\sigma}_{R}H$. ∎

######  Example 1.7.

(Left twisted tensor product) ([6]): Let $H$, $A$ be two algebras and
$R:H\otimes A\longrightarrow A\otimes H$ a linear map. Then $A\\#_{R}H$
(=$A\otimes H$ as a vector space) is an associative algebra with unit
$1_{A}\otimes{1_{H}}$ and multiplication

$(a\otimes x)(a^{\prime}\otimes x^{\prime})=a{a^{\prime}_{R}}\otimes
x_{R}x^{\prime}$

for all $a,a^{\prime}\in A$ and $x,x^{\prime}\in H$ if and only if Eqs.(LB1),
(LB2) and the following condition hold:

$\displaystyle(LT3)$
$\displaystyle{a_{R}}\otimes(xx^{\prime})_{R}={a}_{Rr}\otimes
x_{r}{x^{\prime}_{R}},$

where $a,a^{\prime}\in A$, $x,x^{\prime}\in H$ and $r=R$.

###### Proof.

Let $G$ be trivial, i.e., $G(x\otimes x^{\prime})=1_{A}\otimes xx^{\prime}$ in
$A\\#^{G}_{R}H$. ∎

######  Remark 1.8.

(1) The well known Drinfeld double is also a special case of left Brzeziński’s
crossed product.
Remarks (2) From above we know that the left Brzeziński’s crossed product is a
very general product structure.

######  Definition 1.9.

(Majid’s double biproduct) We recall, from [17, 18], the construction of the
so-called double biproduct. Let $H$ be a bialgebra, $A$ a bialgebra in
${}^{H}_{H}{\mathbb{YD}}$, and $B$ a bialgebra in ${\mathbb{YD}}{}^{H}_{H}$.
Adopt the following notation for the structure maps: the counits are
$\varepsilon_{A}$ and $\varepsilon_{B}$, the comultiplications are
$\Delta_{A}(a)=a_{1}\otimes a_{2}$ and $\Delta_{B}(b)=b_{1}\otimes b_{2}$, and
the actions and coactions are

$\displaystyle H\otimes A\longrightarrow A,x\otimes~{}~{}~{}a\mapsto
x\triangleright a,~{}~{}~{}A\longrightarrow H\otimes A,~{}~{}~{}a\mapsto
a_{-1}\otimes a_{0},$ $\displaystyle B\otimes H\longrightarrow
B,~{}~{}~{}b\otimes x\mapsto b\triangleleft x,~{}~{}~{}~{}B\longrightarrow
B\otimes H,~{}~{}~{}b\mapsto b_{[0]}\otimes b_{[1]}$

for all $x\in H$, $a\in A$, $b\in B$. The vector space $A\otimes H\otimes B$
becomes an algebra (called the two-sided smash product, $A\\#H\\#B$) with unit
$1_{A}\otimes 1_{H}\otimes 1_{B}$ and multiplication

$\displaystyle(a\otimes x\otimes b)(a^{\prime}\otimes x^{\prime}\otimes
b^{\prime})=a(x_{1}\triangleright a^{\prime})\otimes
x_{2}x^{\prime}_{1}\otimes(b\triangleleft x^{\prime}_{2})b^{\prime},$ (4)

and a coalgebra (called the two-sided smash coproduct, $A\times H\times B$)
with counit $\varepsilon(a\otimes x\otimes
b)=\varepsilon_{A}(a)\varepsilon_{H}(x)\varepsilon_{B}(b)$ and
comultiplication

$\displaystyle\Delta(a\otimes x\otimes b)=a_{1}\otimes a_{2-1}x_{1}\otimes
b_{1{[0]}}\otimes a_{20}\otimes x_{2}b_{1{[1]}}\otimes b_{2}.$ (5)

Moreover, assume that Eq.(1) holds, it follows that $A\otimes H\otimes B$ with
two-sided smash product algebra and two-sided smash product coalgebra is a
bialgebra, called the Majid’s double biproduct, denoted by $A\star H\star B$.

######  Remark 1.10.

(1) Replacing $A$ (or $B$) with $K$, the double biproduct is exactly the right
(or left) variant of Radford’s biproduct.

(2) The double biproduct here is actually the case of Majid’s in [17, Theorem
A.1] with a trivial pairing.

##  2\. Brzeziński’s two-sided crossed product

In this section, we will first introduce the notion of Brzeziński’s two-sided
crossed product generalizing the two-sided smash product. First we need to
list the right version of some crossed products.

###  2.1. Right Brzeziński’s crossed product and its special cases

######  Proposition 2.1.

(Right Brzeziński’s crossed product) ([19, Theorem 2.1]) Let $H$ be a vector
space with a distinguished element $1_{H}\in H$ and $B$ an algebra. Let
$F:H\otimes H\longrightarrow H\otimes B$ (write $F(x\otimes
x^{\prime})=x_{F}\otimes x^{\prime F}$ for all $x,x^{\prime}\in H$) and
$T:B\otimes H\longrightarrow H\otimes B$ (write $T(b\otimes x)=x_{T}\otimes
b_{T}$ for all $x\in H$ and $b\in B$) be two linear maps such that

$\displaystyle(RB1)$ $\displaystyle 1_{HT}\otimes b_{T}=1_{H}\otimes
b,~{}x_{T}\otimes 1_{BT}=x\otimes 1_{B}$

holds. Then $H^{F}_{T}\\#B$ (=$H\otimes B$ as a vector space) is an
associative algebra with unit $1_{H}\otimes{1_{B}}$ and multiplication

$\displaystyle(x\otimes b)(x^{\prime}\otimes
b^{\prime})=x_{F}\otimes{x^{\prime}_{T}}^{F}b_{T}b^{\prime}$

for all $b,b^{\prime}\in B$ and $x,x^{\prime}\in H$ if and only if the
following conditions hold:

$\displaystyle(RB2)$ $\displaystyle
x_{T}\otimes(bb^{\prime})_{T}=x_{Tt}\otimes b_{t}b^{\prime}_{T};$
$\displaystyle(RB3)$ $\displaystyle 1_{HF}\otimes
x^{F}=x_{F}\otimes{1_{H}}^{F}=x\otimes 1_{B};$ $\displaystyle(RB4)$
$\displaystyle x_{Ff}\otimes{x^{\prime\prime}_{T}}^{f}{x^{\prime
F}}_{T}=x_{f}\otimes{x^{\prime}_{F}}^{f}x^{\prime\prime F};$
$\displaystyle(RB5)$ $\displaystyle
x_{TF}\otimes{x^{\prime}_{t}}^{F}{b}_{Tt}=x_{FT}\otimes{b_{T}}x^{\prime F},$

where $b,b^{\prime}\in B$, $x,x^{\prime},x^{\prime\prime}\in H$ and $f=F,t=T$.

######  Example 2.2.

(1) (Right twisted crossed product) Let $B$ be an algebra, $H$ a bialgebra,
$\tau:H\otimes H\longrightarrow B$ and $T:B\otimes H\longrightarrow H\otimes
B$ two linear maps such that Eq.(RB1) holds. Then $H^{\tau}_{T}\\#B$
(=$H\otimes B$ as a vector space) is an associative algebra with unit
$1_{H}\otimes{1_{B}}$ and multiplication

$(x\otimes b)(x^{\prime}\otimes
b^{\prime})=x_{1}x^{\prime}_{T1}\otimes\tau(x_{2},x^{\prime}_{T2})b_{T}b^{\prime}$

for all $a,a^{\prime}\in A$ and $x,x^{\prime}\in H$ if and only if Eq.(RB2)
and the following conditions hold:

$\displaystyle(RTC2)$
$\displaystyle\tau(x,1_{H})=\tau(1_{H},x)=\varepsilon(x)1_{B};$
$\displaystyle(RTC3)$ $\displaystyle
x_{1}x^{\prime}_{1}x^{\prime\prime}_{T1}\otimes\tau(x_{2}x^{\prime}_{2},x^{\prime\prime}_{T2})\tau(x_{3},x^{\prime}_{3})_{T}=x_{1}x^{\prime}_{1}x^{\prime\prime}_{1}\otimes\tau(x_{2},x^{\prime}_{2}x^{\prime\prime}_{2})\tau(x^{\prime}_{3},x^{\prime\prime}_{3});$
$\displaystyle(RTC4)$ $\displaystyle
x_{T1}x^{\prime}_{t1}\otimes\tau(x_{T2},x^{\prime}_{t2}){b}_{Tt}=(x_{1}x^{\prime}_{1})_{T}\otimes{b_{T}}\tau(x_{2},x^{\prime}_{2}),$

where $b,b^{\prime}\in B$, $x,x^{\prime},x^{\prime\prime}\in H$ and $t=T$.

###### Proof.

Let $F(x\otimes
x^{\prime})=x_{1}x^{\prime}_{1}\otimes\tau(x_{2},x^{\prime}_{2})$ in
Proposition 2.1. ∎

(2) (Right unified product) Let $B$, $H$ be two bialgebras and
$\blacktriangleright:B\otimes H\longrightarrow H$,
$\blacktriangleleft:B\otimes H\longrightarrow B$, $\tau:H\otimes
H\longrightarrow B$ linear maps such that

$(RU1)~{}~{}~{}~{}b_{1}\blacktriangleright{1_{H}}\otimes
b_{2}\blacktriangleleft{1_{H}}={1_{H}}\otimes b;~{}~{}1_{B}\blacktriangleright
x_{1}\otimes 1_{B}\blacktriangleleft x_{2}=x\otimes 1_{B}$

holds. Then $H^{\tau}_{\blacktriangleleft\blacktriangleright}\\#B$ (=$H\otimes
B$ as a vector space) is an associative algebra with unit ${1_{H}}\otimes
1_{B}$ and multiplication

$(x\otimes b)(x^{\prime}\otimes b^{\prime})=x_{1}(b_{1}\blacktriangleright
x^{\prime}_{1})_{1}\otimes\tau(x_{2},(b_{1}\blacktriangleright
x^{\prime}_{1})_{2})(b_{2}\blacktriangleleft x^{\prime}_{2})b^{\prime}$

for all $x,x^{\prime}\in H$ and $b,b^{\prime}\in B$ if and only if the
following conditions hold:

$\displaystyle(RU2)$ $\displaystyle b_{1}b^{\prime}_{1}\blacktriangleright
x_{1}\otimes b_{2}b^{\prime}_{2}\blacktriangleleft
x_{2}=b_{1}\blacktriangleright(b^{\prime}_{1}\blacktriangleright
x_{1})_{1}\otimes(b_{2}\blacktriangleleft(b^{\prime}_{1}\blacktriangleright
x_{1})_{2})(b^{\prime}_{2}\blacktriangleleft x_{2});$ $\displaystyle(RU3)$
$\displaystyle
x_{1}\otimes\tau(1_{H},x_{2})=x_{1}\otimes\tau(x_{2},1_{H})=x\otimes 1_{B};$
$\displaystyle(RU4)$
$\displaystyle~{}~{}x_{1}x^{\prime}_{1}(\tau(x_{3},x^{\prime}_{3})_{1}\blacktriangleright
x^{\prime\prime}_{1})_{1}\otimes\tau(x_{2}x^{\prime}_{2},(\tau(x_{3},x^{\prime}_{3})_{1}\blacktriangleright
x^{\prime\prime}_{1})_{2})(\tau(x_{3},x^{\prime}_{3})_{2}\blacktriangleright
x^{\prime\prime}_{2})$
$\displaystyle=x_{1}x^{\prime}_{1}x^{\prime\prime}_{1}\otimes\tau(x_{2},x^{\prime}_{2}x^{\prime\prime}_{2})\tau(x^{\prime}_{3},x^{\prime\prime}_{3});$
$\displaystyle(RU5)$ $\displaystyle(b_{1}\blacktriangleright
x_{1})_{1}((b_{2}\blacktriangleleft x_{2})_{1}\blacktriangleright
x^{\prime}_{1})_{1}\otimes\tau((b_{1}\blacktriangleright
x_{1})_{2},((b_{2}\blacktriangleleft x_{2})_{1}\blacktriangleright
x^{\prime}_{1})_{2})$ $\displaystyle\times((b_{2}\blacktriangleleft
x_{2})_{2}\blacktriangleleft x^{\prime}_{2})=b_{1}\blacktriangleright
x_{1}x^{\prime}_{1}\otimes(b_{2}\blacktriangleleft
x_{2}x^{\prime}_{2})\tau(x_{3},x^{\prime}_{3}),$

where $b,b^{\prime}\in B$, $x,x^{\prime},x^{\prime\prime}\in H$.

###### Proof.

Let $T(b\otimes x)=b_{1}\blacktriangleright x_{1}\otimes
b_{2}\blacktriangleleft x_{2}$ in $H^{\tau}_{T}\\#B$. ∎

(3) (Right $F$-twist product) Let $H$ be a vector space with a distinguished
element $1_{H}\in H$ and $B$ an algebra. Let $F:H\otimes H\longrightarrow
A\otimes H$ (write $F(x\otimes x^{\prime})=x_{F}\otimes x^{\prime F}$ for all
$x,x^{\prime}\in H$) be a linear maps. Then $H^{F}\\#B$ (=$H\otimes B$ as a
vector space) is an associative algebra with unit $1_{H}\otimes{1_{B}}$ and
multiplication

$(x\otimes b)(x^{\prime}\otimes
b^{\prime})=x_{F}\otimes{x^{\prime}}^{F}bb^{\prime}$

for all $b,b^{\prime}\in B$ and $x,x^{\prime}\in H$ if and only if Eq.(RB3)
and the following conditions hold:

$\displaystyle(RF2)$ $\displaystyle
x_{Ff}\otimes{x^{\prime\prime}}^{f}{x^{\prime
F}}=x_{f}\otimes{x^{\prime}_{F}}^{f}x^{\prime\prime F};$ $\displaystyle(RF3)$
$\displaystyle x_{F}\otimes{x^{\prime}}^{F}{b}=x_{F}\otimes{b}x^{\prime F},$

where $b,b^{\prime}\in B$, $x,x^{\prime},x^{\prime\prime}\in H$ and $f=F$.

###### Proof.

Let $T$ be trivial in Proposition 2.1. ∎

Here we remark that if we set
$F(x,x^{\prime})=x_{1}x^{\prime}_{1}\otimes\tau(x_{2},x^{\prime}_{2})$ in
$H^{F}\\#B$, then we get the right twist product $H^{\tau}\\#B$ and the
conditions $(RB3),(RF2)$, $(RF3)$ become the conditions $(RTC2)$ and

$\displaystyle(R2)$
$\displaystyle\tau(x_{1}x^{\prime}_{1},x^{\prime\prime})\tau(x_{2},x^{\prime}_{2})=\tau(x,x^{\prime}_{1}x^{\prime\prime}_{1})\tau(x^{\prime}_{2},x^{\prime\prime}_{2});$
$\displaystyle(R3)$ $\displaystyle\tau(x,x^{\prime})b=b\tau(x,x^{\prime}),$

which is the right version of $(LTC3),(L2)$, $(L3)$.

(4) (Right twisted tensor product) Let $H$, $B$ be two algebras and
$T:B\otimes H\longrightarrow H\otimes B$ (write $T(b\otimes x)=x_{T}\otimes
b_{T}$ for all $x\in H$ and $b\in B$) a linear map. Then $H_{T}\\#B$
(=$H\otimes B$ as a vector space) is an associative algebra with unit
$1_{H}\otimes{1_{B}}$ and multiplication

$(x\otimes b)(x^{\prime}\otimes b^{\prime})=x{x^{\prime}_{T}}\otimes
b_{T}b^{\prime}$

for all $b,b^{\prime}\in B$ and $x,x^{\prime}\in H$ if and only if Eqs.(RB1),
(RB2) and the following condition hold:

$\displaystyle(RT3)$ $\displaystyle
x_{T}{x^{\prime}_{t}}\otimes{b}_{Tt}=(xx^{\prime})_{T}\otimes{b_{T}},$

where $b,b^{\prime}\in B$, $x,x^{\prime}\in H$ and $t=T$.

###### Proof.

Let $F$ be trivial, i.e., $F(x\otimes x^{\prime})=xx^{\prime}\otimes 1_{B}$ in
Proposition 2.1. ∎

(5) (Right crossed product) ([14]) Let $H$ be a Hopf algebra with a right weak
action $\triangleleft$ on the algebra $B$. Let $\tau:H\otimes H\longrightarrow
B$ be a $K$-linear map. Let $H^{\tau}_{\triangleleft}\\#B$ be the (usually
nonassociative) algebra whose underlying space is $H\otimes B$ and whose
multiplication is given by

$(x\otimes b)(x^{\prime}\otimes
b^{\prime})=x_{1}x^{\prime}_{1}\otimes\tau(x_{2},x^{\prime}_{2})(a\triangleleft
x^{\prime}_{3})b^{\prime}$

for all $b,b^{\prime}\in B$ and $x,x^{\prime}\in H$. This
$H^{\tau}_{\triangleleft}\\#B$ is an associative algebra with unit
$1_{H}\otimes{1_{B}}$ if and only if $\tau$ satisfies Eq.(RTC2) and the
following conditions:

$\displaystyle(RC2)$
$\displaystyle\tau(x_{1}x^{\prime}_{1},x^{\prime\prime}_{1})(\tau(x_{2},x^{\prime}_{2})\triangleleft
x^{\prime\prime}_{2})=\tau(x,x^{\prime}_{1}x^{\prime\prime}_{1})\tau(x^{\prime}_{2},x^{\prime\prime}_{2});$
$\displaystyle(RC3)$
$\displaystyle\tau(x^{\prime}_{1},x^{\prime\prime}_{1})((b\triangleleft
x^{\prime}_{2})\triangleleft x^{\prime\prime}_{2})=(b\triangleleft
x^{\prime}_{1}x^{\prime\prime}_{1})\tau(x^{\prime}_{2},x^{\prime\prime}_{2})$

for all $b\in B$ and $x,x^{\prime},x^{\prime\prime}\in H$.

###### Proof.

Let $T(b\otimes x)=x_{1}\otimes b\triangleleft x_{2}$ in $H^{\tau}_{T}\\#B$. ∎

###  2.2. Brzeziński’s two-sided crossed product algebra and its special cases

Next we give the main result in this section.

######  Theorem 2.3.

Let $H$ be a coalgebra with an element $1_{H}$ such that
$\Delta(1_{H})=1_{H}\otimes 1_{H}$ and $A$, $B$ two algebras. Let $G:H\otimes
H\longrightarrow A\otimes H$, $R:H\otimes A\longrightarrow A\otimes H$,
$T:B\otimes H\longrightarrow H\otimes B$ and $\tau:H\otimes H\longrightarrow
B$ be linear maps, such that Eqs. $(LB1)$-$(LB3)$, $(RB1)$, $(RB2)$, $(RTC2)$
hold. Then $A\\#{{}_{R}^{G}H^{\tau}_{T}}\\#B$ $(=A\otimes H\otimes B$ as
vector space) is an associative algebra with unit $1_{A}\otimes
1_{H}\otimes{1_{B}}$ and multiplication

$(a\otimes x\otimes b)(a^{\prime}\otimes x^{\prime}\otimes
b^{\prime})=aa^{\prime}_{R}{x_{R1}}^{G}\otimes
x^{\prime}_{T1G}\otimes\tau(x_{R2},x^{\prime}_{T2})b_{T}b^{\prime}$

for all $a,a^{\prime}\in A$, $x,x^{\prime}\in H$ and $b,b^{\prime}\in B$ if
and only if the following conditions hold:

$\displaystyle(BT1)$
$\displaystyle~{}~{}{x_{1}}^{G}{x^{\prime}_{1G1}}^{g}\otimes
x^{\prime\prime}_{T1g}\otimes\tau(x^{\prime}_{1G2},x^{\prime\prime}_{T2}){\tau(x_{2},x^{\prime}_{2})}_{T}$
$\displaystyle~{}~{}~{}~{}~{}~{}~{}={{x^{\prime}_{1}}^{G}}_{R}{x_{R1}}^{g}\otimes
x^{\prime\prime}_{1G1g}\otimes\tau(x_{R2},x^{\prime\prime}_{1G2})\tau(x^{\prime}_{2},x^{\prime\prime}_{2});$
$\displaystyle(BT2)$ $\displaystyle~{}~{}a_{Rr}{x_{r1}}^{G}\otimes
x^{\prime}_{R1G}\otimes\tau(x_{r2},x^{\prime}_{R2})$
$\displaystyle~{}~{}~{}~{}~{}~{}~{}={x_{1}}^{G}a_{R}\otimes
x^{\prime}_{1GR}\otimes\tau(x_{2},x^{\prime}_{2});$ $\displaystyle(BT3)$
$\displaystyle~{}~{}{x_{1}}^{G}\otimes x^{\prime}_{1GT}\otimes
b_{T}\tau(x_{2},x^{\prime}_{2})$
$\displaystyle~{}~{}~{}~{}~{}~{}~{}={x_{T1}}^{G}\otimes
x^{\prime}_{t1G}\otimes\tau(x_{T2},x^{\prime}_{t2})b_{Tt};$
$\displaystyle(BT4)$ $\displaystyle~{}~{}a_{R}\otimes x_{RT}\otimes
b_{T}=a_{R}\otimes x_{TR}\otimes b_{T}.$

where $a\in A$, $x,x^{\prime},x^{\prime\prime}\in H$, $b\in B$ and
$R=r,G=g,T=t$. In this case, we call the algebra
$A\\#{{}_{R}^{G}H^{\tau}_{T}}\\#B$ Brzeziński’s two-sided crossed product
algebra.

###### Proof.

$(\Longleftarrow)$ We check the associativity as follows: for all
$a,a^{\prime},a^{\prime\prime}\in A$, $x,x^{\prime},x^{\prime\prime}\in H$ and
$b,b^{\prime},b^{\prime\prime}\in B$,

$\displaystyle((a\otimes x\otimes b)(a^{\prime}\otimes x^{\prime}\otimes
b^{\prime}))(a^{\prime\prime}\otimes x^{\prime\prime}\otimes
b^{\prime\prime})$ $\displaystyle=$ $\displaystyle
aa^{\prime}_{R}{x_{R1}}^{g}a^{\prime\prime}_{r}{x^{\prime}_{T1gr1}}^{G}\otimes
x^{\prime\prime}_{t1G}\otimes\tau(x^{\prime}_{T1gr2},x^{\prime\prime}_{t2})(\tau(x_{R2},x^{\prime}_{T2})b_{T}b^{\prime})_{t}b^{\prime\prime}$
$\displaystyle\stackrel{{\scriptstyle(RB2)}}{{=}}$ $\displaystyle
aa^{\prime}_{R}{x_{R1}}^{g}a^{\prime\prime}_{r}{x^{\prime}_{T1gr1}}^{G}\otimes
x^{\prime\prime}_{t\bar{T}\bar{t}1G}\otimes\tau(x^{\prime}_{T1gr2},x^{\prime\prime}_{t\bar{T}\bar{t}2}){\tau(x_{R2},x^{\prime}_{T2})}_{\bar{t}}b_{T\bar{T}}b^{\prime}_{t}b^{\prime\prime}$
$\displaystyle\stackrel{{\scriptstyle(BT2)}}{{=}}$ $\displaystyle
aa^{\prime}_{R}a^{\prime\prime}_{r\bar{R}}{x_{R\bar{R}1}}^{g}{x^{\prime}_{Tr1g1}}^{G}\otimes
x^{\prime\prime}_{t\bar{T}\bar{t}1G}\otimes\tau(x^{\prime}_{Tr1g2},x^{\prime\prime}_{t\bar{T}\bar{t}2}){\tau(x_{R\bar{R}2},x^{\prime}_{Tr2})}_{\bar{t}}b_{T\bar{T}}b^{\prime}_{t}b^{\prime\prime}$
$\displaystyle\stackrel{{\scriptstyle(BT4)}}{{=}}$ $\displaystyle
aa^{\prime}_{R}a^{\prime\prime}_{r\bar{R}}{x_{R\bar{R}1}}^{g}{x^{\prime}_{rT1g1}}^{G}\otimes
x^{\prime\prime}_{t\bar{T}\bar{t}1G}\otimes\tau(x^{\prime}_{rT1g2},x^{\prime\prime}_{t\bar{T}\bar{t}2})\tau(x_{R\bar{R}2},x^{\prime}_{rT2})_{\bar{t}}b_{T\bar{T}}b^{\prime}_{t}b^{\prime\prime}$

and

$\displaystyle(a\otimes x\otimes b)((a^{\prime}\otimes x^{\prime}\otimes
b^{\prime})(a^{\prime\prime}\otimes x^{\prime\prime}\otimes
b^{\prime\prime}))$ $\displaystyle=$ $\displaystyle
a(a^{\prime}a^{\prime\prime}_{r}{x^{\prime}_{r1}}^{g})_{R}{x_{R1}}^{G}\otimes
x^{\prime\prime}_{t1gT1G}\otimes\tau(x_{R2},x^{\prime\prime}_{t1gT2})b_{T}\tau(x^{\prime}_{r2},x^{\prime\prime}_{t2})b^{\prime}_{t}b^{\prime\prime}$
$\displaystyle\stackrel{{\scriptstyle(LB2)}}{{=}}$ $\displaystyle
aa^{\prime}_{R}(a^{\prime\prime}_{r}{x^{\prime}_{r1}}^{g})_{\bar{R}}{x_{R\bar{R}1}}^{G}\otimes
x^{\prime\prime}_{t1gT1G}\otimes\tau(x_{R\bar{R}2},x^{\prime\prime}_{t1gT2})b_{T}\tau(x^{\prime}_{r2},x^{\prime\prime}_{t2})b^{\prime}_{t}b^{\prime\prime}$
$\displaystyle\stackrel{{\scriptstyle(BT3)}}{{=}}$ $\displaystyle
aa^{\prime}_{R}(a^{\prime\prime}_{r}{x^{\prime}_{r\bar{T}1}}^{g})_{\bar{R}}{x_{R\bar{R}1}}^{G}\otimes
x^{\prime\prime}_{tT1g1G}\otimes\tau(x_{R\bar{R}2},x^{\prime\prime}_{tT1g2})\tau(x^{\prime}_{r\bar{T}2},x^{\prime\prime}_{tT2})b_{\bar{T}T}b^{\prime}_{t}b^{\prime\prime}$
$\displaystyle\stackrel{{\scriptstyle(LB2)}}{{=}}$ $\displaystyle
aa^{\prime}_{R}a^{\prime\prime}_{r\bar{R}}{{x^{\prime}_{r\bar{T}1}}^{g}}_{\bar{r}}{x_{R\bar{R}\bar{r}1}}^{G}\otimes
x^{\prime\prime}_{tT1g1G}\otimes\tau(x_{R\bar{R}\bar{r}2},x^{\prime\prime}_{tT1g2})\tau(x^{\prime}_{r\bar{T}2},x^{\prime\prime}_{tT2})b_{\bar{T}T}b^{\prime}_{t}b^{\prime\prime}$
$\displaystyle\stackrel{{\scriptstyle(BT1)}}{{=}}$ $\displaystyle
aa^{\prime}_{R}a^{\prime\prime}_{r\bar{R}}{x_{R\bar{R}1}}^{g}{x^{\prime}_{r\bar{T}1g1}}^{G}\otimes
x^{\prime\prime}_{tT\bar{t}1G}\otimes\tau(x^{\prime}_{r\bar{T}1g2},x^{\prime\prime}_{tT\bar{t}2}){\tau(x_{R\bar{R}2},x^{\prime}_{r\bar{T}2})}_{\bar{t}}b_{\bar{T}T}b^{\prime}_{t}b^{\prime\prime},$

so

$((a\otimes x\otimes b)(a^{\prime}\otimes x^{\prime}\otimes
b^{\prime}))(a^{\prime\prime}\otimes x^{\prime\prime}\otimes
b^{\prime\prime})=(a\otimes x\otimes b)((a^{\prime}\otimes x^{\prime}\otimes
b^{\prime})(a^{\prime\prime}\otimes x^{\prime\prime}\otimes
b^{\prime\prime})).$

We check the unit as follows: for all $a\in A$, $x\in H$ and $b\in B$,

$\displaystyle(a\otimes x\otimes b)(1_{A}\otimes 1_{H}\otimes 1_{B})$
$\displaystyle=$ $\displaystyle a1_{AR}{x_{R1}}^{G}\otimes
1_{HT1G}\otimes\tau(x_{R2},1_{HT2})b_{T}$
$\displaystyle\stackrel{{\scriptstyle(LB1)}}{{=}}$ $\displaystyle
a{x_{1}}^{G}\otimes 1_{HT1G}\otimes\tau(x_{2},1_{HT2})b_{T}$
$\displaystyle\stackrel{{\scriptstyle(RB1)}}{{=}}$ $\displaystyle
a{x_{1}}^{G}\otimes 1_{HG}\otimes\tau(x_{2},1_{H})b$
$\displaystyle\stackrel{{\scriptstyle(LB3)}}{{=}}$ $\displaystyle a\otimes
x_{1}\otimes\tau(x_{2},1_{H})b$
$\displaystyle\stackrel{{\scriptstyle(RTC2)}}{{=}}$ $\displaystyle a\otimes
x\otimes b.$ $\displaystyle(1_{A}\otimes 1_{H}\otimes 1_{B})(a\otimes x\otimes
b)$ $\displaystyle=$ $\displaystyle a_{R}{1_{HR1}}^{G}\otimes
x_{T1G}\otimes\tau(1_{HR2},x_{T2})1_{BT}b$
$\displaystyle\stackrel{{\scriptstyle(LB1)}}{{=}}$ $\displaystyle
a{1_{H}}^{G}\otimes x_{T1G}\otimes\tau(1_{H},x_{T2})1_{BT}b$
$\displaystyle\stackrel{{\scriptstyle(RB1)}}{{=}}$ $\displaystyle
a{1_{H}}^{G}\otimes x_{1G}\otimes\tau(1_{H},x_{2})b$
$\displaystyle\stackrel{{\scriptstyle(LB3)}}{{=}}$ $\displaystyle a\otimes
x_{1}\otimes\tau(1_{H},x_{2})b$
$\displaystyle\stackrel{{\scriptstyle(RTC2)}}{{=}}$ $\displaystyle a\otimes
x\otimes b.$

$(\Longrightarrow)$ It is straightforward. ∎

######  Remark 2.4.

Taking either $A=K$ or $B=K$, we obtain the right Brzeziński crossed product
and left twisted crossed product, respectively.

######  Example 2.5.

(1) (Two-sided twisted crossed product) Let $H$ be a bialgebra and $A$, $B$
two algebras. Let $\sigma:H\otimes H\longrightarrow A$, $\tau:H\otimes
H\longrightarrow B$, $R:H\otimes A\longrightarrow A\otimes H$ and $T:B\otimes
H\longrightarrow H\otimes B$ be linear maps such that Eqs.$(LB1)$, $(LB2)$,
$(RB1)$, $(RB2)$, $(LTC3)$, $(RTC3)$ hold. Then $A\\#{{}_{R}^{\
\sigma}H^{\tau}_{T}}\\#B$ $(=A\otimes H\otimes B$ as vector space) is an
associative algebra with unit $1_{A}\otimes 1_{H}\otimes{1_{B}}$ and
multiplication

$(a\otimes x\otimes b)(a^{\prime}\otimes x^{\prime}\otimes
b^{\prime})=aa^{\prime}_{R}\sigma(x_{R1},x^{\prime}_{T1})\otimes
x_{R2}x^{\prime}_{T2}\otimes\tau(x_{R3},x^{\prime}_{T3})b_{T}b^{\prime}$

for all $a,a^{\prime}\in A$, $x,x^{\prime}\in H$ and $b,b^{\prime}\in B$ if
and only if the conditions $(BT4)$ and

$\displaystyle(TC1)$
$\displaystyle\sigma(x_{1},x^{\prime}_{1})a_{R}\otimes(x_{2}x^{\prime}_{2})_{R}\otimes\tau(x_{3},x^{\prime}_{3})=a_{Rr}\sigma(x_{r1},x^{\prime}_{R1})\otimes
x_{r2}x^{\prime}_{R2}\otimes\tau(x_{r3},x^{\prime}_{R3})$ $\displaystyle(TC2)$
$\displaystyle\sigma(x_{1},x^{\prime}_{1})\sigma(x_{2}x^{\prime}_{2},x^{\prime\prime}_{T1})\otimes
x_{3}x^{\prime}_{3}x^{\prime\prime}_{T2}\otimes\tau(x_{4}x^{\prime}_{4},x^{\prime\prime}_{T3})\tau(x_{5},x^{\prime}_{5})_{T}$
$\displaystyle~{}~{}~{}~{}~{}~{}~{}~{}=\sigma(x^{\prime}_{1},x^{\prime\prime}_{1})_{R}\sigma(x_{R1},x^{\prime}_{2}x^{\prime\prime}_{2})\otimes
x_{R2}x^{\prime}_{3}x^{\prime\prime}_{3}\otimes\tau(x_{R3},x^{\prime}_{4}x^{\prime\prime}_{4})\tau(x^{\prime}_{5},x^{\prime\prime}_{5})$
$\displaystyle(TC3)$ $\displaystyle\sigma(x_{T1},x^{\prime}_{t1})\otimes
x_{T2}x^{\prime}_{t2}\otimes\tau(x_{T3},x^{\prime}_{t3})b_{Tt}=\sigma(x_{1},x^{\prime}_{1})\otimes(x_{2}x^{\prime}_{2})_{T}\otimes
b_{T}\tau(x_{3},x^{\prime}_{3})$

are satisfied for all $a\in A$, $x,x^{\prime},x^{\prime\prime}\in H$,
$b,b^{\prime}\in B$ and $R=r,T=t$.

###### Proof.

Let $G(x\otimes x^{\prime})=\sigma(x_{1},x^{\prime}_{1})\otimes
x_{2}x^{\prime}_{2}$ in Theorem 2.3. ∎

(2) (Two-sided twisted product) Let $H$ be a bialgebra and $A$, $B$ two
algebras. Let $\sigma:H\otimes H\longrightarrow A$, $\tau:H\otimes
H\longrightarrow B$ be linear maps and $A\\#^{\sigma}H$ left twisted product,
$H^{\tau}\\#B$ right twisted product. Then $A\\#{{}^{\sigma}H^{\tau}}\\#B$
$(=A\otimes H\otimes B$ as vector space) is an associative algebra with unit
$1_{A}\otimes 1_{H}\otimes{1_{B}}$ and multiplication

$(a\otimes x\otimes b)(a^{\prime}\otimes x^{\prime}\otimes
b^{\prime})=aa^{\prime}\sigma(x_{1},x^{\prime}_{1})\otimes
x_{2}x^{\prime}_{2}\otimes\tau(x_{3},x^{\prime}_{3})bb^{\prime}$

for all $a,a^{\prime}\in A$, $x,x^{\prime}\in H$ and $b,b^{\prime}\in B$.

###### Proof.

Let $R$ and $T$ be trivial in $A\\#{{}_{R}^{\ \sigma}H^{\tau}_{T}}\\#B$. And
the rest is straightforward. ∎

(3) (Two-sided twisted tensor product) Let $H$, $A$, $B$ be three algebras.
Let $R:H\otimes A\longrightarrow A\otimes H$, $T:B\otimes H\longrightarrow
H\otimes B$ be linear maps and $A\\#_{R}H$ left twisted tensor product,
$H_{T}\\#B$ right twisted tensor product. Then $A\\#{{}_{R}H_{T}}\\#B$
$(=A\otimes H\otimes B$ as vector space) is an associative algebra with unit
$1_{A}\otimes 1_{H}\otimes{1_{B}}$ and multiplication

$(a\otimes x\otimes b)(a^{\prime}\otimes x^{\prime}\otimes
b^{\prime})=aa^{\prime}_{R}\otimes x_{R}x^{\prime}_{T}\otimes b_{T}b^{\prime}$

for all $a,a^{\prime}\in A$, $x,x^{\prime}\in H$ and $b,b^{\prime}\in B$ if
and only if Eq.(BT4) holds.

###### Proof.

We only prove the necessity of $(BT4)$ as follows. And the rest is
straightforward by letting $\sigma$ and $\tau$ be trivial in $A\\#{{}_{R}^{\
\sigma}H^{\tau}_{T}}\\#B$. By the associativity in $A\\#{{}_{R}H_{T}}\\#B$, we
can get

$\displaystyle
aa^{\prime}_{R}a^{\prime\prime}_{r}\otimes(x_{R}x^{\prime}_{T})_{r}x^{\prime\prime}_{t}\otimes(b_{T}b^{\prime})_{t}b^{\prime\prime}=a(a^{\prime}a^{\prime\prime}_{R})_{r}\otimes
x_{r}(x^{\prime}_{R}x^{\prime\prime}_{T})_{t}\otimes
b_{t}b^{\prime}_{T}b^{\prime\prime}.$

Let $a=a^{\prime}=1_{A}$, $b^{\prime}=b^{\prime\prime}=1_{B}$ and
$x^{\prime}=x^{\prime\prime}=1_{H}$ in the above equation. Then we obtain
$(BT4)$. ∎

(4) (Two-sided crossed product) ([14]) Let $H$ be a bialgebra with a left weak
action $\triangleright$ on the algebra $A$ and a right weak action
$\triangleleft$ on the algebra $B$. Let $\sigma:H\otimes H\rightarrow A$ and
$\tau:H\otimes H\rightarrow B$ be linear maps and
$A\\#_{\triangleright}^{\sigma}H$ left crossed product,
$H^{\tau}_{\triangleleft}\\#B$ right crossed product. Then
$A\\#{{}_{\triangleright}^{\sigma}H^{\tau}_{\triangleleft}}\\#B$ $(=A\otimes
H\otimes B$ as vector space) is an associative algebra with unit $1_{A}\otimes
1_{H}\otimes{1_{B}}$ and multiplication

$(a\otimes x\otimes b)(a^{\prime}\otimes x^{\prime}\otimes
b^{\prime})=a(x_{1}\triangleright
a^{\prime})\sigma(x_{2},x^{\prime}_{1})\otimes
x_{3}x^{\prime}_{2}\otimes\tau(x_{4},x^{\prime}_{3})(b\triangleleft
x^{\prime}_{4})b^{\prime}$

for all $a,a^{\prime}\in A$, $x,x^{\prime}\in H$ and $b,b^{\prime}\in B$.

###### Proof.

The result can be proved by letting $R(a\otimes x)=x_{1}\triangleright
a\otimes x_{2}$ and $T(b\otimes x)=x_{1}\otimes b\triangleleft x_{2}$ in
Example 2.5 (1). ∎

######  Remark 2.6.

(1) When $\sigma$ and $\tau$ are trivial in
$A\\#{{}_{\triangleright}^{\sigma}H^{\tau}_{\triangleleft}}\\#B$, we can
obtain the two-sided smash product $A\\#H\\#B$ in [17, 18].

(2) Two-sided twisted tensor product $A\\#{{}_{R}H_{T}}\\#B$ in Example 2.5
(3) is exactly the iterated twisted tensor product in [9, Theorem 2.1] with
the map $B\otimes A\longrightarrow A\otimes B$ trivial. In this case, Eq.(BT4)
is exactly [9, Eq.(2.1)]. We note that Eq.(2.1) in [9, Theorem 2.1] is only a
sufficient condition for $A\\#{{}_{R}H_{T}}\\#B$ to be an associative algebra.
Here Eq.(BT4) is also necessary.

(3) Two-sided twisted tensor product $A\\#{{}_{R}H_{T}}\\#B$ in Example 2.5
(3) can be seen as the iterated crossed product in [19, Theorem 2.3] with the
structure maps $Q:B\otimes A\longrightarrow A\otimes H\otimes B,b\otimes
a\mapsto a\otimes 1_{H}\otimes b$, $\upsilon:A\otimes A\longrightarrow
A\otimes H,a\otimes a^{\prime}\mapsto aa^{\prime}\otimes 1_{H}$ and
$\sigma:B\otimes B\longrightarrow H\otimes B,b\otimes b^{\prime}\mapsto
1_{H}\otimes bb^{\prime}$. In this case, Eq.(BT4) is exactly [19, Eq.(2.9)]
and [19, Eqs.(2.10) and (2.11)] hold automatically. Eq.(2.9) in [19, Theorem
2.3] is just one sufficient condition for $A\\#{{}_{R}H_{T}}\\#B$ to be an
associative algebra.

###  2.3. Brzeziński’s two-sided crossed product bialgebra

######  Theorem 2.7.

Let $A$, $B$ be bialgebras and $H$ a coalgebra with an element $1_{H}$ such
that $\Delta(1_{H})=1_{H}\otimes 1_{H}$. Let $G:H\otimes H\longrightarrow
A\otimes H$, $R:H\otimes A\longrightarrow A\otimes H$, $T:B\otimes
H\longrightarrow H\otimes B$ and $\tau:H\otimes H\longrightarrow B$ be linear
maps such that
$\varepsilon_{B}(\tau(x,x^{\prime}))=\varepsilon_{H}(x)\varepsilon_{H}(x^{\prime})$.
Then $A\\#{{}_{R}^{G}H^{\tau}_{T}}\\#B$ equipped with the two-sided tensor
product coalgebra $A\otimes H\otimes B$ becomes a bialgebra if and only if the
following conditions hold:

$\displaystyle(G1)$
$\displaystyle\varepsilon_{A}(x^{G})\varepsilon_{H}({x^{\prime}}_{G})=\varepsilon_{H}(x)\varepsilon_{H}(x^{\prime});$
$\displaystyle(G2)$
$\displaystyle\varepsilon_{A}(a_{R})\varepsilon_{H}(x_{R})=\varepsilon_{A}(a)\varepsilon_{H}(x);\varepsilon_{B}(b_{T})\varepsilon_{H}(x_{T})=\varepsilon_{B}(b)\varepsilon_{H}(x);$
$\displaystyle(G3)$ $\displaystyle{a}_{R1}\otimes x_{R1}\otimes{a}_{R2}\otimes
x_{R2}={a}_{1R}\otimes x_{1R}\otimes{a}_{2r}\otimes x_{2r};$
$\displaystyle(G4)$ $\displaystyle x_{T1}\otimes b_{T1}\otimes x_{T2}\otimes
b_{T2}=x_{1T}\otimes b_{1T}\otimes x_{2t}\otimes b_{2t};$ $\displaystyle(G5)$
$\displaystyle{{x_{1}}^{G}}_{1}\otimes
x^{\prime}_{1G1}\otimes{\tau(x_{2},x^{\prime}_{2})}_{1}\otimes{{x_{1}}^{G}}_{2}\otimes
x^{\prime}_{1G2}\otimes{\tau(x_{2},x^{\prime}_{2})}_{2}$
$\displaystyle~{}~{}~{}~{}~{}~{}~{}={x_{1}}^{G}\otimes
x^{\prime}_{1G}\otimes\tau(x_{2},x^{\prime}_{2})\otimes{x_{3}}^{g}\otimes
x^{\prime}_{3g}\otimes\tau(x_{4},x^{\prime}_{4}),$

where $a\in A$, $x,x^{\prime}\in H$ and $b\in B$. In this case, we call this
bialgebra Brzeziński’s two-sided crossed product bialgebra and denoted by
$A\star{{}_{R}^{G}H^{\tau}_{T}}\star B$.

###### Proof.

$(\Longleftarrow)$ First by Eqs.(G1) and (G2), we can show that
$\varepsilon_{A\otimes H\otimes B}$ is an algebra map. Next, we check that
$\Delta_{A\otimes H\otimes B}$ is an algebra map. In fact, for all
$a,a^{\prime}\in A$, $x,x^{\prime}\in H$ and $b,b^{\prime}\in B$, we have

$\displaystyle\Delta_{A\otimes H\otimes B}((a\otimes x\otimes
b)(a^{\prime}\otimes x^{\prime}\otimes b^{\prime}))$ $\displaystyle=$
$\displaystyle\Delta_{A\otimes H\otimes B}(aa^{\prime}_{R}{x_{R1}}^{G}\otimes
x^{\prime}_{T1G}\otimes\tau(x_{R2},x^{\prime}_{T2})b_{T}b^{\prime})$
$\displaystyle=$ $\displaystyle a_{1}a^{\prime}_{R1}{{x_{R1}}^{G}}_{1}\otimes
x^{\prime}_{T1G1}\otimes{\tau(x_{R2},x^{\prime}_{T2})}_{1}b_{T1}b^{\prime}_{1}\otimes
a_{2}a^{\prime}_{R2}{{x_{R1}}^{G}}_{2}\otimes
x^{\prime}_{T1G2}\otimes{\tau(x_{R2},x^{\prime}_{T2})}_{2}b_{T2}b^{\prime}_{2}$
$\displaystyle\stackrel{{\scriptstyle(G5)}}{{=}}$ $\displaystyle
a_{1}a^{\prime}_{R1}{x_{R11}}^{G}\otimes
x^{\prime}_{T11G}\otimes\tau(x_{R12},x^{\prime}_{T12})b_{T1}b^{\prime}_{1}\otimes
a_{2}a^{\prime}_{R2}{x_{R21}}^{g}\otimes
x^{\prime}_{T21g}\otimes\tau(x_{R22},x^{\prime}_{T22})b_{T2}b^{\prime}_{2}$
$\displaystyle\stackrel{{\scriptstyle(G3)}}{{=}}$ $\displaystyle
a_{1}a^{\prime}_{1R}{x_{1R1}}^{G}\otimes
x^{\prime}_{T11G}\otimes\tau(x_{1R2},x^{\prime}_{T12})b_{T1}b^{\prime}_{1}\otimes
a_{2}a^{\prime}_{2r}{x_{2r1}}^{g}\otimes
x^{\prime}_{T21g}\otimes\tau(x_{2r2},x^{\prime}_{T22})b_{T2}b^{\prime}_{2}$
$\displaystyle\stackrel{{\scriptstyle(G4)}}{{=}}$ $\displaystyle
a_{1}a^{\prime}_{1R}{x_{1R1}}^{G}\otimes
x^{\prime}_{1T1G}\otimes\tau(x_{1R2},x^{\prime}_{1T2})b_{1T}b^{\prime}_{1}\otimes
a_{2}a^{\prime}_{2r}{x_{2r1}}^{g}\otimes
x^{\prime}_{2t1g}\otimes\tau(x_{2r2},x^{\prime}_{2t2})b_{2t}b^{\prime}_{2}$
$\displaystyle=$ $\displaystyle\Delta_{A\otimes H\otimes B}(a\otimes x\otimes
b)\Delta_{A\otimes H\otimes B}(a^{\prime}\otimes x^{\prime}\otimes
b^{\prime})$

and

$\displaystyle\Delta_{A\otimes H\otimes B}(1_{A}\otimes 1_{H}\otimes 1_{B})$
$\displaystyle=$ $\displaystyle 1_{A1}\otimes 1_{H1}\otimes 1_{B1}\otimes
1_{A2}\otimes 1_{H2}\otimes 1_{B2}$ $\displaystyle=$ $\displaystyle
1_{A}\otimes 1_{H}\otimes 1_{B}\otimes 1_{A}\otimes 1_{H}\otimes 1_{B}.$

$(\Longrightarrow)$ Since $\varepsilon_{A\times H\times B}$ is an algebra map,
then we have

$\displaystyle(BA2)$
$\displaystyle~{}~{}\varepsilon_{A}(aa^{\prime}_{R}{x_{R1}}^{G})\varepsilon_{H}(x^{\prime}_{T1G})\varepsilon_{B}(\tau(x_{R2},x^{\prime}_{T2})b_{T}b^{\prime})=\varepsilon_{A}(a)\varepsilon_{H}(x)\varepsilon_{B}(b)\varepsilon_{A}(a^{\prime})\varepsilon_{H}(x^{\prime})\varepsilon_{B}(b^{\prime}).$

Let $a=a^{\prime}=1_{A}$,$b=b^{\prime}=1_{B}$ in Eq.$(BA2)$, we obtain $(G1)$.
Similarly, $(G2)$ holds.

By $\Delta((a\otimes x\otimes b)(a^{\prime}\otimes x^{\prime}\otimes
b^{\prime}))=\Delta(a\otimes x\otimes b)\Delta(a^{\prime}\otimes
x^{\prime}\otimes b^{\prime})$, we have

$\displaystyle(BA3)$
$\displaystyle~{}~{}(aa^{\prime}_{R}{x_{R1}}^{G})_{1}\otimes
x^{\prime}_{T1G1}\otimes(\tau(x_{R2},x^{\prime}_{T2})b_{T}b^{\prime})_{1}\otimes(aa^{\prime}_{R}{x_{R1}}^{G})_{2}\otimes
x^{\prime}_{T1G2}\otimes(\tau(x_{R2},x^{\prime}_{T2})b_{T}b^{\prime})_{2}$
$\displaystyle~{}~{}~{}~{}~{}~{}~{}=a_{1}a^{\prime}_{1R}{x_{1R1}}^{G}\otimes
x^{\prime}_{1T1G}\otimes\tau(x_{1R2},x^{\prime}_{1T2})b_{1T}b^{\prime}_{1}\otimes
a_{2}a^{\prime}_{2r}{x_{2r1}}^{g}\otimes
x^{\prime}_{2t1g}\otimes\tau(x_{2r2},x^{\prime}_{2t2})b_{2t}b^{\prime}_{2}.$

Let $a=1_{A},b=b^{\prime}=1_{B},x^{\prime}=1_{H}$ in Eq.$(BA3)$, we have

$\displaystyle(a^{\prime}_{R}{x_{R1}}^{G})_{1}\otimes
1_{HT1G1}\otimes(\tau(x_{R2},1_{HT2})1_{BT})_{1}\otimes(a^{\prime}_{R}{x_{R1}}^{G})_{2}\otimes
1_{HT1G2}$
$\displaystyle\otimes(\tau(x_{R2},1_{HT2})1_{BT})_{2}=a^{\prime}_{1R}{x_{1R1}}^{G}\otimes
1_{HT1G}\otimes\tau(x_{1R2},1_{HT2})1_{BT}$ $\displaystyle\otimes
a^{\prime}_{2r}{x_{2r1}}^{g}\otimes
1_{Ht1g}\otimes\tau(x_{2r2},1_{Ht2})1_{Bt}.$

Then we can obtain $(G3)$ by $(LB1),(LB3)$ and $(RB1),(RTC2)$. Liewise, one
has $(G4)$ and $(G5)$.∎

######  Definition 2.8.

Let $A,B$ be two algebras, and $H$ be a coalgebra with a distinguished element
$1_{H}\in H$. Let $G:H\otimes H\longrightarrow A\otimes H$ , $\tau:H\otimes
H\longrightarrow B$ and $S:H\longrightarrow H$ be linear maps. Then $S$ is
called a $(G,\tau)$-antipode of $H$ if for all $x\in H$, the following
conditions hold:

$\displaystyle(I1)~{}~{}{S(x_{1})_{1}}^{G}\otimes
x_{2G}\otimes\tau(S(x_{1})_{2},x_{3})=1_{A}\otimes 1_{H}\otimes
1_{B}\varepsilon_{H}(x);$ $\displaystyle(I2)~{}~{}{x_{1}}^{G}\otimes
S(x_{3})_{1G}\otimes\tau(x_{2},S(x_{3})_{2})=1_{A}\otimes 1_{H}\otimes
1_{B}\varepsilon_{H}(x).$

In this case, we call $H$ a $(G,\tau)$-Hopf algebra.

######  Remark 2.9.

(1) Let $G$ and $\tau$ be trivial, i.e., $x^{G}\otimes y_{G}=1_{A}\otimes xy$
and $\tau(x,y)=1_{B}\varepsilon(x)v(y)$ hold for all $x,y\in H$. Then $(I1)$
and $(I2)$ can ensure that $S$ is the antipode of $H$. Therefore in this case,
the Hopf algebra $H$ is a $(G,\tau)$-Hopf algebra.
Remarks (2) Let $\sigma:H\otimes H\longrightarrow A$ be linear map. When
$G(x,x^{\prime})=\sigma(x_{1},x^{\prime}_{1})\otimes x_{2}x^{\prime}_{2}$ in
Definition 2.8, we call $H$ a $(\sigma,\tau)$-Hopf algebra.

######  Proposition 2.10.

Let $A,B$ be two Hopf algebras with antipode $S_{A},S_{B}$, $H$ a coalgebra
with an element $1_{H}$ such that $\Delta(1_{H})=1_{H}\otimes 1_{H}$ and
$S_{H}:H\longrightarrow H$ a linear map. Suppose that
$A\star{{}_{R}^{G}H^{\tau}_{T}}\star B$ is a Brzeziński’s two-sided crossed
product bialgebra. Then $A\star{{}_{R}^{G}H^{\tau}_{T}}\star B$ is a Hopf
algebra with antipode $\overline{S}$ defined by

$\displaystyle\overline{S}(a\otimes x\otimes b)=(1_{A}\otimes 1_{H}\otimes
S_{B}(b))(1_{A}\otimes S_{H}(x)\otimes 1_{B})(S_{A}(a)\otimes 1_{H}\otimes
1_{B})$ (6)

if and only if $H$ is a $(G,\tau)$-Hopf algebra.

###### Proof.

$(\Longleftarrow)$ For all $a\in A,b\in B,x\in H$, and $R=r,T=t$, we have

$\displaystyle(\overline{S}\ast id_{A\star{{}_{R}^{\ \sigma}H^{\tau}_{T}}\star
B})(a\otimes x\otimes b)$ $\displaystyle=$ $\displaystyle
S_{A}(a_{1})_{R}a_{2r}{S_{H}(x_{1})_{TRr1}}^{G}\otimes
x_{2t1G}\otimes\tau(S_{H}(x_{1})_{TRr2},x_{2t2})S_{B}(b_{1})_{Tt}b_{2}$
$\displaystyle\stackrel{{\scriptstyle(LB2)}}{{=}}$
$\displaystyle(S_{A}(a_{1})a_{2})_{R}{S_{H}(x_{1})_{TR1}}^{G}\otimes
x_{2t1G}\otimes\tau(S_{H}(x_{1})_{TR2},x_{2t2})S_{B}(b_{1})_{Tt}b_{2}$
$\displaystyle\stackrel{{\scriptstyle(BT3)}}{{=}}$
$\displaystyle{S_{H}(x_{1})_{1}}^{G}\otimes x_{2GT}\otimes
S(b_{1})_{T}\tau(S(x_{1})_{2},x_{3})b_{2}\varepsilon(a)$
$\displaystyle\stackrel{{\scriptstyle(I1)}}{{=}}$ $\displaystyle 1_{A}\otimes
1_{HT}\otimes S(b_{1})_{T}b_{2}\varepsilon(a)\varepsilon(x)$ $\displaystyle=$
$\displaystyle 1_{A}\otimes 1_{H}\otimes
1_{B}\varepsilon(a)\varepsilon(x)\varepsilon(b)$

and

$\displaystyle(id_{A\star{{}_{R}^{\ \sigma}H^{\tau}_{T}}\star
B}\ast\overline{S})(a\otimes x\otimes b)$ $\displaystyle=$ $\displaystyle
a_{1}S_{A}(a_{2})_{Rr}{x_{1r1}}^{G}\otimes
S_{H}(x_{2})_{TRt1G}\otimes\tau(x_{1r2},S_{H}(x_{2})_{TRt2})b_{1t}S_{B}(b_{2})_{T}$
$\displaystyle\stackrel{{\scriptstyle(BT4)}}{{=}}$ $\displaystyle
a_{1}S_{A}(a_{2})_{Rr}{x_{1r1}}^{G}\otimes
S_{H}(x_{2})_{RTt1G}\otimes\tau(x_{1r2},S_{H}(x_{2})_{RTt2})b_{1t}S_{B}(b_{2})_{T}$
$\displaystyle\stackrel{{\scriptstyle(RB2)}}{{=}}$ $\displaystyle
a_{1}S_{A}(a_{2})_{Rr}{x_{1r1}}^{G}\otimes
S_{H}(x_{2})_{RT1G}\otimes\tau(x_{1r2},S_{H}(x_{2})_{RT2})(b_{1}S_{B}(b_{2}))_{T}$
$\displaystyle=$ $\displaystyle a_{1}S_{A}(a_{2})_{Rr}{x_{1r1}}^{G}\otimes
S_{H}(x_{2})_{R1G}\otimes\tau(x_{1r2},S_{H}(x_{2})_{R2})\varepsilon(b)$
$\displaystyle\stackrel{{\scriptstyle(BT2)}}{{=}}$ $\displaystyle
a_{1}{x_{1}}^{G}S_{A}(a_{2})_{R}\otimes
S_{H}(x_{3})_{1GR}\otimes\tau(x_{2},S_{H}(x_{3})_{2})\varepsilon(b)$
$\displaystyle\stackrel{{\scriptstyle(I2)}}{{=}}$ $\displaystyle
a_{1}S_{A}(a_{2})_{R}\otimes 1_{HR}\otimes 1_{B}\varepsilon(x)\varepsilon(b)$
$\displaystyle=$ $\displaystyle 1_{A}\otimes 1_{H}\otimes
1_{B}\varepsilon(a)\varepsilon(x)\varepsilon(b).$

Thus, $\overline{S}$ is the antipode of $A\\#{{}_{R}^{\
\sigma}H^{\tau}_{T}}\\#B$.

$(\Longrightarrow)$ Obvious. ∎

Next we only list two special cases. Other cases can be given similarly.

######  Corollary 2.11.

Let $H$, $A$, $B$ be bialgebras. Let $\sigma:H\otimes H\longrightarrow
A\otimes H$, $R:H\otimes A\longrightarrow A\otimes H$, $T:B\otimes
H\longrightarrow H\otimes B$ and $\tau:H\otimes H\longrightarrow B$ be linear
maps such that
$\varepsilon_{B}(\tau(x,x^{\prime}))=\varepsilon_{H}(x)\varepsilon_{H}(x^{\prime})$.
Then $A\\#{{}_{R}^{\sigma}H^{\tau}_{T}}\\#B$ equipped with the two-sided
tensor product coalgebra $A\otimes H\otimes B$ becomes a bialgebra if and only
if Eqs. $(G2)$-$(G4)$ and the following conditions hold:

$\displaystyle(J1)$
$\displaystyle\varepsilon_{A}(\sigma(x,x^{\prime}))=\varepsilon_{H}(x)\varepsilon_{H}(x^{\prime});$
$\displaystyle(J2)$
$\displaystyle{\sigma(x,x^{\prime})}_{1}\otimes{\sigma(x,x^{\prime})}_{2}={\sigma(x_{1},x^{\prime}_{1})}\otimes\sigma(x_{2},x^{\prime}_{2});$
$\displaystyle(J3)$
$\displaystyle{\tau(x,x^{\prime})}_{1}\otimes{\tau(x,x^{\prime})}_{2}={\tau(x_{1},x^{\prime}_{1})}\otimes\tau(x_{2},x^{\prime}_{2});$
$\displaystyle(J4)$ $\displaystyle{\tau(x_{1},x^{\prime}_{1})}\otimes
x_{2}x^{\prime}_{2}={\tau(x_{2},x^{\prime}_{2})}\otimes x_{1}x^{\prime}_{1};$
$\displaystyle(J5)$ $\displaystyle{\sigma(x_{1},x^{\prime}_{1})}\otimes
x_{2}x^{\prime}_{2}={\sigma(x_{2},x^{\prime}_{2})}\otimes x_{1}x^{\prime}_{1}$
$\displaystyle(J6)$
$\displaystyle\tau(x_{1},x^{\prime}_{1})\otimes\sigma(x_{2},x^{\prime}_{2})=\tau(x_{2},x^{\prime}_{2})\otimes\sigma(x_{1},x^{\prime}_{1}),$

where $x,x^{\prime}\in H$. In this case, we call this bialgebra two-sided
twisted crossed product bialgebra and denoted by
$A\star{{}_{R}^{\sigma}H^{\tau}_{T}}\star B$. Furthermore,
$A\star{{}_{R}^{\sigma}H^{\tau}_{T}}\star B$ is a Hopf algebra with antipode
Eq.(6) if and only if $H$ is a $(\sigma,\tau)$-Hopf algebra.

###### Proof.

Note here Eqs.$(J2)$-$(J6)$ are equivalent to Eq.$(G5)$ for
$G(x,x^{\prime})=\sigma(x_{1},x^{\prime}_{1})\otimes x_{2}x^{\prime}_{2}$. The
rest is obvious. ∎

######  Corollary 2.12.

Let $H$ be a bialgebra. Let $\sigma:H\otimes H\rightarrow A$ be a linear map,
where $A$ is a bialgebra with a left $H$-weak action such that
$\varepsilon_{A}(\sigma(x,x^{\prime}))=\varepsilon_{H}(x)\varepsilon_{H}(x^{\prime})$.
Let $\tau:H\otimes H\rightarrow B$ be a linear map, where $B$ is a bialgebra
with a right $H$-weak action such that
$\varepsilon_{B}(1_{B})=1,\varepsilon_{B}(\tau(x,x^{\prime}))=\varepsilon_{H}(x)\varepsilon_{H}(x^{\prime})$.
The two-sided crossed product $A\\#^{\sigma}H~{}^{\tau}\\#B$ equipped with the
two-sided tensor product coalgebra $A\otimes H\otimes B$ becomes a bialgebra
if and only if Eqs.$(J2)$-$(J6)$, $(D4)$, $(F3)$, $(F4)$ and the following
condition holds:

$\displaystyle(P1)$ $\displaystyle x_{2}\otimes(x_{1}\triangleright
a)=x_{1}\otimes(x_{2}\triangleright a),~{}~{}~{}(b\triangleleft x_{2})\otimes
x_{1}=(b\triangleleft x_{1})\otimes x_{2},$

where $x\in H,a\in A$. In this case, we call this bialgebra two-sided crossed
product bialgebra and denoted by $A\star{{}^{\sigma}H^{\tau}}\star B$.
Furthermore, $A\star{{}^{\sigma}H^{\tau}}\star B$ is a Hopf algebra with
antipode Eq.(6) if and only if $H$ is a $(\sigma,\tau)$-Hopf algebra.

###### Proof.

It is straightforward by setting the left and right coactions are trivial in
Theorem 3.2. ∎

######  Example 2.13.

Let $H=KC_{2}=K\\{1,a\\}$ and $A=B=KC_{4}=K\\{1,x,x^{2},x^{3}\\}$ be group
Hopf algebras ([22]), where $C_{i}$ is cyclic group with order $i$, $i=2,4$.
Define the linear maps $\triangleright:H\otimes A\longrightarrow A$,
$\triangleleft:B\otimes H\longrightarrow B$, $\sigma:H\otimes H\longrightarrow
A$ and $\tau:H\otimes H\longrightarrow B$ by

$\displaystyle 1_{H}\triangleright 1_{A}=1_{A},~{}1_{H}\triangleright
x=x,~{}1_{H}\triangleright x^{2}=x^{2},~{}1_{H}\triangleright x^{3}=x^{3},$
$\displaystyle a\triangleright 1_{A}=1_{A},~{}a\triangleright
x=x^{3},~{}a\triangleright x^{2}=x^{2},~{}a\triangleright x^{3}=x;$
$\displaystyle 1_{B}\triangleleft 1_{H}=1_{B},~{}x\triangleleft
1_{H}=x,~{}x^{2}\triangleleft 1_{H}=x^{2},~{}x^{3}\triangleleft 1_{H}=x^{3},$
$\displaystyle 1_{B}\triangleleft a=1_{B},~{}x\triangleleft
a=x^{3},~{}x^{2}\triangleleft a=x^{2},~{}x^{3}\triangleleft a=x;$

and

$\displaystyle\sigma(1_{H},1_{H})=1_{A},~{}\sigma(1_{H},a)=1_{A},~{}\sigma(a,1_{H})=1_{A},~{}\sigma(a,a)=x;$
$\displaystyle\tau(1_{H},1_{H})=1_{A},~{}\tau(1_{H},a)=1_{A},~{}\tau(a,1_{H})=1_{A},~{}\tau(a,a)=x,$

then by [1, Example 2.11], we know that $\triangleright$ is a weak action of
$H$ on $A$ and $\triangleleft$ is a weak action of $H$ on $B$. Thus we can get
two-sided crossed product $A\\#{{}^{\sigma}H^{\tau}}\\#B$. Also by Corollary
2.12, two-sided crossed product $A\\#^{\sigma}H~{}^{\tau}\\#B$ equipped with
the two-sided tensor product coalgebra $A\otimes H\otimes B$ becomes a
bialgebra, since $H$ and $A,B$ are cocommucative. But unfortunately $\bar{S}$
defined by Eq.(6) is not an antipode of $A\star{{}^{\sigma}H^{\tau}}\star B$,
because Eq.(I1) doesn’t hold for $a\in H$.

##  3\. An extended version of Majid’s double biproduct

Firstly, we recall the construction of double crossed biproduct in [14].

######  Lemma 3.1.

([14]) Let $H$ be a bialgebra. Let $\sigma:H\otimes H\rightarrow A$ be a
linear map, where $A$ is a left $H$-comodule coalgebra such that
$\varepsilon_{A}(1_{A})=1,\varepsilon_{A}(\sigma(x,x^{\prime}))=\varepsilon_{H}(x)\varepsilon_{H}(x^{\prime})$,
and an algebra with a left $H$-weak action. Let $\tau:H\otimes H\rightarrow B$
be a linear map, where $B$ is a right $H$-comodule coalgebra such that
$\varepsilon_{B}(1_{B})=1,\varepsilon_{B}(\tau(x,x^{\prime}))=\varepsilon_{H}(x)\varepsilon_{H}(x^{\prime})$,
and an algebra with a right $H$-weak action. The two-sided crossed product
$A\\#^{\sigma}H~{}^{\tau}\\#B$ equipped with the two-sided smash coproduct
$A\times H\times B$ becomes a bialgebra if and only if the following
conditions hold:

$\displaystyle(C1)$
$\displaystyle\varepsilon_{A},~{}\varepsilon_{B}\hbox{~{}are algebra maps};$
$\displaystyle(C2)$ $\displaystyle\Delta_{A}(1_{A})=1_{A}\otimes
1_{A},~{}\Delta_{B}(1_{B})=1_{B}\otimes 1_{B};$ $\displaystyle(C3)$
$\displaystyle{1_{A}}_{-1}\otimes{1_{A}}_{0}=1_{H}\otimes
1_{A},~{}{1_{B}}_{[0]}\otimes{1_{B}}_{[1]}=1_{B}\otimes 1_{H};$
$\displaystyle(C4)$ $\displaystyle\varepsilon_{A}(x\triangleright
a)=\varepsilon_{A}(a)\varepsilon_{H}(x),~{}\varepsilon_{B}(b\triangleleft
x)=\varepsilon_{B}(b)\varepsilon_{H}(x);$ $\displaystyle(C5)$ $\displaystyle
a_{1}\otimes a_{2-1}x\otimes a_{20}=a_{1}\sigma(a_{2-11},x_{1})\otimes
a_{2-12}x_{2}\otimes a_{20};$ $\displaystyle(C6)$ $\displaystyle
b_{1[0]}\otimes xb_{1[1]}\otimes b_{2}=b_{1[0]}\otimes
x_{1}b_{1[1]}\otimes\tau(x_{2},b_{1[2]})b_{2};$ $\displaystyle(C7)$
$\displaystyle(aa^{\prime})_{1}\otimes(aa^{\prime})_{2-1}\otimes(aa^{\prime})_{20}$
$\displaystyle~{}~{}~{}~{}~{}~{}~{}=a_{1}(a_{2-11}\triangleright
a^{\prime}_{1})\sigma(a_{2-12},a^{\prime}_{2-11})\otimes
a_{2-13}a^{\prime}_{2-12}\otimes a_{20}a^{\prime}_{20};$ $\displaystyle(C8)$
$\displaystyle(bb^{\prime})_{1[0]}\otimes(bb^{\prime})_{1[1]}\otimes(bb^{\prime})_{2}=b_{1[0]}b^{\prime}_{1[0]}\otimes
b_{1[1]}b^{\prime}_{1[1]}\otimes\tau(b_{1[2]},b^{\prime}_{1[2]})(b_{2}\triangleleft
b^{\prime}_{1[3]})b^{\prime}_{2};$ $\displaystyle(C9)$
$\displaystyle\sigma(x_{1},x^{\prime}_{1})_{1}\otimes\sigma(x_{1},x^{\prime}_{1})_{2-1}x_{2}x^{\prime}_{2}\otimes\sigma(x_{1},x^{\prime}_{1})_{20}=\sigma(x_{1},x^{\prime}_{1})\otimes
x_{2}x^{\prime}_{2}\otimes\sigma(x_{3},x^{\prime}_{3});$ $\displaystyle(C10)$
$\displaystyle\tau(x_{2},x^{\prime}_{2})_{1[0]}\otimes
x_{1}x^{\prime}_{1}\tau(x_{2},x^{\prime}_{2})_{1[1]}\otimes\tau(x_{2},x^{\prime}_{2})_{2}=\tau(x_{1},x^{\prime}_{1})\otimes
x_{2}x^{\prime}_{2}\otimes\tau(x_{3},x^{\prime}_{3});$ $\displaystyle(C11)$
$\displaystyle(x_{1}\triangleright a)_{1}\otimes(x_{1}\triangleright
a)_{2-1}x_{2}\otimes(x_{1}\triangleright a)_{20}$
$\displaystyle~{}~{}~{}~{}~{}~{}~{}=(x_{1}\triangleright
a_{1})\sigma(x_{2},a_{2-11})\otimes x_{3}a_{2-12}\otimes x_{4}\triangleright
a_{20};$ $\displaystyle(C12)$ $\displaystyle(b\triangleleft
x^{\prime}_{2})_{1[0]}\otimes x^{\prime}_{1}(b\triangleleft
x^{\prime}_{2})_{1[1]}\otimes(b\triangleleft
x^{\prime}_{2})_{2}=b_{1[0]}\triangleleft x^{\prime}_{1}\otimes
b_{1[1]}x^{\prime}_{2}\otimes\tau(b_{1[2]},x^{\prime}_{3})(b_{2}\triangleleft
x^{\prime}_{4});$ $\displaystyle(C13)$ $\displaystyle
a_{-1}x_{1}a^{\prime}_{-1}x^{\prime}_{1}\otimes\tau(x_{4},x^{\prime}_{3})(b_{[0]}\triangleleft
x^{\prime}_{4})b^{\prime}_{[0]}\otimes a_{0}(x_{2}\triangleright
a^{\prime}_{0})\sigma(x_{3},x^{\prime}_{2})\otimes
x_{5}b_{[1]}x^{\prime}_{5}b^{\prime}_{[1]}$
$\displaystyle~{}~{}~{}~{}~{}~{}~{}=a_{-11}x_{1}a^{\prime}_{-11}x^{\prime}_{1}\otimes\tau(a_{-12}x_{2},a^{\prime}_{-12}x^{\prime}_{2})(b_{[0]}\triangleleft
a^{\prime}_{-13}x^{\prime}_{3})b^{\prime}_{[0]}$
$\displaystyle~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\otimes
a_{0}(x_{3}b_{[1]}\triangleright
a^{\prime}_{0})\sigma(x_{4}b_{[2]},x^{\prime}_{4}b^{\prime}_{[1]})\otimes
x_{5}b_{[3]}x^{\prime}_{5}b^{\prime}_{[2]}.$

In this case, we call this bialgebra the double crossed biproduct denoted by
$A\star^{\sigma}H~{}{{}^{\tau}\star}B$.

Next we give an improved version of the above result.

######  Theorem 3.2.

Under the assumption of Lemma 3.1. The two-sided crossed product
$A\\#^{\sigma}H~{}^{\tau}\\#B$ equipped with the two-sided smash coproduct
$A\times H\times B$ becomes a bialgebra if and only if $(C1)-(C6)$ and the
following conditions hold:

$\displaystyle(D7)$
$\displaystyle(aa^{\prime})_{1}\otimes(aa^{\prime})_{2}=a_{1}(a_{2-11}\triangleright
a^{\prime}_{1})\sigma(a_{2-12},a^{\prime}_{2-1})\otimes
a_{20}a^{\prime}_{20};$ $\displaystyle(D8)$
$\displaystyle(aa^{\prime})_{-1}\otimes(aa^{\prime})_{0}=a_{-1}a^{\prime}_{-1}\otimes
a_{0}a^{\prime}_{0};$ $\displaystyle(D9)$
$\displaystyle(bb^{\prime})_{1}\otimes(bb^{\prime})_{2}=b_{1[0]}b^{\prime}_{1[0]}\otimes\tau(b_{1[1]},b^{\prime}_{1[1]})(b_{2}\triangleleft
b^{\prime}_{1[2]})b^{\prime}_{2};$ $\displaystyle(D10)$
$\displaystyle(bb^{\prime})_{[0]}\otimes(bb^{\prime})_{[1]}=b_{[0]}b^{\prime}_{[0]}\otimes
b_{[1]}b^{\prime}_{[1]};$ $\displaystyle(D11)$
$\displaystyle\sigma(x_{1},x^{\prime}_{1})_{-1}x_{2}x^{\prime}_{2}\otimes\sigma(x_{1},x^{\prime}_{1})_{0}=x_{1}x^{\prime}_{1}\otimes\sigma(x_{2},x^{\prime}_{2});$
$\displaystyle(D12)$
$\displaystyle\sigma(x,x^{\prime})_{1}\otimes\sigma(x,x^{\prime})_{2}=\sigma(x_{1},x^{\prime}_{1})\otimes\sigma(x_{2},x^{\prime}_{2});$
$\displaystyle(D13)$ $\displaystyle\tau(x_{2},x^{\prime}_{2})_{[0]}\otimes
x_{1}x^{\prime}_{1}\tau(x_{2},x^{\prime}_{2})_{[1]}=\tau(x_{1},x^{\prime}_{1})\otimes
x_{2}x^{\prime}_{2};$ $\displaystyle(D14)$
$\displaystyle\tau(x,x^{\prime})_{1}\otimes\tau(x,x^{\prime})_{2}=\tau(x_{1},x^{\prime}_{1})\otimes\tau(x_{2},x^{\prime}_{2});$
$\displaystyle(D15)$ $\displaystyle(x\triangleright
a)_{1}\otimes(x\triangleright a)_{2}=(x_{1}\triangleright
a_{1})\sigma(x_{2},a_{2-1})\otimes x_{3}\triangleright a_{20};$
$\displaystyle(D16)$ $\displaystyle(x_{1}\triangleright
a)_{-1}x_{2}\otimes(x_{1}\triangleright a)_{0}=x_{1}a_{-1}\otimes
x_{2}\triangleright a_{0};$ $\displaystyle(D17)$ $\displaystyle(b\triangleleft
x_{2})_{[0]}\otimes x_{1}(b\triangleleft x_{2})_{[1]}=b_{[0]}\triangleleft
x_{1}\otimes b_{[1]}x_{2};$ $\displaystyle(D18)$ $\displaystyle(b\triangleleft
x)_{1}\otimes(b\triangleleft x)_{2}=b_{1[0]}\triangleleft
x_{1}\otimes\tau(b_{1[1]},x_{2})(b_{2}\triangleleft x_{3});$
$\displaystyle(D19)$ $\displaystyle a_{-1}x\otimes 1_{B}\otimes
a_{0}=a_{-11}x_{1}\otimes\tau(a_{-12},x_{2})\otimes a_{0};$
$\displaystyle(D20)$ $\displaystyle b_{[0]}\otimes 1_{A}\otimes
xb_{[1]}=b_{[0]}\otimes\sigma(x_{1},b_{[1]1})\otimes x_{2}b_{[1]2};$
$\displaystyle(D21)$ $\displaystyle b_{[1]}\triangleright a_{0}\otimes
b_{[0]}\triangleleft a_{-1}=a\otimes b;$ $\displaystyle(D22)$
$\displaystyle\tau(x_{3},x^{\prime}_{2})(b\triangleleft
x^{\prime}_{3})b^{\prime}\otimes a(x_{1}\triangleright
a^{\prime})\sigma(x_{2},x^{\prime}_{1})$
$\displaystyle=\tau(a_{-1}x_{1},a^{\prime}_{-1}x^{\prime}_{1})(b_{[0]}\triangleleft
x^{\prime}_{2})b^{\prime}_{[0]}\otimes a_{0}(x_{2}\triangleright
a^{\prime}_{0})\sigma(x_{3}b_{[1]},x^{\prime}_{3}b^{\prime}_{[1]}).$

###### Proof.

By Lemma 3.1, we only need to prove that the conditions $(C7)$-$(C13)$ are
equivalent to $(D7)$-$(D22)$.

  1. (a)

Step 1. Firstly, apply $id\otimes\varepsilon\otimes id$ to Eq.$(C7)$, we can
get Eq.$(D7)$; apply $\varepsilon\otimes id\otimes id$ to Eq.$(C7)$, then
Eq.$(D8)$ holds. On the other hand, by Eqs.$(D7)$ and $(D8)$, we have

$\displaystyle(aa^{\prime})_{1}\otimes(aa^{\prime})_{2-1}\otimes(aa^{\prime})_{20}$
$\displaystyle\stackrel{{\scriptstyle(D7)}}{{=}}$ $\displaystyle
a_{1}(a_{2-11}\triangleright
a^{\prime}_{1})\sigma(a_{2-12},a^{\prime}_{2-1})\otimes(a_{20}a^{\prime}_{20})_{-1}\otimes(a_{20}a^{\prime}_{20})_{0}$
$\displaystyle\stackrel{{\scriptstyle(D8)}}{{=}}$ $\displaystyle
a_{1}(a_{2-11}\triangleright
a^{\prime}_{1})\sigma(a_{2-12},a^{\prime}_{2-1})\otimes
a_{20-1}a^{\prime}_{20-1}\otimes a_{200}a^{\prime}_{200}$ $\displaystyle=$
$\displaystyle a_{1}(a_{2-11}\triangleright
a^{\prime}_{1})\sigma(a_{2-12},a^{\prime}_{2-11})\otimes
a_{2-13}a^{\prime}_{2-12}\otimes a_{20}a^{\prime}_{20},$

i.e., Eq.$(C7)$ holds. Thus Eq.$(C7)$ $\Leftrightarrow$ Eqs.$(D7)$ and $(D8)$.

Similarly, we can obtain that Eq.$(C8)$ $\Leftrightarrow$ Eqs.$(D9)$ and
$(D10)$; Eq.$(C9)$ $\Leftrightarrow$ Eqs.$(D11)$ and $(D12)$; Eq.$(C10)$
$\Leftrightarrow$ Eqs.$(D13)$ and $(D14)$; Eq.$(C11)$ $\Leftrightarrow$
Eqs.$(D15)$ and $(D16)$; Eq.$(C12)$ $\Leftrightarrow$ Eqs.$(D17)$ and $(D18)$.

  2. (b)

Step 2. Secondly, let $x=1,a^{\prime}=1,b=b^{\prime}=1$ in Eq.$(C13)$, we have

$\displaystyle a_{-1}x^{\prime}_{1}\otimes 1_{B}\otimes a_{0}\otimes
x^{\prime}_{2}=a_{-11}x^{\prime}_{1}\otimes\tau(a_{-12},x^{\prime}_{2})\otimes
a_{0}\otimes x^{\prime}_{3}$ (7)

Using $id\otimes id\otimes id\otimes\varepsilon$ to Eq.$(\ref{eq:c13-1})$, one
can get Eq.$(D19)$. Applying $\varepsilon\otimes id\otimes id$ to Eq.(D19), we
have

$\displaystyle a\otimes 1_{B}\varepsilon(x)=a_{0}\otimes\tau(a_{-1},x).$ (8)

Set $a=a^{\prime}=1,b=1,x^{\prime}=1$ in Eq.$(C13)$, we have

$\displaystyle x_{1}\otimes b^{\prime}_{[0]}\otimes 1_{A}\otimes
x_{2}b^{\prime}_{[1]}=x_{1}\otimes
b^{\prime}_{[0]}\otimes\sigma(x_{2},b^{\prime}_{[1]})\otimes
x_{3}b^{\prime}_{[2]}$ (9)

Applying $\varepsilon\otimes id\otimes id\otimes id$ to Eq.$(\ref{eq:c13-2})$,
one can get Eq.$(D20)$. Applying $id\otimes id\otimes\varepsilon$ to Eq.(D20),
we have

$\displaystyle b\otimes 1_{A}\varepsilon(x)=b_{[0]}\otimes\sigma(x,b_{[1]}).$
(10)

Setting $a=1,b^{\prime}=1,x=x^{\prime}=1$ in Eq.$(C13)$, we have

$\displaystyle a^{\prime}_{-1}\otimes b_{[0]}\otimes a^{\prime}_{0}\otimes
b_{[1]}=a^{\prime}_{-11}\otimes(b_{[0]}\triangleleft
a^{\prime}_{-12})\otimes(b_{[1]}\triangleright a^{\prime}_{0})\otimes
b_{[2]}.$ (11)

Applying $\varepsilon\otimes id\otimes id\otimes\varepsilon$ to
Eq.$(\ref{eq:c13-4})$, we have Eq.$(D21)$.

Applying $\varepsilon\otimes id\otimes id\otimes\varepsilon$ to Eq.$(C13)$, we
have

$\displaystyle\tau(x_{3},x^{\prime}_{2})(b\triangleleft
x^{\prime}_{3})b^{\prime}\otimes a(x_{1}\triangleright
a^{\prime})\sigma(x_{2},x^{\prime}_{1})$ (12)
$\displaystyle~{}~{}~{}~{}=\tau(a_{-1}x_{1},a^{\prime}_{-11}x^{\prime}_{1})(b_{[0]}\triangleleft
a^{\prime}_{-12}x^{\prime}_{2})b^{\prime}_{[0]}\otimes
a_{0}(x_{2}b_{[1]1}\triangleright
a^{\prime}_{0})\sigma(x_{3}b_{[1]2},x^{\prime}_{3}b^{\prime}_{[1]}).$

While

$\displaystyle\hbox{RHS~{} of~{}Eq}.(\ref{eq:c13-3})$
$\displaystyle\stackrel{{\scriptstyle(D19)(D20)}}{{=}}$
$\displaystyle\tau(a_{-1}x_{1},a^{\prime}_{-11}x^{\prime}_{1})(b_{[0]}\triangleleft
a^{\prime}_{-12}x^{\prime}_{2})\tau(a^{\prime}_{-13},x^{\prime}_{3})b^{\prime}_{[0]}$
$\displaystyle\otimes a_{0}\sigma(x_{2},b_{[1]1})(x_{3}b_{[1]2}\triangleright
a^{\prime}_{0})\sigma(x_{4}b_{[1]3},x^{\prime}_{4}b^{\prime}_{[1]})$
$\displaystyle\stackrel{{\scriptstyle(LC3)(RC3)}}{{=}}$
$\displaystyle\tau(a_{-1}x_{1},a^{\prime}_{-11}x^{\prime}_{1})\tau(a^{\prime}_{-12},x^{\prime}_{2})[(b_{[0]}\triangleleft
a^{\prime}_{-13})\triangleleft x^{\prime}_{3}]b^{\prime}_{[0]}$
$\displaystyle\otimes a_{0}(x_{2}\triangleright(b_{[1]1}\triangleright
a^{\prime}_{0}))\sigma(x_{3},b_{[1]2})\sigma(x_{4}b_{[1]3},x^{\prime}_{4}b^{\prime}_{[1]})$
$\displaystyle\stackrel{{\scriptstyle}}{{=}}$
$\displaystyle\tau(a_{-1}x_{1},a^{\prime}_{-11}x^{\prime}_{1})\tau(a^{\prime}_{-12},x^{\prime}_{2})((b_{[0][0]}\triangleleft
a^{\prime}_{0-1})\triangleleft x^{\prime}_{3})b^{\prime}_{[0]}$
$\displaystyle\otimes a_{0}(x_{2}\triangleright(b_{[0][1]}\triangleright
a^{\prime}_{00}))\sigma(x_{3},b_{[1]1})\sigma(x_{4}b_{[1]2},x^{\prime}_{4}b^{\prime}_{[1]})$
$\displaystyle\stackrel{{\scriptstyle(D21)}}{{=}}$
$\displaystyle\tau(a_{-1}x_{1},a^{\prime}_{-11}x^{\prime}_{1})\tau(a^{\prime}_{-12},x^{\prime}_{2})(b_{[0]}\triangleleft
x^{\prime}_{3})b^{\prime}_{[0]}$ $\displaystyle\otimes
a_{0}(x_{2}\triangleright
a^{\prime}_{0})\sigma(x_{3},b_{[1]1})\sigma(x_{4}b_{[1]2},x^{\prime}_{4}b^{\prime}_{[1]})$
$\displaystyle\stackrel{{\scriptstyle(D19)(D20)}}{{=}}$
$\displaystyle\tau(a_{-1}x_{1},a^{\prime}_{-1}x^{\prime}_{1})(b_{[0]}\triangleleft
x^{\prime}_{2})b^{\prime}_{[0]}\otimes a_{0}(x_{2}\triangleright
a^{\prime}_{0})\sigma(x_{3}b_{[1]},x^{\prime}_{3}b^{\prime}_{[1]})$
$\displaystyle=$ $\displaystyle\hbox{RHS~{} of~{}Eq}.(D22)$

and LHS of Eq.(12) is exactly LHS of Eq.$(D22)$, so we obtain Eq.$(D22)$.

On the other hand, we have

$\displaystyle\hbox{RHS~{} of~{}Eq}.(C13)$
$\displaystyle\stackrel{{\scriptstyle(D20)}}{{=}}$ $\displaystyle
a_{-11}x_{1}a^{\prime}_{-11}x^{\prime}_{1}\otimes\tau(a_{-12}x_{2},a^{\prime}_{-12}x^{\prime}_{2})(b_{[0]}\triangleleft
a^{\prime}_{-13}x^{\prime}_{3})b^{\prime}_{[0]}$ $\displaystyle\otimes
a_{0}\sigma(x_{3},b_{[1]1})(x_{4}b_{[1]2}\triangleright
a^{\prime}_{0})\sigma(x_{5}b_{[1]3},x^{\prime}_{4}b^{\prime}_{[1]1})\otimes
x_{6}b_{[1]4}x^{\prime}_{5}b^{\prime}_{[1]2}$
$\displaystyle\stackrel{{\scriptstyle(D19)}}{{=}}$ $\displaystyle
a_{-11}x_{1}a^{\prime}_{-11}x^{\prime}_{1}\otimes\tau(a_{-12}x_{2},a^{\prime}_{-12}x^{\prime}_{2})(b_{[0]}\triangleleft
a^{\prime}_{-13}x^{\prime}_{3})\tau(a^{\prime}_{-14},x^{\prime}_{4})b^{\prime}_{[0]}$
$\displaystyle\otimes a_{0}\sigma(x_{3},b_{[1]1})(x_{4}b_{[1]2}\triangleright
a^{\prime}_{0})\sigma(x_{5}b_{[1]3},x^{\prime}_{5}b^{\prime}_{[1]1})\otimes
x_{6}b_{[1]4}x^{\prime}_{6}b^{\prime}_{[1]2}$
$\displaystyle\stackrel{{\scriptstyle(LC3)(RC3)}}{{=}}$ $\displaystyle
a_{-11}x_{1}a^{\prime}_{-11}x^{\prime}_{1}\otimes\tau(a_{-12}x_{2},a^{\prime}_{-12}x^{\prime}_{2})\tau(a^{\prime}_{-13},x^{\prime}_{3})((b_{[0]}\triangleleft
a^{\prime}_{-14})\triangleleft x^{\prime}_{4})b^{\prime}_{[0]}$
$\displaystyle\otimes a_{0}(x_{3}\triangleright(b_{[1]1}\triangleright
a^{\prime}_{0}))\sigma(x_{4},b_{[1]2})\sigma(x_{5}b_{[1]3},x^{\prime}_{5}b^{\prime}_{[1]1})\otimes
x_{6}b_{[1]4}x^{\prime}_{6}b^{\prime}_{[1]2}$
$\displaystyle\stackrel{{\scriptstyle}}{{=}}$ $\displaystyle
a_{-11}x_{1}a^{\prime}_{-11}x^{\prime}_{1}\otimes\tau(a_{-12}x_{2},a^{\prime}_{-12}x^{\prime}_{2})\tau(a^{\prime}_{-13},x^{\prime}_{3})((b_{[0][0]}\triangleleft
a^{\prime}_{0-1})\triangleleft x^{\prime}_{4})b^{\prime}_{[0]}$
$\displaystyle\otimes a_{0}(x_{3}\triangleright(b_{[0][1]}\triangleright
a^{\prime}_{00}))\sigma(x_{4},b_{[1]1})\sigma(x_{5}b_{[1]2},x^{\prime}_{5}b^{\prime}_{[1]1})\otimes
x_{6}b_{[1]3}x^{\prime}_{6}b^{\prime}_{[1]2}$
$\displaystyle\stackrel{{\scriptstyle(D21)}}{{=}}$ $\displaystyle
a_{-11}x_{1}a^{\prime}_{-11}x^{\prime}_{1}\otimes\tau(a_{-12}x_{2},a^{\prime}_{-12}x^{\prime}_{2})\tau(a^{\prime}_{-13},x^{\prime}_{3})(b_{[0]}\triangleleft
x^{\prime}_{4})b^{\prime}_{[0]}$ $\displaystyle\otimes
a_{0}(x_{3}\triangleright
a^{\prime}_{0})\sigma(x_{4},b_{[1]1})\sigma(x_{5}b_{[1]2},x^{\prime}_{5}b^{\prime}_{[1]1})\otimes
x_{6}b_{[1]3}x^{\prime}_{6}b^{\prime}_{[1]2}$
$\displaystyle\stackrel{{\scriptstyle}}{{=}}$ $\displaystyle
a_{-11}x_{1}a^{\prime}_{-11}x^{\prime}_{1}\otimes\tau(a_{-12}x_{2},a^{\prime}_{-12}x^{\prime}_{2})\tau(a^{\prime}_{0-1},x^{\prime}_{3})(b_{[0][0]}\triangleleft
x^{\prime}_{4})b^{\prime}_{[0]}$ $\displaystyle\otimes
a_{0}(x_{3}\triangleright
a^{\prime}_{00})\sigma(x_{4},b_{[0][1]})\sigma(x_{5}b_{[1]1},x^{\prime}_{5}b^{\prime}_{[1]1})\otimes
x_{6}b_{[1]2}x^{\prime}_{6}b^{\prime}_{[1]2}$
$\displaystyle\stackrel{{\scriptstyle(\ref{eq:d19-1})(\ref{eq:d20-1})}}{{=}}$
$\displaystyle
a_{-11}x_{1}a^{\prime}_{-11}x^{\prime}_{1}\otimes\tau(a_{-12}x_{2},a^{\prime}_{-12}x^{\prime}_{2})(b_{[0]}\triangleleft
x^{\prime}_{3})b^{\prime}_{[0]}$ $\displaystyle\otimes
a_{0}(x_{3}\triangleright
a^{\prime}_{0})\sigma(x_{4}b_{[1]1},x^{\prime}_{4}b^{\prime}_{[1]1})\otimes
x_{5}b_{[1]2}x^{\prime}_{5}b^{\prime}_{[1]2}$
$\displaystyle\stackrel{{\scriptstyle}}{{=}}$ $\displaystyle
a_{-1}x_{1}a^{\prime}_{-11}x^{\prime}_{1}\otimes\tau(a_{0-1}x_{2},a^{\prime}_{-12}x^{\prime}_{2})(b_{[0]}\triangleleft
x^{\prime}_{3})b^{\prime}_{[0][0]}$ $\displaystyle\otimes
a_{00}(x_{3}\triangleright
a^{\prime}_{0})\sigma(x_{4}b_{[1]1},x^{\prime}_{4}b^{\prime}_{[0][1]})\otimes
x_{5}b_{[1]2}x^{\prime}_{5}b^{\prime}_{[1]}$
$\displaystyle\stackrel{{\scriptstyle}}{{=}}$ $\displaystyle
a_{-1}x_{1}a^{\prime}_{-1}x^{\prime}_{1}\otimes\tau(a_{0-1}x_{2},a^{\prime}_{0-1}x^{\prime}_{2})(b_{[0][0]}\triangleleft
x^{\prime}_{3})b^{\prime}_{[0][0]}$ $\displaystyle\otimes
a_{00}(x_{3}\triangleright
a^{\prime}_{00})\sigma(x_{4}b_{[0][1]},x^{\prime}_{4}b^{\prime}_{[0][1]})\otimes
x_{5}b_{[1]}x^{\prime}_{5}b^{\prime}_{[1]}$
$\displaystyle\stackrel{{\scriptstyle(D22)}}{{=}}$ $\displaystyle\hbox{LHS~{}
of~{}Eq}.(C13).$

Thus Eq.$(C13)$ $\Leftrightarrow$ Eqs.$(D19)$-$(D22)$. These finish the proof.

∎

######  Remark 3.3.

(1) Eq.$(D21)$ is exactly Eq.(1) in the Majid’s double biproduct.
Remarks (2) The first half parts in Eqs.$(C1)$-$(C4)$ and Eqs.$(D5)$, $(D7)$,
$(D8)$, $(D11)$, $(D12)$, $(D15)$, $(D16)$ are the conditions
$A_{1})$-$A_{9})$ and twisted comodule cocycle in [24, Theorem 2.5].
Remarks (3) The second half parts in Eqs.$(C1)$-$(C4)$ and Eqs.$(D6)$, $(D9)$,
$(D10)$, $(D13)$, $(D14)$, $(D17)$, $(D18)$ are the conditions $(G1)$-$(G9)$
in [14, Remark 2.3] and $(C6)$ in [14, Theorem 2.1]. These are the necessary
and sufficient conditions for the right crossed product $H~{}^{\tau}\\#B$
equipped with the right smash coproduct $H\times B$ becomes a bialgebra.

######  Proposition 3.4.

Let $H$ be a bialgebra. Let $\sigma:H\otimes H\rightarrow A$ be a linear map,
where $A$ is a left $H$-comodule coalgebra such that
$\varepsilon_{A}(1_{A})=1$, and an algebra with a left $H$-weak action. Let
$B$ be a right $H$-module algebra and also a comodule coalgebra such that
$\varepsilon_{B}(1_{B})=1$. The one-sided crossed product $A\\#^{\sigma}H\\#B$
equipped with the two-sided smash coproduct $A\times H\times B$ becomes a
bialgebra if and only if Eqs.$(C1)$-$(C5)$, $(D7)$, $(D8)$, $(D10)$-$(D12)$,
$(D15)$-$(D17)$, $(D20)$, $(D21)$ and the following conditions hold:

$\displaystyle(E1)$
$\displaystyle(bb^{\prime})_{1}\otimes(bb^{\prime})_{2}=b_{1}b^{\prime}_{1[0]}\otimes(b_{2}\triangleleft
b^{\prime}_{1[1]})b^{\prime}_{2};$ $\displaystyle(E2)$
$\displaystyle(b\triangleleft x)_{1}\otimes(b\triangleleft
x)_{2}=(b_{1}\triangleleft x_{1})\otimes(b_{2}\triangleleft x_{2});$
$\displaystyle(E3)$ $\displaystyle(b_{[0]}\triangleleft
x^{\prime}_{2})b^{\prime}\otimes(x_{1}\triangleright
a)\sigma(x_{2},x^{\prime}_{1})=(b_{[0]}\triangleleft
x^{\prime}_{1})b^{\prime}_{[0]}\otimes(x_{1}\triangleright
a)\sigma(x_{2}b_{[1]},x^{\prime}_{2}b^{\prime}_{[1]}).$

###### Proof.

Let $\tau$ be trivial, i.e., $\tau(x,y)=\varepsilon(x)\varepsilon(y)1_{B}$ in
Theorem 3.2. ∎

######  Remark 3.5.

(1) By Eq.$(D21)$, Eq.(1) in the Majid’s double biproduct is not only
sufficient but also necessary.
Remarks (2) The conditions in Proposition 3.4 are equivalent to the ones in
[10, Theorem 3.2].

###### Proof.

By Eq.$(D22)$, we have

$\displaystyle(b_{[0]}\triangleleft x^{\prime}_{1})b^{\prime}_{[0]}\otimes
a(x_{1}\triangleright
a^{\prime})\sigma(x_{2}b_{[1]},x^{\prime}_{2}b^{\prime}_{[1]})=(b\triangleleft
x^{\prime}_{2})b^{\prime}\otimes a(x_{1}\triangleright
a^{\prime})\sigma(x_{2},x^{\prime}_{1}).$ (13)

Setting $a=1=a^{\prime},b^{\prime}=1$ in Eq.(13), we have

$\displaystyle(b_{[0]}\triangleleft
x^{\prime}_{1})\otimes\sigma(xb_{[1]},x^{\prime}_{2})=(b\triangleleft
x^{\prime}_{2})\otimes\sigma(x,x^{\prime}_{1}).$ (14)

Setting $a=1=a^{\prime},b=1$ in Eq.(13), we get

$\displaystyle
b^{\prime}_{[0]}\otimes\sigma(x,x^{\prime}b^{\prime}_{[1]})=b^{\prime}\otimes\sigma(x,x^{\prime}).$
(15)

On the other hand, we have

$\displaystyle(b_{[0]}\triangleleft x^{\prime}_{1})b^{\prime}_{[0]}\otimes
a(x_{1}\triangleright
a^{\prime})\sigma(x_{2}b_{[1]},x^{\prime}_{2}b^{\prime}_{[1]})$
$\displaystyle\stackrel{{\scriptstyle(\ref{eq:d22-3})}}{{=}}$
$\displaystyle(b_{[0]}\triangleleft x^{\prime}_{1})b^{\prime}\otimes
a(x_{1}\triangleright a^{\prime})\sigma(x_{2}b_{[1]},x^{\prime}_{2})$
$\displaystyle\stackrel{{\scriptstyle(\ref{eq:d22-2})}}{{=}}$
$\displaystyle(b\triangleleft x^{\prime}_{2})b^{\prime}\otimes
a(x_{1}\triangleright a^{\prime})\sigma(x_{2},x^{\prime}_{1}).$

Therefore Eq.(13)$\Leftrightarrow$ Eqs.(14) and (15).

$(\Longrightarrow)$ We only check that Eq.$(C14)$ in [10, Theorem 3.2] holds.
In fact, one can compute as follows.

$\displaystyle\hbox{LHS~{} of~{}Eq}.(C14)$
$\displaystyle\stackrel{{\scriptstyle}}{{=}}$
$\displaystyle(b_{[0]}\triangleleft
x^{\prime}_{2})b^{\prime}_{[0]}\otimes(x_{1}\triangleright
a^{\prime})\sigma(x_{2},x^{\prime}_{1})\otimes
x_{3}b_{[1]}x^{\prime}_{3}b^{\prime}_{[1]}$
$\displaystyle\stackrel{{\scriptstyle(\ref{eq:d22-2})}}{{=}}$
$\displaystyle(b_{[0][0]}\triangleleft
x^{\prime}_{1})b^{\prime}_{[0]}\otimes(x_{1}\triangleright
a^{\prime})\sigma(x_{2}b_{[0][1]},x^{\prime}_{2})\otimes
x_{3}b_{[1]}x^{\prime}_{3}b^{\prime}_{[1]}$
$\displaystyle\stackrel{{\scriptstyle(\ref{eq:d22-3})}}{{=}}$
$\displaystyle(b_{[0][0]}\triangleleft
x^{\prime}_{1})b^{\prime}_{[0][0]}\otimes(x_{1}\triangleright
a^{\prime})\sigma(x_{2}b_{[0][1]},x^{\prime}_{2}b^{\prime}_{[0][1]})\otimes
x_{3}b_{[1]}x^{\prime}_{3}b^{\prime}_{[1]}$
$\displaystyle\stackrel{{\scriptstyle}}{{=}}$
$\displaystyle(b_{[0]}\triangleleft
x^{\prime}_{1})b^{\prime}_{[0]}\otimes(x_{1}\triangleright
a^{\prime})\sigma(x_{2}b_{[1]1},x^{\prime}_{2}b^{\prime}_{[1]1})\otimes
x_{3}b_{[1]2}x^{\prime}_{3}b^{\prime}_{[1]2}$
$\displaystyle\stackrel{{\scriptstyle(\ref{eq:d20-1})}}{{=}}$
$\displaystyle(b_{[0][0]}\triangleleft
x^{\prime}_{1})b^{\prime}_{[0]}\otimes(x_{1}\triangleright
a^{\prime})\sigma(x_{2},b_{[0][1]})\sigma(x_{3}b_{[1]1},x^{\prime}_{2}b^{\prime}_{[1]1})\otimes
x_{4}b_{[1]2}x^{\prime}_{3}b^{\prime}_{[1]2}$
$\displaystyle\stackrel{{\scriptstyle}}{{=}}$
$\displaystyle(b_{[0]}\triangleleft
x^{\prime}_{1})b^{\prime}_{[0]}\otimes(x_{1}\triangleright
a^{\prime})\sigma(x_{2},b_{[1]1})\sigma(x_{3}b_{[1]2},x^{\prime}_{2}b^{\prime}_{[1]1})\otimes
x_{4}b_{[1]3}x^{\prime}_{3}b^{\prime}_{[1]2}$
$\displaystyle\stackrel{{\scriptstyle(D21)}}{{=}}$
$\displaystyle(b_{[0][0]}\triangleleft
a^{\prime}_{-1}x^{\prime}_{1})b^{\prime}_{[0]}\otimes[x_{1}\triangleright(b_{[0][1]}\triangleright
a^{\prime}_{0})]\sigma(x_{2},b_{[1]1})\sigma(x_{3}b_{[1]2},x^{\prime}_{2}b^{\prime}_{[1]1})$
$\displaystyle\otimes x_{4}b_{[1]3}x^{\prime}_{3}b^{\prime}_{[1]2}$
$\displaystyle\stackrel{{\scriptstyle}}{{=}}$
$\displaystyle(b_{[0]}\triangleleft
a^{\prime}_{-1}x^{\prime}_{1})b^{\prime}_{[0]}\otimes[x_{1}\triangleright(b_{[1]1}\triangleright
a^{\prime}_{0})]\sigma(x_{2},b_{[1]2})\sigma(x_{3}b_{[1]3},x^{\prime}_{2}b^{\prime}_{[1]1})$
$\displaystyle\otimes x_{4}b_{[1]4}x^{\prime}_{3}b^{\prime}_{[1]2}$
$\displaystyle\stackrel{{\scriptstyle(LC3)}}{{=}}$
$\displaystyle(b_{[0]}\triangleleft
a^{\prime}_{-1}x^{\prime}_{1})b^{\prime}_{[0]}\otimes\sigma(x_{1},b_{[1]1})(x_{2}b_{[1]2}\triangleright
a^{\prime}_{0})\sigma(x_{3}b_{[1]3},x^{\prime}_{2}b^{\prime}_{[1]1})$
$\displaystyle\otimes x_{4}b_{[1]4}x^{\prime}_{3}b^{\prime}_{[1]2}$
$\displaystyle\stackrel{{\scriptstyle(D20)}}{{=}}$
$\displaystyle(b_{[0]}\triangleleft
a^{\prime}_{-1}x^{\prime}_{1})b^{\prime}_{[0]}\otimes(x_{1}b_{[1]1}\triangleright
a^{\prime}_{0})\sigma(x_{2}b_{[1]2},x^{\prime}_{2}b^{\prime}_{[1]1})\otimes
x_{3}b_{[1]3}x^{\prime}_{3}b^{\prime}_{[1]2}$ $\displaystyle=$
$\displaystyle\hbox{RHS~{} of~{}Eq}.(C14)$

$(\Longleftarrow)$ Let $b=1,a^{\prime}=1,x^{\prime}=1$ in Eq.$(C14)$, then we
can get Eq.$(D20)$; Applying $id\otimes id\otimes\varepsilon$ to Eq.$(C14)$,
one can obtain

$\displaystyle(b\triangleleft
x^{\prime}_{2})b^{\prime}\otimes(x_{1}\triangleright
a^{\prime})\sigma(x_{2},x^{\prime}_{1})=(b_{[0]}\triangleleft
a^{\prime}_{-1}x^{\prime}_{1})b^{\prime}_{[0]}\otimes(x_{1}b_{[1]1}\triangleright
a^{\prime}_{0})\sigma(x_{2}b_{[1]2},x^{\prime}_{2}b^{\prime}_{[1]}).$ (16)

Let $a^{\prime}=1,b^{\prime}=1$ in Eq.(16), we get Eq.(14). Likewise, we can
obtain Eqs.(15) and $(D21)$.

Up to now, we have checked that Eqs.$(D20),(D21),(E3)$ $\Leftrightarrow$
Eq.$(C14)$ in [10, Theorem 3.2]. And the rest is obvious. ∎

######  Corollary 3.6.

Let $H$ be a bialgebra. Let $A$ be a left $H$-comodule coalgebra such that
$\varepsilon_{A}(1_{A})=1$, and an algebra with a left $H$-action
$\triangleright$. Let $B$ be a right $H$-comodule coalgebra such that
$\varepsilon_{B}(1_{B})=1$, and an algebra with a right $H$-action
$\triangleleft$. The two-sided smash product $A\\#H\\#B$ equipped with the
two-sided smash coproduct $A\times H\times B$ becomes a bialgebra if and only
if Eqs.$(C1)$-$(C4)$, $(D8)$,$(D10)$,$(D16)$,$(D17)$,$(D21)$ and the following
conditions hold:

$\displaystyle(F1)$
$\displaystyle(aa^{\prime})_{1}\otimes(aa^{\prime})_{2}=a_{1}(a_{2-1}\triangleright
a^{\prime}_{1})\otimes a_{20}a^{\prime}_{2};$ $\displaystyle(F2)$
$\displaystyle(bb^{\prime})_{1}\otimes(bb^{\prime})_{2}=b_{1}b^{\prime}_{1[0]}\otimes(b_{2}\triangleleft
b^{\prime}_{1[1]})b^{\prime}_{2};$ $\displaystyle(F3)$
$\displaystyle(x\triangleright a)_{1}\otimes(x\triangleright
a)_{2}=(x_{1}\triangleright a_{1})\otimes(x_{2}\triangleright a_{2});$
$\displaystyle(F4)$ $\displaystyle(b\triangleleft x)_{1}\otimes(b\triangleleft
x)_{2}=(b_{1}\triangleleft x_{1})\otimes(b_{2}\triangleleft x_{2}).$

###### Proof.

Let $\sigma$ and $\tau$ be trivial, i.e.,
$\sigma(x,y)=\varepsilon(x)\varepsilon(y)1_{A}$ and
$\tau(x,y)=\varepsilon(x)\varepsilon(y)1_{B}$ in Theorem 3.2. ∎

######  Remark 3.7.

(1) By Eq.$(D21)$, Eq.(1) in the Majid’s double biproduct is not only
sufficient but also necessary.
Remarks (2) The first half parts in Eqs.$(C1)$-$(C4)$ and Eqs.$(F1)$,
$(D8)$,$(F3)$, $(D16)$ and the assumption of left module algebra and left
comodule coalgebra are equivalent to $A$ is a bialgebra in
${}^{H}_{H}{\mathbb{YD}}$. Similarly, the second half parts in
Eqs.$(C1)$-$(C4)$ and Eqs.$(F2)$, $(D10)$, $(D17)$, $(F4)$ and the assumption
of right module algebra and right comodule coalgebra are equivalent to $B$ is
a bialgebra in ${\mathbb{YD}}{}^{H}_{H}$.

######  Corollary 3.8.

([24]) Let $H$ be a bialgebra. Let $\sigma:H\otimes H\rightarrow A$ be a
linear map, where $A$ is a left $H$-comodule coalgebra, and an algebra with a
left $H$-weak action. The left crossed product $A\\#^{\sigma}H$ equipped with
the left smash coproduct $A\times H$ becomes a bialgebra if and only if the
first half parts in Eqs.$(C1)$-$(C4)$ and Eqs.$(D5)$, $(D7)$, $(D8)$, $(D11)$,
$(D12)$, $(D15)$, $(D16)$ hold.

###### Proof.

Let $B=K$ in Theorem 3.2. ∎

######  Remark 3.9.

(1) Note that condition $(D5)$ is exactly the condition of the twisted
comodule cocycle in [24]. Although this condition is the only prerequisite for
the bialgebra $A\star^{\sigma}H$ in [24], here it is just one of the necessary
conditions.
Remarks (2) The conditions in Corollary 3.8 are simpler than the ones in [10,
Corollary 3.3], but obvious they are equivalent.

######  Corollary 3.10.

Let $H$ be a bialgebra. Let $\tau:H\otimes H\rightarrow B$ be a linear map,
where $B$ is a right $H$-comodule coalgebra, and an algebra with a right
$H$-weak action. The right crossed product $H~{}^{\tau}\\#B$ equipped with the
right smash coproduct $H\times B$ becomes a bialgebra if and only if the
second half parts in Eqs.$(C1)$-$(C4)$ and Eqs.$(D6)$, $(D9)$, $(D10)$,
$(D13)$, $(D14)$, $(D17)$, $(D18)$ hold. In this case, we denote this
bialgebra by $H~{}^{\tau}\star B$.

###### Proof.

Let $A=K$ in Theorem 3.2. ∎

######  Remark 3.11.

The conditions in Corollary 3.10 are simpler than the ones in [14, Proposition
2.2], but obvious they are equivalent.

##  4\. Conclusion

We end this paper by three questions:

######  Question 4.1.

Majid realized a categorical interpretation of Radford’s biproduct ([15, 16]):
$A\star H$ is a Radford’s biproduct if and only if $A$ is a bialgebra in the
Yetter-Drinfel’d category ${}^{H}_{H}{\mathbb{YD}}$. This ensure that $A\star
H$ can play a central role in the classification of finite-dimensional pointed
Hopf algebras. By Theorem 3.2 and Remark 3.3, we know that the first half
parts in Eqs.$(C1)$-$(C4)$ and Eqs.$(D5)$, $(D7)$, $(D8)$, $(D11)$, $(D12)$,
$(D15)$, $(D16)$ are the necessary and sufficient conditions for
$A\star^{\sigma}H$ to be bialgebra. These conditions correspond to the
conditions for Radford’s biproduct one by one given in [21]. Here we note that
Eq.$(D16)$ is compatible condition for Yetter-Drinfeld module. Then whether
there exists an appropriate category ${}^{H}_{H}{\mathbb{YD}}^{\sigma}$ such
that the conditions above for $A\star^{\sigma}H$ hold if and only if $A$ is a
bialgebra in the category ${}^{H}_{H}{\mathbb{YD}}^{\sigma}$.

######  Question 4.2.

In Theorem 3.2, we give the necessary and sufficient conditions for the two-
sided crossed product $A\\#^{\sigma}H~{}^{\tau}\\#B$ equipped with the two-
sided smash coproduct $A\times H\times B$ becomes a bialgebra. In Theorem 2.3,
a more general two-sided crossed product algebra
$A\\#{{}_{R}^{G}H^{\tau}_{T}}\\#B$ is constructed. Replaced
$A\\#^{\sigma}H~{}^{\tau}\\#B$ by $A\\#{{}_{R}^{G}H^{\tau}_{T}}\\#B$, what are
the new conditions for the new bialgebra?

######  Question 4.3.

In Theorem 3.2, we only obtain the bialgebra structure
$A\star^{\sigma}H~{}{{}^{\tau}\star}B$. It is natural to ask when
$A\star^{\sigma}H~{}{{}^{\tau}\star}B$ is a Hopf algebra, i.e., how to
construct the antipode for $A\star^{\sigma}H~{}{{}^{\tau}\star}B$. Although we
provide the antipode in Proposition 2.10 for
$A\star{{}_{R}^{F}H^{\tau}_{T}}\star B$, unfortunately, the coalgebra
structure of $A\star{{}_{R}^{F}H^{\tau}_{T}}\star B$ is just the two-sided
tensor product coalgebra. Recently, the authors in [23] give the antipode for
a class of Majid’s double biproduct by (co)triangular Hopf quasigroups.

Acknowledgments. The authors are deeply indebted to the referee for his/her
very useful suggestions and some improvements to the original manuscript. This
work was partially supported by Natural Science Foundation of Henan Province
(No. 212300410365) and Research and innovation funding project for
Postgraduates of Henan Normal University (No. YL202019).

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