# Reconstruction of weak bialgebra maps and its applications

Michihisa Wakui
Department of Mathematics, Faculty of Engineering Science,
Kansai University, Suita-shi, Osaka 564-8680, Japan
E-mail address<EMAIL_ADDRESS>

###### Abstract

In this note we give a precise statement and a detailed proof for
reconstruction problem of weak bialgebra maps. As an application we
characterize indecomposability of weak algebras in categorical setting.

##  1 Introduction

The concept of weak bialgebras and weak Hopf algebras was introduced by Böhm,
Nill and Szlachányi [1] as a generalization of bialgebras and Hopf algebras.
In some literatures they are called quantum groupoids [13]. As shown by
Etingof, Nikshych and Ostrik [5] the language of weak Hopf algebras is
convenient to visualize various categorical construction, including
(multi-)fusion categories. At present, a lot of concepts and results for the
ordinal bialgebras and Hopf algebras are generalized or extended to weak
versions.

By the classical Tannaka-Krein reconstruction theorem, it is well known that
any finite-dimensional bialgebra $A$ over a fixed field $\boldsymbol{k}$ is
determined up to isomorphism by its comodule category $\mathbb{M}^{A}$ whose
objects are of finite dimension [4, 9, 15, 19]. More precisely, let $A$ and
$B$ be two finite-dimensional bialgebras over $\boldsymbol{k}$, and
$F:\mathbb{M}^{A}\longrightarrow\mathbb{M}^{B}$ be a $\boldsymbol{k}$-linear
monoidal functor. If $F$ is fibered, then there is a bialgebra map
$\varphi:A\longrightarrow B$ such that $F=\mathbb{M}^{\varphi}$, where
$\mathbb{M}^{\varphi}$ is the induced monoidal functor from $f$. This
statement can be found in Majid’s book [10, Theorem 2.2] and a detailed proof
is given in Franco’s lecture note [6, p.80–84].

In this note we treat the weak bialgebra version of the above result. It is
noted that as is well known, the reconstruction theorem for weak bialgebras
has been established by several researchers [3, 7, 11, 2] (see [20] for other
variants). However, it seems that there is no statement on a reconstruction
theorem for “weak bialgebra map” in any papers though it is very fundamental.
This note is devoted to give a precise statement of it and prove it. In fact,
although the same statement with the classical one holds, the proof is a
rather complicated since the unit object in the comodule category over a weak
bialgebra $A$ is not necessary to the base field $\boldsymbol{k}$. Actually,
it is a subalgebra, which is called the source counital subalgebra of $A$. For
that reason, the proof of the reconstruction theorem for weak bialgebra maps
is accomplished by re-examining the method used in the proof of the
reconstruction of coalgebra map.

Recently, the author study on indecomposability of weak bialgebras, and find
some interesting results and unsolved problems [21]. As an application of the
reconstruction theorem of weak bialgebra maps, we derive a categorical
interpretation of indecomposability of a weak bialgebra.

This paper is organized as follows. In Section 2 we recall the definition and
basic properties of weak bialgebras, and also the comodule structure over
them. In Section 3 after we overview the proof of the reconstruction theorem
of coalgebra maps, we state and prove the reconstruction theorem of bialgebra
maps. In Section 4, that is the final section, we apply the theorem to
characterize indecomposability of finite-dimensional weak bialgebras.

Throughout this note, $\boldsymbol{k}$ denotes a field, and
$\text{Vect}_{\boldsymbol{k}}^{\text{f.d.}}$ stands for the monoidal category
whose objects are finite-dimensional vector spaces over $\boldsymbol{k}$ and
morphisms are $\boldsymbol{k}$-linear maps between them. For a weak bialgebra
or a weak Hopf algebra $H$, we denote by $\Delta_{H}$, $\varepsilon_{H}$ and
$S_{H}$ the comultiplication, the counit and the antipode of $H$,
respectively. When it is clear that they are for $H$, they are simply denoted
by $\Delta$, $\varepsilon$ and $S$, respectively. The notation $\Delta^{(2)}$
means the composition
$(\Delta\otimes\text{id})\circ\Delta=(\text{id}\otimes\Delta)\circ\Delta$. We
use Sweedler’s notation such as $\Delta(x)=x_{(1)}\otimes x_{(2)}$ for $x\in
H$, and for a right $H$-comodule $M$ with coaction $\rho$ we also use the
notation $\rho(m)=m_{(0)}\otimes m_{(1)}$ for $m\in M$.

A monoidal category is described as
$\mathcal{C}=(\mathcal{C},\otimes,I,a,l,r)$, as in MacLane’s book [8], and a
monoidal functor between monoidal categories $\mathcal{C}$ and $\mathcal{D}$
is described as $F=(F,\Phi^{F},\omega^{F})$, where $F$ is a covariant functor
from $\mathcal{C}$ to $\mathcal{D}$, and $\Phi^{F}$ is a natural
transformation obtained by collecting morphisms $\phi^{F}_{M,N}:F(M)\otimes
F(N)\longrightarrow F(M\otimes N)$ for all $M,N\in\mathcal{C}$, and
$\omega^{F}:F(I_{\mathcal{C}})\longrightarrow I_{\mathcal{D}}$ is a morphism,
where $I_{\mathcal{C}}$ and $I_{\mathcal{D}}$ are the unit objects in
$\mathcal{C}$, and $\mathcal{D}$, respectively. If $\Phi^{F}$ is a natural
equivalence and $\omega^{F}$ is an isomorphism, then the monoidal functor
$(F,\Phi^{F},\omega^{F})$ is called strong. As in the classical case, every
weak bialgebra map $\varphi:H\longrightarrow K$ induces a covariant functor
$\mathbb{M}^{\varphi}:\mathbb{M}^{H}\longrightarrow\mathbb{M}^{K}$. We note
that the functor $\mathbb{M}^{\varphi}$ is not monoidal but is comonoidal. By
a comonoidal functor we mean a triplet
$F=(F,\bar{\Phi}^{F},\bar{\omega}^{F})$, which all arrows in $\bar{\Phi}^{F}$
and $\bar{\omega}^{F}$ are in reverse in monoidal categories, that is,
$\bar{\Phi}^{F}$ is a natural transformation obtained by collecting morphisms
$\bar{\phi}^{F}_{M,N}:F(M\otimes N)\longrightarrow F(M)\otimes F(N)$ for all
$M,N\in\mathcal{C}$, and $\bar{\omega}^{F}:I_{\mathcal{D}}\longrightarrow
F(I_{\mathcal{C}})$ is a morphism. By the same condition for a monoidal
functor, the concept called strong is defined for a comonoidal functor. In
some literatures terminologies “colax” or “op-monoidal” are used for
“comonoidal”.

For general facts on Hopf algebras, we refer the reader to Montgomery’s book
[12].

##  2 Definition of weak bialgebras and structures of their comodule
categories

In this section we recall the definition and basic properties of weak
bialgebras, and also the comodule structure over them mainly following [1] and
[13].

Let $H$ be a vector space over $\boldsymbol{k}$, and $(H,\mu,\eta)$ is an
algebra and $(H,\Delta,\varepsilon)$ is a coalgebra over $\boldsymbol{k}$. The
$5$-tuple $(H,\mu,\eta,\Delta,\varepsilon)$ is said to be a weak bialgebra
over $\boldsymbol{k}$ if the following three conditions are satisfied.

  1. (WH1)

$\Delta(xy)=\Delta(x)\Delta(y)$ for all $x,y\in H$.

  2. (WH2)

$\Delta^{(2)}(1)=(\Delta(1)\otimes
1)(1\otimes\Delta(1))=(1\otimes\Delta(1))(\Delta(1)\otimes 1)$, where
$1=\eta(1)$ is the identity element of the algebra $(H,\mu,\eta)$.

  3. (WH3)

For all $x,y,z\in H$

    1. (i)

$\varepsilon(xyz)=\varepsilon(xy_{(1)})\varepsilon(y_{(2)}z)$.

    2. (ii)

$\varepsilon(xyz)=\varepsilon(xy_{(2)})\varepsilon(y_{(1)}z)$.

Let $S:H\longrightarrow H$ be a $\boldsymbol{k}$-linear map. The $6$-tuple
$(H,\mu,\eta,\Delta,\varepsilon,S)$ is said to be a weak Hopf algebra over
$\boldsymbol{k}$ if the above three conditions and the following additional
condition are satisfied.

  1. (WH4)

For all $x\in H$

    1. (i)

$x_{(1)}S(x_{(2)})=\varepsilon(1_{(1)}x)1_{(2)}$.

    2. (ii)

$S(x_{(1)})x_{(2)}=1_{(1)}\varepsilon(x1_{(2)})$.

    3. (iii)

$S(x_{(1)})x_{(2)}S(x_{(3)})=S(x)$.

The above $S$ is called the antipode of $(H,\mu,\eta,\Delta,\varepsilon)$ and
$(H,\mu,\eta,\Delta,\varepsilon,S)$. We note that it is unique if it exists.

For a weak bialgebra $H=(H,\mu,\eta,\Delta,\varepsilon)$, by the condition
(WH4)(i),(ii), the following two $\boldsymbol{k}$-linear maps
$\varepsilon_{t},\varepsilon_{s}:H\longrightarrow H$ are defined:

$\displaystyle\varepsilon_{t}(x)=\varepsilon(1_{(1)}x)1_{(2)},$ (2.1)
$\displaystyle\varepsilon_{s}(x)=1_{(1)}\varepsilon(x1_{(2)}).$ (2.2)

The maps $\varepsilon_{t}$ and $\varepsilon_{s}$ are called the target
counital map and the source counital map, respectively. These maps satisfy the
following properties.

######  Lemma 2.1.

  1. $(1)$

$\varepsilon_{t}^{2}=\varepsilon_{t},\ \varepsilon_{s}^{2}=\varepsilon_{s}$.

  2. $(2)$

For all $x\in H$

    1. $(i)$

$((\mathrm{id}\otimes\varepsilon_{t})\circ\Delta)(x)=1_{(1)}x\otimes 1_{(2)}$.

    2. $(ii)$

$((\varepsilon_{s}\otimes\mathrm{id})\circ\Delta)(x)=1_{(1)}\otimes x1_{(2)}$.

In particular,

$1_{(1)}\otimes\varepsilon_{t}(1_{(2)})=1_{(1)}\otimes
1_{(2)}=\varepsilon_{s}(1_{(1)})\otimes 1_{(2)}.$ (2.3)

  3. $(3)$

For all $x\in H$

    1. $(i)$

$\varepsilon_{t}(x)=x\ \ \Longleftrightarrow\ \ \Delta(x)=1_{(1)}x\otimes
1_{(2)}$.

    2. $(ii)$

$\varepsilon_{s}(x)=x\ \ \Longleftrightarrow\ \ \Delta(x)=1_{(1)}\otimes
x1_{(2)}$.

Especially, by Part $(2)$$(i)$

$\displaystyle 1_{(1)}1_{[1]}\otimes 1_{(2)}\otimes 1_{[2]}$
$\displaystyle=1_{(1)}\otimes\varepsilon_{t}(1_{(2)})\otimes 1_{(3)},$
$\displaystyle 1_{(1)}\otimes 1_{[1]}\otimes 1_{(2)}1_{[2]}$
$\displaystyle=1_{(1)}\otimes\varepsilon_{s}(1_{(2)})\otimes 1_{(3)},$

where $\Delta^{(2)}(1)=1_{(1)}\otimes 1_{(2)}\otimes 1_{(3)}$ and
$\Delta(1)=1_{[1]}\otimes 1_{[2]}$.

######  Lemma 2.2.

Let $H$ be a weak bialgebra over $\boldsymbol{k}$. For all $x,y\in H$

  1. $(1)$

$\varepsilon_{t}(x\varepsilon_{t}(y))=\varepsilon_{t}(xy)$,
$\varepsilon_{s}(\varepsilon_{s}(x)y)=\varepsilon_{s}(xy)$.

  2. $(2)$

$\varepsilon(x\varepsilon_{t}(y))=\varepsilon(xy)$,
$\varepsilon(\varepsilon_{s}(x)y)=\varepsilon(xy)$.

  3. $(3)$

$\varepsilon\circ\varepsilon_{t}=\varepsilon=\varepsilon\circ\varepsilon_{s}$.

  4. $(4)$

$x=\varepsilon_{t}(x_{(1)})x_{(2)}=x_{(1)}\varepsilon_{s}(x_{(2)})$.

  5. $(5)$

$x\varepsilon_{t}(y)=\varepsilon_{t}(x_{(1)}y)x_{(2)}$,
$\varepsilon_{s}(x)y=y_{(1)}\varepsilon_{s}(xy_{(2)})$.

######  Lemma 2.3.

Let $H$ be a weak bialgebra over $\boldsymbol{k}$, and set
$H_{t}:=\varepsilon_{t}(H),\ H_{s}:=\varepsilon_{s}(H)$. Then

  1. $(1)$

$x\varepsilon_{t}(y)=\varepsilon_{t}(xy)$ for all $x\in H_{t}$ and $y\in H$.

  2. $(2)$

$\varepsilon_{s}(x)y=\varepsilon_{s}(xy)$ for all $x\in H$ and $y\in H_{s}$.

  3. $(3)$

    1. $(i)$

An element of $H_{t}$ and an element of $H_{s}$ commute, and

    2. $(ii)$

$H_{t}$ and $H_{s}$ are a left coideal and a right coideal subalgebras of $H$,
respectively.

The subalgebras $H_{t}$ and $H_{s}$ are called the target and source
subalgebras of $H$, respectively. By (2.3) we have

$\Delta(1)\in H_{s}\otimes H_{t}.$ (2.4)

  1. $(4)$

For all $x\in H$, $z\in H_{t}$ and $y\in H_{s}$

$\displaystyle\Delta(xz)=x_{(1)}z\otimes x_{(2)},\quad$
$\displaystyle\Delta(zx)=zx_{(1)}\otimes x_{(2)},$
$\displaystyle\Delta(xy)=x_{(1)}\otimes x_{(2)}y,\quad$
$\displaystyle\Delta(yx)=x_{(1)}\otimes yx_{(2)}.$

In particular

$\displaystyle xz=\varepsilon(x_{(1)}z)x_{(2)},\quad$ $\displaystyle
zx=\varepsilon(zx_{(1)})x_{(2)},$ $\displaystyle
xy=x_{(1)}\varepsilon(x_{(2)}y),\quad$ $\displaystyle
yx=x_{(1)}\varepsilon(yx_{(2)}).$

For a weak bialgebra $H$, one can also consider two $\boldsymbol{k}$-linear
maps $\varepsilon_{t}^{\prime},\varepsilon_{s}^{\prime}:H\longrightarrow H$
defined by

$\displaystyle\varepsilon_{t}^{\prime}(x)$
$\displaystyle=\varepsilon(x1_{(1)})1_{(2)},$ (2.5)
$\displaystyle\varepsilon_{s}^{\prime}(x)$
$\displaystyle=1_{(1)}\varepsilon(1_{(2)}x)$ (2.6)

for all $x\in H$. Then,
$H^{\mathrm{op}}=(H,\mu^{\mathrm{op}},\eta,\Delta,\varepsilon)$,
$H^{\mathrm{cop}}=(H,\mu,\eta,\Delta^{\mathrm{cop}},\varepsilon)$,
$H^{\mathrm{opcop}}=(H,\mu^{\mathrm{op}},\eta,\Delta^{\mathrm{cop}}$,
$\varepsilon)$ are weak bialgebras over $\boldsymbol{k}$, where
$\mu^{\mathrm{op}}$ and $\Delta^{\mathrm{cop}}$ are the opposite
multiplication and comultiplication, respectively. The target and the source
subalgebras of them are given by $(H^{\mathrm{op}})_{t}=H_{t},\
(H^{\mathrm{op}})_{s}=H_{s}$, $(H^{\mathrm{cop}})_{t}=H_{s},\
(H^{\mathrm{cop}})_{s}=H_{t}$, $(H^{\mathrm{opcop}})_{t}=H_{s},\
(H^{\mathrm{opcop}})_{s}=H_{t}$. The target and the source counital maps of
them are given by
$(\varepsilon_{H^{\mathrm{op}}})_{t}=\varepsilon_{t}^{\prime},\
(\varepsilon_{H^{\mathrm{op}}})_{s}=\varepsilon_{s}^{\prime}$,
$(\varepsilon_{H^{\mathrm{cop}}})_{t}=\varepsilon_{s}^{\prime},\
(\varepsilon_{H^{\mathrm{cop}}})_{s}=\varepsilon_{t}^{\prime}$,
$(\varepsilon_{H^{\mathrm{opcop}}})_{t}=\varepsilon_{s},\
(\varepsilon_{H^{\mathrm{opcop}}})_{s}=\varepsilon_{t}$. If $S$ is the
antipode of $H$, then it is also of $H^{\mathrm{opcop}}$.

For a weak bialgebra $H$ over $\boldsymbol{k}$, we denote by
$\text{\boldmath{$\mathsf{M}$}}^{H}$ the $\boldsymbol{k}$-linear category
whose objects are right $H$-comodules and morphisms are $H$-comodule maps
between them. The comodule category $\text{\boldmath{$\mathsf{M}$}}^{H}$ has a
monoidal structure [13, Section 4] as the following lemma.

######  Lemma 2.4 ([14, Lemma 4.2]).

Let $H$ be a weak bialgebra over $\boldsymbol{k}$.

  1. $(1)$

For two right $H$-comodules $(M,\rho_{M}),\ (N,\rho_{N})$, the pair
$(M,\rho_{M})\circledast(N,\rho_{N}):=(M\otimes_{H_{s}}N,\rho)$ is also a
right $H$-comodule, where
$\rho:M\otimes_{H_{s}}N\longrightarrow(M\otimes_{H_{s}}N)\otimes H$ is a
$\boldsymbol{k}$-linear map defined by

$\rho(m\otimes_{H_{s}}n)=(m_{(0)}\otimes_{H_{s}}n_{(0)})\otimes
m_{(1)}n_{(1)}\qquad(m\in M,\ n\in N).$ (2.7)

The source algebra $H_{s}$ can be regarded as a right $H$-comodule with the
coaction $\Delta_{s}:=\Delta|_{H_{s}}:H_{s}:\longrightarrow H_{s}\otimes H$,
and for all right $H$-comodules $L,M,N$ there are natural isomorphisms

    1. $(i)$

$H_{s}\circledast M\cong M\cong M\circledast H_{s}$ as right $H$-comodules,

    2. $(ii)$

$(L\circledast M)\circledast N\cong L\circledast(M\circledast N)$ as right
$H$-comodules.

Here the natural isomorphisms in $(i)$ are given as follows: For a right
$H$-comodule $M$

$\displaystyle l_{M}:$ $\displaystyle\ H_{s}\otimes_{H_{s}}M\longrightarrow
M,\ \quad l_{M}(y\otimes_{H_{s}}m)=y\cdot m\qquad(y\in H_{s},\ m\in M),$
$\displaystyle l_{M}^{-1}:$ $\displaystyle\ M\longrightarrow
H_{s}\otimes_{H_{s}}M,\ \quad l_{M}^{-1}(m)=1\otimes_{H_{s}}m\qquad\quad(m\in
M),$ $\displaystyle r_{M}:$ $\displaystyle\
M\otimes_{H_{s}}H_{s}\longrightarrow M,\ \quad r_{M}(m\otimes_{H_{s}}y)=m\cdot
y\qquad(m\in M,\ y\in H_{s}),$ $\displaystyle r_{M}^{-1}:$ $\displaystyle\
M\longrightarrow M\otimes_{H_{s}}H_{s},\ \quad
r_{M}^{-1}(m)=m\otimes_{H_{s}}1\qquad\quad(m\in M).$

The isomorphism in $(ii)$ is induced from a usual isomorphism between vector
spaces.

  2. $(2)$

Let $f:(M,\rho_{M})\longrightarrow(N,\rho_{N})$,
$g:(M^{\prime},\rho_{M}^{\prime})\longrightarrow(N^{\prime},\rho_{N}^{\prime})$
be right $H$-comodule maps. Then
$f\otimes_{H_{s}}g:M\otimes_{H_{s}}M^{\prime}\longrightarrow
N\otimes_{H_{s}}N^{\prime}$ is also a right $H$-comodule map with respect to
the comodule structure given in $(1)$. We denote the map $f\otimes_{H_{s}}g$
by $f\circledast g$.

By Parts $(1)$, $(2)$ the abelian category
$\text{\boldmath{$\mathsf{M}$}}^{H}$ becomes a $\boldsymbol{k}$-linear
monoidal category whose unit object is $(H_{s},\Delta_{s})$.

######  Lemma 2.5 ([14, Proposition 4.1]).

Let $H$ be a weak bialgebra over $\boldsymbol{k}$, and $(M,\rho_{M})$ be a
right $H$-comodule. For $y\in H_{s}$ and $m\in M$, the elements $y\cdot m$ and
$m\cdot y$ in $M$ are defined as follows:

$\displaystyle y\cdot m$ $\displaystyle:=m_{(0)}\varepsilon(ym_{(1)}),$ (2.8)
$\displaystyle m\cdot y$ $\displaystyle:=m_{(0)}\varepsilon(m_{(1)}y).$ (2.9)

  1. $(1)$

$M$ becomes an $(H_{s},H_{s})$-bimodule equipped with the above actions.

  2. $(2)$

$M\otimes H$ becomes an $(H_{s},H_{s})$-bimodule equipped with the following
actions: For $y\in H_{s},\ m\in M,\ x\in H$,

$\displaystyle y\cdot(m\otimes x)$ $\displaystyle:=(1_{(1)}\cdot
m)\otimes(y1_{(2)}x)=m_{(0)}\otimes\varepsilon_{t}(m_{(1)})yx,$ (2.10)
$\displaystyle(m\otimes x)\cdot y$ $\displaystyle:=(m\cdot
1_{(1)})\otimes(xy1_{(2)})=m_{(0)}\otimes
xy\varepsilon_{t}^{\prime}(m_{(1)}).$ (2.11)

  3. $(3)$

The following equations hold for all $y\in H_{s}$ and $m\in M$:

    1. $(i)$

$\rho_{M}(y\cdot m)=m_{(0)}\otimes ym_{(1)}=y\cdot\rho_{M}(m)$.

    2. $(ii)$

$\rho_{M}(m\cdot y)=m_{(0)}\otimes m_{(1)}y=\rho_{M}(m)\cdot y$.

    3. $(iii)$

$\varepsilon_{s}^{\prime}(m_{(1)})\cdot
m_{(0)}=m=m_{(0)}\cdot\varepsilon_{s}(m_{(1)})$.

In particular, $\rho_{M}:M\longrightarrow M\otimes H$ is an
$(H_{s},H_{s})$-bimodule map by $(i)$ and $(ii)$.

  4. $(4)$

Let $(N,\rho_{N})$ be a right $H$-comodule and
$f:(M,\rho_{M})\longrightarrow(N,\rho_{N})$ be an $H$-comodule map. Then $f$
becomes an $(H_{s},H_{s})$-bimodule map with respect to the bimodule
structures given by $(1)$.

Let $H$ be a weak bialgebra over $\boldsymbol{k}$, and denote by
${}_{H_{s}}\text{\boldmath{$\mathsf{M}$}}_{H_{s}}$ the $\boldsymbol{k}$-linear
category consisting of whose objects are $(H_{s},H_{s})$-bimodules and
morphisms are $(H_{s},H_{s})$-bimodule maps between them. By Lemma 2.5 for a
right $H$-comodule $(M,\rho_{M})$, the underlying vector space $M$ has an
$(H_{s},H_{s})$-bimodule structure, and $\rho_{M}:M\longrightarrow M\otimes H$
becomes an $(H_{s},H_{s})$-bimodule map. Thus $(M,\rho_{M})$ can be regarded
as a right $H$-comodule in ${}_{H_{s}}\text{\boldmath{$\mathsf{M}$}}_{H_{s}}$.
Then any $H$-comodule map $f:M\longrightarrow N$ always an
$(H_{s},H_{s})$-bimodule map.

We denote by ${}_{H_{s}}\text{\boldmath{$\mathsf{M}$}}_{H_{s}}^{H}$ the
$\boldsymbol{k}$-linear category whose objects are right $H$-comodules in
${}_{H_{s}}\text{\boldmath{$\mathsf{M}$}}_{H_{s}}$ and morphisms are right
$H$-comodule and $(H_{s},H_{s})$-bimodule maps between them. The category
${}_{H_{s}}\text{\boldmath{$\mathsf{M}$}}_{H_{s}}^{H}$ has a
$\boldsymbol{k}$-linear monoidal structure whose tensor product is given by
$\otimes_{H_{s}}$.

As a special case of [18, Theorem 2.2] we have:

######  Lemma 2.6.

Let $H$ be a weak bialgebra over $\boldsymbol{k}$. By Lemma 2.5 for a right
$H$-comodule $(M,\rho_{M})$, there is an $(H_{s},H_{s})$-bimodule structure on
the underlying vector space $M$, and $\rho_{M}:M\longrightarrow M\otimes H$
becomes an $(H_{s},H_{s})$-bimodule map. Thus $(M,\rho_{M})$ can be regarded
as a right $H$-comodule in ${}_{H_{s}}\text{\boldmath{$\mathsf{M}$}}_{H_{s}}$.
Then any $H$-comodule map $f:M\longrightarrow N$ is always an
$(H_{s},H_{s})$-bimodule map. This correspondence gives rise to a
$\boldsymbol{k}$-linear monoidal equivalence
$\Xi^{H}:\text{\boldmath{$\mathsf{M}$}}^{H}\longrightarrow{}_{H_{s}}\text{\boldmath{$\mathsf{M}$}}_{H_{s}}^{H}$
between $\boldsymbol{k}$-linear monoidal categories.

We set

$\hat{U}^{H}:=U^{H}\circ\Xi^{H}:\text{\boldmath{$\mathsf{M}$}}^{H}\longrightarrow{}_{H_{s}}\text{\boldmath{$\mathsf{M}$}}_{H_{s}},$

where
$U^{H}:{}_{H_{s}}\text{\boldmath{$\mathsf{M}$}}_{H_{s}}^{H}\longrightarrow{}_{H_{s}}\text{\boldmath{$\mathsf{M}$}}_{H_{s}}$
is the forgetful monoidal functor. The composition $\hat{U}^{H}$ of monoidal
functors becomes a monoidal functor, whose structure is given as follows:

  1. $\bullet$

$\phi_{M,N}^{\hat{U}^{H}}=\text{id}_{M\otimes_{H_{s}}N}$ for each
$M,N\in\text{\boldmath{$\mathsf{M}$}}^{H}$,

  2. $\bullet$

$\omega^{\hat{U}^{H}}:H_{s}\longrightarrow\hat{U}^{H}(H_{s})=H_{s}$ is the
identity map.

######  Lemma 2.7.

Let $H,K$ be weak bialgebras over $\boldsymbol{k}$, and
$\varphi:H\longrightarrow K$ be a weak bialgebra map, namely, an algebra map
and coalgebra map. Then $\varphi(H_{s})\subset K_{s}$ and
$\varphi(H_{t})\subset K_{t}$. The former inclusion induces an algebra map
$\varphi_{s}:=\varphi|_{H_{s}}:H_{s}\longrightarrow K_{s}$.

  1. $(1)$

For a right $H$-comodule $(M,\rho_{M})$

$\text{\boldmath{$\mathsf{M}$}}^{\varphi}(M,\rho_{M}):=(M,(\text{id}_{M}\otimes\varphi)\circ\rho_{M})$

is a right $K$-comodule, and for a right $H$-comodule map
$f:(M,\rho_{M})\longrightarrow(N,\rho_{N})$

$\text{\boldmath{$\mathsf{M}$}}^{\varphi}(f):=f:\text{\boldmath{$\mathsf{M}$}}^{\varphi}(M,\rho_{M})\longrightarrow\text{\boldmath{$\mathsf{M}$}}^{\varphi}(N,\rho_{N})$

is a right $K$-comodule map. These correspondences define a covariant functor
$\text{\boldmath{$\mathsf{M}$}}^{\varphi}:\text{\boldmath{$\mathsf{M}$}}^{H}\longrightarrow\text{\boldmath{$\mathsf{M}$}}^{K}$.

  2. $(2)$

The functor $\text{\boldmath{$\mathsf{M}$}}^{\varphi}$ is a
$\boldsymbol{k}$-linear comonoidal functor. If $\varphi_{s}$ is bijective,
then the $\boldsymbol{k}$-linear comonoidal functor
$\text{\boldmath{$\mathsf{M}$}}^{\varphi}$ is strong. In this case it can be
regarded as a $\boldsymbol{k}$-linear monoidal functor.

  3. $(3)$

The algebra map $\varphi_{s}$ induces a $\boldsymbol{k}$-linear monoidal
functor
${}_{\varphi_{s}}\text{\boldmath{$\mathsf{M}$}}_{\varphi_{s}}:{}_{K_{s}}\text{\boldmath{$\mathsf{M}$}}_{K_{s}}\longrightarrow{}_{H_{s}}\text{\boldmath{$\mathsf{M}$}}_{H_{s}}$,
and if $\varphi_{s}$ is bijective, then
$\hat{U}^{K}\circ\text{\boldmath{$\mathsf{M}$}}^{\varphi}={}_{\varphi_{s}^{-1}}\text{\boldmath{$\mathsf{M}$}}_{\varphi_{s}^{-1}}\circ\hat{U}^{H}:\text{\boldmath{$\mathsf{M}$}}^{H}\longrightarrow{}_{K_{s}}\text{\boldmath{$\mathsf{M}$}}_{K_{s}}$
as monoidal functors.

###### Proof..

(2) For right $H$-comodules $(M,\rho_{M}),(N,\rho_{N})$ we set

$\displaystyle\text{\boldmath{$\mathsf{M}$}}^{\varphi}(M,\rho_{M})\circledast\text{\boldmath{$\mathsf{M}$}}^{\varphi}(N,\rho_{N})$
$\displaystyle=(M\otimes_{K_{s}}N,\rho),$
$\displaystyle\text{\boldmath{$\mathsf{M}$}}^{\varphi}\bigl{(}(M,\rho_{M})\circledast(N,\rho_{N})\bigr{)}$
$\displaystyle=(M\otimes_{H_{s}}N,\rho^{\prime}),$

where $\rho$ and $\rho^{\prime}$ are given as follows:

$\displaystyle\rho:M\otimes_{K_{s}}N\longrightarrow(M\otimes_{K_{s}}N)\otimes
K,\quad$
$\displaystyle\rho(m\otimes_{K_{s}}n)=m_{(0)}\otimes_{K_{s}}n_{(0)}\otimes\varphi(m_{(1)})\varphi(n_{(1)}),$
$\displaystyle\rho^{\prime}:M\otimes_{H_{s}}N\longrightarrow(M\otimes_{H_{s}}N)\otimes
K,\quad$
$\displaystyle\rho^{\prime}(m\otimes_{H_{s}}n)=m_{(0)}\otimes_{H_{s}}n_{(0)}\otimes\varphi(m_{(1)}n_{(1)}).$

Since $\varphi$ is an algebra map, the identity map $\text{id}_{M\otimes N}$
induces a surjection $\iota_{M,N}:M\otimes_{H_{s}}N\longrightarrow
M\otimes_{K_{s}}N$ which is a right $K$-comodule map.

The restriction $\varphi_{s}:H_{s}\longrightarrow K_{s}$ can be regarded as a
right $K$-comodule map from
$\text{\boldmath{$\mathsf{M}$}}^{\varphi}(H_{s},(\Delta_{H})_{s})=(H_{s},\
(\text{id}_{H_{s}}\otimes\varphi)\circ(\Delta_{H})_{s})$ to
$(K_{s},(\Delta_{K})_{s})$. Thus, the triplet
$(\text{\boldmath{$\mathsf{M}$}}^{\varphi},\iota,\varphi_{s})$ is a comonoidal
functor from $\text{\boldmath{$\mathsf{M}$}}^{H}$ to
$\text{\boldmath{$\mathsf{M}$}}^{K}$, where
$\iota:=\\{\iota_{M,N}\\}_{M,N\in\text{\boldmath{$\mathsf{M}$}}^{H}}$.

If $\varphi_{s}$ is bijective, then $\iota_{M,N}$ for all
$M,N\in\text{\boldmath{$\mathsf{M}$}}^{H}$ is an isomorphism. Thus,
$(\text{\boldmath{$\mathsf{M}$}}^{\varphi},\iota^{-1},\varphi_{s}^{-1}):\text{\boldmath{$\mathsf{M}$}}^{H}\longrightarrow\text{\boldmath{$\mathsf{M}$}}^{K}$
is a strong monoidal functor.

(3) For a $(K_{s},K_{s})$-bimodule $(M,\alpha_{l},\alpha_{r})$ we set

${}_{\varphi_{s}}\text{\boldmath{$\mathsf{M}$}}_{\varphi_{s}}(M,\alpha_{l},\alpha_{r})=\bigl{(}M,\alpha_{l}\circ(\varphi_{s}\otimes\text{id}_{M}),\alpha_{r}\circ(\text{id}_{M}\otimes\varphi_{s})\bigr{)},$

and a $(K_{s},K_{s})$-bimodule map $f:M\longrightarrow N$ we set

${}_{\varphi_{s}}\text{\boldmath{$\mathsf{M}$}}_{\varphi_{s}}(f)=f.$

Then, ${}_{\varphi_{s}}\text{\boldmath{$\mathsf{M}$}}_{\varphi_{s}}$ is a
$\boldsymbol{k}$-linear covariant functor.

For $(K_{s},K_{s})$-bimodules $M=(M,\alpha_{l}^{M},\alpha_{r}^{M})$ and
$N=(N,\alpha_{l}^{N},\alpha_{r}^{N})$, let

$\jmath_{M,N}:{}_{\varphi_{s}}\text{\boldmath{$\mathsf{M}$}}_{\varphi_{s}}(M)\otimes_{H_{s}}{}_{\varphi_{s}}\text{\boldmath{$\mathsf{M}$}}_{\varphi_{s}}(N)\longrightarrow{}_{\varphi_{s}}\text{\boldmath{$\mathsf{M}$}}_{\varphi_{s}}(M\otimes_{K_{s}}N)$

be the induced $\boldsymbol{k}$-linear map from the identity map
$\text{id}_{M_{\otimes}N}$. This is an $(H_{s},H_{s})$-module map which is
natural with respect to $M,N$. So, we have a monoidal functor
$({}_{\varphi_{s}}\text{\boldmath{$\mathsf{M}$}}_{\varphi_{s}},\jmath,\varphi_{s})$
by setting
$\jmath=\\{\jmath_{M,N}\\}_{M,N\in{}_{K_{s}}\text{\boldmath{$\mathsf{M}$}}_{K_{s}}}$.
If $\varphi_{s}$ is bijective, then one can show that
$\hat{U}^{K}\circ\text{\boldmath{$\mathsf{M}$}}^{\varphi}={}_{\varphi_{s}^{-1}}\text{\boldmath{$\mathsf{M}$}}_{\varphi_{s}^{-1}}\circ\hat{U}^{H}$
as $\boldsymbol{k}$-linear monoidal functors. ∎

##  3 Reconstruction of a weak bialgebra map

First of all, we will recall the proof of the reconstruction theorem of
coalgebra maps, that is a classical and fundamental theorem for Tannakian
reconstruction theorem.

For a coalgebra $C$ over $\boldsymbol{k}$ we denote by $\mathbb{M}^{C}$ the
$\boldsymbol{k}$-linear category consisting of whose objects are right
$C$-comodules of finite dimension and morphisms are $C$-comodule maps between
them, and denote by $U^{C}$ the forgetful functor from $\mathbb{M}^{C}$ to
$\mathrm{Vect}_{\boldsymbol{k}}^{\mathrm{f.d.}}$.

######  Theorem 3.1.

Let $C,D$ be coalgebras over $\boldsymbol{k}$, and
$F:\mathbb{M}^{C}\longrightarrow\mathbb{M}^{D}$ be a $\boldsymbol{k}$-linear
covariant functor. If $U^{D}\circ F=U^{C}$, then there is a unique coalgebra
map $\varphi:C\longrightarrow D$ such that
$F=\mathbb{M}^{\varphi}:\mathbb{M}^{C}\longrightarrow\mathbb{M}^{D}$. Here,
$\mathbb{M}^{\varphi}$ denotes the $\boldsymbol{k}$-linear functor induced
from $\varphi$.

###### Proof..

The proof follows from Franco’s lecture note [6, p.81–84].

Let $(M,\rho_{M})$ be a finite-dimensional $C$-comodule. By the assumption
$U^{D}\circ F=U^{C}$, one can set $F(M,\rho_{M})=(M,\rho_{M}^{F})$, where
$\rho_{M}^{F}:M\longrightarrow M\otimes D$ is a right coaction of $D$.

Let $P$ be a finite-dimensional subcoalgebra of $C$. It can be regarded as a
right $C$-comodule by the coaction

$\rho_{P}:P\xrightarrow{\ \ \Delta_{P}\ \ }P\otimes P\xrightarrow{\ \
\text{id}\otimes\iota_{P}\ \ }P\otimes C,$

where $\iota_{P}$ stands for the inclusion. Since $P$ is finite-dimensional,
it gives an object of $\mathbb{M}^{C}$, and therefore we have
$F(P,\rho_{P})=(P,\rho_{P}^{F})\in\mathbb{M}^{D}$.

Let us consider the composition

$\varphi_{P}:P\xrightarrow{\ \ \rho_{P}^{F}\ \ }P\otimes D\xrightarrow{\
\varepsilon_{P}\otimes\text{id}\ }\boldsymbol{k}\otimes D\cong D.$

Then $\varphi_{P}:P\longrightarrow D$ is a coalgebra map. This fact can be
verified as follows.

$\bullet$ The equation $\varepsilon_{D}\circ\varphi_{P}=\varepsilon_{P}$ comes
from the following commutative diagram.

$P$$P\otimes D$$\boldsymbol{k}\otimes
D$$\cong$$D$$P$$\cong$$P\otimes\boldsymbol{k}$$\boldsymbol{k}\otimes\boldsymbol{k}$$\cong$$\boldsymbol{k}$$\rho_{P}^{F}$$\varepsilon_{P}\otimes\text{id}$$\varepsilon_{P}\otimes\text{id}$id$\text{id}\otimes\varepsilon_{D}$$\text{id}\otimes\varepsilon_{D}$$\varepsilon_{D}$

$\bullet$ To show the equation
$\Delta_{D}\circ\varphi_{P}=(\varphi_{P}\otimes\varphi_{P})\circ\Delta_{P}$,
it is enough to verify the following diagram commutes:

$P$$P\otimes P$$P\otimes D\otimes P\otimes D$$P\otimes D$$P\otimes P\otimes
D$$P\otimes D\otimes D$$\cong$$P\otimes D\otimes\boldsymbol{k}\otimes
D$$\boldsymbol{k}\otimes D\otimes\boldsymbol{k}\otimes
D$$\boldsymbol{k}\otimes D\otimes D$$P\otimes D$$\boldsymbol{k}\otimes D\cong
D$$D\otimes
D$(A)(B)(C)$\Delta_{P}$$\rho_{P}^{F}\otimes\rho_{P}^{F}$$\text{id}\otimes\rho_{P}^{F}$$\rho_{P}^{F}$$\Delta_{P}\otimes\text{id}$$\rho_{P}^{F}\otimes\text{id}$$\text{id}\otimes\Delta_{D}$$\rho_{P}^{F}$$\varepsilon_{P}\otimes\text{id}$$\Delta_{P}$$\rho_{P}^{F}\otimes\text{id}$$\text{id}\otimes\varepsilon_{P}\otimes\text{id}$$\varepsilon_{P}\otimes\text{id}$$\cong$$\varepsilon_{P}\otimes\text{id}\otimes\varepsilon_{P}\otimes\text{id}$$\cong$

Part (A) is commutative since $\rho_{P}^{F}$ is a right $D$-coaction, and Part
(C) is commutative since the following diagram is so.

$P$$P\otimes P$$P\otimes D\otimes P\otimes D$$P\otimes D$$P\otimes D$$P\otimes
D$$D\otimes
D$$\Delta_{P}\otimes\text{id}$$\rho_{P}^{F}\otimes\text{id}$$\cong$$\text{id}\otimes\varepsilon_{P}\otimes\text{id}$$\cong$$\rho_{P}^{F}\otimes\text{id}$$\rho_{P}^{F}\otimes\text{id}$$\cong$$\text{id}\otimes\varepsilon_{P}\otimes\text{id}$

The proof the commutativity of Part (B) is a little technical. For any
$\boldsymbol{k}$-linear map $\gamma:P\longrightarrow\boldsymbol{k}$, the
composition

$P\xrightarrow{\ \ \Delta_{P}\ \ }P\otimes P\xrightarrow{\
\gamma\otimes\text{id}_{P}\ }P$

is a right $P$-comodule map, and hence it is also a right $C$-comodule map.
Sending it by $F$ we have a right $D$-comodule map
$(\gamma\otimes\text{id}_{P})\circ\Delta_{P}:P\longrightarrow P$. Thus the
following diagram commutes:

$P$$P\otimes P$$P$$P$$P\otimes P\otimes D$$P\otimes
D$$\Delta_{P}$$\gamma\otimes\text{id}$$\Delta_{P}\otimes\text{id}$$\gamma\otimes\text{id}\otimes\text{id}$$\rho_{P}^{F}$$\rho_{P}^{F}$

Combining the above diagram with
$\rho_{P}^{F}\circ(\gamma\otimes\text{id}_{P})=(\gamma\otimes\text{id}_{P\otimes
D})\circ(\text{id}_{P}\otimes\rho_{P}^{F})$, we have

$(\gamma\otimes\text{id}_{P\otimes
D})\circ(\text{id}_{P}\otimes\rho_{P}^{F})\circ\Delta_{P}=(\gamma\otimes\text{id}_{P\otimes
D})\circ(\Delta_{P}\otimes\text{id}_{P})\circ\rho_{P}^{F}.$

Since this equation holds for all $\boldsymbol{k}$-linear maps $\gamma$, we
see that
$(\text{id}_{P}\otimes\rho_{P}^{F})\circ\Delta_{P}=(\Delta_{P}\otimes\text{id}_{P})\circ\rho_{P}^{F}$.
This implies the commutativity of Part (B).

By the fundamental theorem for coalgebras, $C$ is a union of finite-
dimensional subcoalgebras. Based on the fact, it can be shown that the
coalgebra maps $\varphi_{P}:P\longrightarrow D$ for all finite-dimensional
subcoalgebras $P$ induce a well-defined coalgebra map
$\varphi:C\longrightarrow D$. In fact it is easily verified that
$(\varphi_{Q})|_{P}=\varphi_{P}$ for two finite-dimensional subcoalgebras $P$
and $Q$ satisfying with $P\subset Q$. In this way, it is proved that there is
a coalgebra map $\varphi:C\longrightarrow D$ such that
$\varphi|_{P}=\varphi_{P}$ for all finite-dimensional subcoalgebra $P$ of $C$.

The coalgebra map $\varphi$ satisfies $F=\mathbb{M}^{\varphi}$. This fact can
be derived as follows.

$\bullet$ Let $(M,\rho_{M})$ be a finite-dimensional right $C$-comodule. Then
there is a finite-dimensional subcoalgebra $P$ of $C$ such that
$\rho_{M}(M)\subset M\otimes P$. Since $(M\otimes
P,\text{id}_{M}\otimes\rho_{P})$ is a right $C$-comodule, we have a right
$D$-comodule $(M\otimes P,\ (\text{id}_{M}\otimes\rho_{P})^{F})$. Since
$M\otimes P$ is decomposed to a direct sum of finite copies of $P$ as a right
$C$-comodule and $F$ is a $\boldsymbol{k}$-linear functor satisfying
$U^{C}=U^{D}\circ F$, it follows that
$(\text{id}_{M}\otimes\rho_{P})^{F}=\text{id}_{M}\otimes\rho_{P}^{F}:M\otimes
P\longrightarrow M\otimes P\otimes D$.

Let $\rho_{M}^{\prime}:M\longrightarrow M\otimes P$ be the restriction of
$\rho_{M}$. Then the map $\rho_{M}^{\prime}$ is a right $C$-comodule map from
$(M,\rho_{M})$ to $(M\otimes P,\text{id}\otimes\rho_{P})$. Thus,
$F(\rho_{M}^{\prime})=\rho_{M}^{\prime}:M\longrightarrow M\otimes P$ is a
right $D$-comodule map from $(M,\rho_{M}^{F})$ to $(M\otimes
P,(\text{id}\otimes\rho_{P})^{F})=(M\otimes
P,\text{id}_{M}\otimes\rho_{P}^{F})$, and hence the following diagram
commutes.

$M$$M\otimes P$$M\otimes D$$M\otimes P\otimes
D$$\rho_{M}^{\prime}$$\rho_{M}^{\prime}\otimes\text{id}$$\rho_{M}^{F}$$\text{id}\otimes\rho_{P}^{F}$

It follows that we have the commutative diagram:

$M$$M\otimes C$$M\otimes P$$M\otimes P\otimes D$$M\otimes\boldsymbol{k}\otimes
D$$M\otimes D$$M\otimes
D$$\rho_{M}$$\rho_{M}^{\prime}$id$\text{id}\otimes\iota_{P}$$\text{id}\otimes\varphi_{P}$$\rho_{M}^{F}$$\text{id}\otimes\varphi$$\text{id}\otimes\rho_{P}^{F}$$\text{id}\otimes\varepsilon_{P}\otimes\text{id}$$\rho_{M}^{\prime}\otimes\text{id}$$\cong$

This implies that $\rho_{M}^{F}=(\text{id}\otimes\varphi)\circ\rho_{M}$, and
hence we see that $F(M,\rho_{M})=\mathbb{M}^{\varphi}(M,\rho_{M})$.

$\bullet$ Let $f:(M,\rho_{M})\longrightarrow(N,\rho_{N})$ be a right
$C$-comodule map between finite-dimensional right $C$-comodules. Then

$\displaystyle F(f)=f$
$\displaystyle:(M,\rho_{M}^{F})\longrightarrow(N,\rho_{N}^{F}),$
$\displaystyle\mathbb{M}^{\varphi}(f)=f$
$\displaystyle:(M,(\text{id}\otimes\varphi)\circ\rho_{M})\longrightarrow(N,(\text{id}\otimes\varphi)\circ\rho_{N}).$

As shown that $\rho_{M}^{F}=(\text{id}\otimes\varphi)\circ\rho_{M}$ and
$\rho_{N}^{F}=(\text{id}\otimes\varphi)\circ\rho_{N}$, we see that
$F(f)=\mathbb{M}^{\varphi}(f)$. Thus $F=\mathbb{M}^{\varphi}$ as
$\boldsymbol{k}$-linear functors.

Finally, we show the uniqueness of $\varphi$. Suppose that a coalgebra map
$\psi:C\longrightarrow D$ satisfies $F=\mathbb{M}^{\psi}$, too. Let $P$ be a
finite-dimensional subcoalgebra of $C$, and regard it as the right
$C$-comodule $(P,\rho_{P})$. Since
$\mathbb{M}^{\varphi}(P,\rho_{P})=\mathbb{M}^{\psi}(P,\rho_{P})$, the
following diagram commutes.

$P$$P\otimes C$$P\otimes C$$P\otimes
D$$\rho_{P}$$\text{id}\otimes\varphi$$\rho_{P}$$\text{id}\otimes\psi$

Since the diagram

$P$$P\otimes C$$P\otimes D$$\boldsymbol{k}\otimes D$$\boldsymbol{k}\otimes D$

$\cong$

$\cong$

$C$$D$$\rho_{P}$$\text{id}\otimes\varphi$$\iota_{P}$$\varepsilon_{P}\otimes\text{id}$$\varepsilon_{P}\otimes\text{id}$$\text{id}\otimes\varphi$$\varphi$

commutes, and the same commutative diagram holds for $\psi$, we have
$\varphi\circ\iota_{P}=\psi\circ\iota_{P}$. This implies that $\varphi=\psi$
since $C$ can be regarded as a union of finite-dimensional subcoalgebras. ∎

######  Remark 3.2.

The coalgebra map $\varphi_{P}:P\longrightarrow D$ in the above proof is a
right $D$-comodule map from $(P,\rho_{P}^{F})$ to $(D,\Delta_{D})$. It follows
from the following commutative diagram.

$P$$D$$P\otimes D$$D\otimes D$$P\otimes D$$\boldsymbol{k}\otimes D$$P\otimes
D\otimes D$$\boldsymbol{k}\otimes D\otimes
D$$\mathrm{(1)}$$\mathrm{(2)}$$\mathrm{(3)}$$\varphi_{P}$$\varepsilon_{P}\otimes\mathrm{id}$$\varepsilon_{P}\otimes\mathrm{id}$$\varphi_{P}\otimes\mathrm{id}$$\rho_{P}^{F}$$\rho_{P}^{F}$$\sim$$\sim$$\rho_{P}^{F}\otimes\mathrm{id}$$\Delta_{D}$$\mathrm{id}\otimes\Delta_{D}$$\mathrm{id}\otimes\Delta_{D}$

Here, commutativity of Parts $\mathrm{(1)}$ and $\mathrm{(2)}$ come from the
definition of $\varphi_{P}$, and Part $\mathrm{(3)}$ comes from what
$(P,\rho_{P}^{F})$ is a right $D$-comodule.

As an application of Theorem 3.1 we have:

######  Corollary 3.3.

Let $C,D$ be two coalgebras over $\boldsymbol{k}$, and
$F:\mathbb{M}^{C}\longrightarrow\mathbb{M}^{D}$ be a $\boldsymbol{k}$-linear
functor. If $F$ is an equivalence of $\boldsymbol{k}$-linear categories and
$U^{D}\circ F=U^{C}$ is satisfied, then the coalgebra map
$\varphi:C\longrightarrow D$ determined by
$F=\mathbb{M}^{\varphi}:\mathbb{M}^{C}\longrightarrow\mathbb{M}^{D}$ in
Theorem 3.1 is an isomorphism.

The above corollary can be easily proved by taking a quasi-inverse $G$ of $F$
such as $U^{C}\circ G=U^{D}$, $G\circ F=1_{\mathbb{M}^{C}}$ and $F\circ
G=1_{\mathbb{M}^{D}}$.

Dualizing of Theorem 3.1 we have also the following corollary.

######  Corollary 3.4.

Let $A,B$ be two finite-dimensional algebras over $\boldsymbol{k}$, and
$F:{}_{B}\mathbb{M}\longrightarrow{}_{A}\mathbb{M}$ be a
$\boldsymbol{k}$-linear functor. If ${}_{A}U\circ F={}_{B}U$ is satisfied for
the forgetful functors ${}_{A}U,{}_{B}U$ to
$\text{Vect}_{\boldsymbol{k}}^{\mathrm{f.d.}}$, then

  1. $(1)$

there is a unique algebra map $\varphi:A\longrightarrow B$ such that
$F={}_{\varphi}\mathbb{M}:{}_{B}\mathbb{M}\longrightarrow{}_{A}\mathbb{M}$,
where ${}_{\varphi}\mathbb{M}$ stands for the $\boldsymbol{k}$-linear functor
induced from $\varphi$,

  2. $(2)$

if $F$ is an equivalence of $\boldsymbol{k}$-linear categories, then the
algebra map $\varphi:A\longrightarrow B$ given by $(1)$ is an isomorphism.

Now, we show the main theorem in the present paper:

######  Theorem 3.5.

Let $A,B$ be weak bialgebras over $\boldsymbol{k}$, and
$F=(F,\bar{\phi}^{F},\bar{\omega}^{F}):\mathbb{M}^{A}\longrightarrow\mathbb{M}^{B}$
be a strong $\boldsymbol{k}$-linear comonoidal functor. Suppose that
$U^{B}\circ F=U^{A}$ as $\boldsymbol{k}$-linear monoidal categories. Then
there is a unique weak bialgebra map $\varphi:A\longrightarrow B$ such that
$F=\mathbb{M}^{\varphi}$ as $\boldsymbol{k}$-linear monoidal categories and
$\bar{\omega}^{F}=\varphi|_{A_{s}}:A_{s}\longrightarrow B_{s}$ is an algebra
isomorphism. Furthermore, $\hat{U}^{B}\circ
F={}_{\varphi_{s}^{-1}}\text{\boldmath{$\mathsf{M}$}}_{\varphi_{s}^{-1}}\circ\hat{U}^{A}$
is satisfied, where
$\hat{U}^{A}:\mathbb{M}^{A}\longrightarrow{}_{A_{s}}\text{\boldmath{$\mathsf{M}$}}_{A_{s}},\
\hat{U}^{B}:\mathbb{M}^{B}\longrightarrow{}_{B_{s}}\text{\boldmath{$\mathsf{M}$}}_{B_{s}}$
are forgetful functors.

###### Proof..

By Theorem 3.1 there is a unique coalgebra map $\varphi:A\longrightarrow B$
such that $F=\mathbb{M}^{\varphi}$ as $\boldsymbol{k}$-linear functors. Since
$U^{B}\circ F=U^{A}$ as $\boldsymbol{k}$-linear monoidal functors, for all
$M,N\in\mathbb{M}^{A}$ the composition

$\displaystyle U^{B}\bigl{(}F(M)\bigr{)}\otimes U^{B}\bigl{(}F(N)\bigr{)}$
$\displaystyle\xrightarrow{\ \phi^{U^{B}}_{F(M),F(N)}\
}U^{B}\bigl{(}F(M)\otimes_{B_{s}}F(N)\bigr{)}$
$\displaystyle\qquad\qquad\xrightarrow{\
U^{B}\bigl{(}(\bar{\phi}^{F}_{M,N})^{-1}\bigr{)}\
}U^{B}\bigl{(}F(M\otimes_{A_{s}}N)\bigr{)}$

coincides with the natural projection $\phi^{U^{A}}_{M,N}:U^{A}(M)\otimes
U^{A}(N)\longrightarrow U^{A}(M\otimes_{A_{s}}N)$. This implies that the
diagram

$M\otimes N$$M\otimes
N$$M\otimes_{A_{s}}A$$M\otimes_{B_{s}}N$$\text{id}_{M\otimes
N}$$\bar{\phi}^{F}_{M,N}$natural proj.natural proj.

commutes. This means that the map
$\bar{\phi}^{F}_{M,N}:F(M\otimes_{A_{s}}N)\longrightarrow
F(M)\otimes_{B_{s}}F(N)$ is induced from the identity map $\text{id}_{M\otimes
N}$.

Let us show that $\bar{\omega}^{F}=\varphi|_{A_{s}}:A_{s}\longrightarrow
B_{s}$ is an algebra isomorphism. Since $U^{B}\circ F=U^{A}$ as
$\boldsymbol{k}$-linear monoidal functors, the composition

$U^{B}\bigl{(}(\bar{\omega}^{F})^{-1}\bigr{)}\circ\omega^{U^{B}}:\boldsymbol{k}\longrightarrow(U^{B}\circ
F)(A_{s})=A_{s}$

coincides with $\omega^{U^{A}}:\boldsymbol{k}\longrightarrow A_{s}$. Thus
$(\bar{\omega}^{F})^{-1}\circ\omega^{U^{B}}=\omega^{U^{A}}$ as maps, and hence
$\bar{\omega}^{F}(1)=1$.

Since $\bar{\omega}^{F}:F(A_{s})\longrightarrow B_{s}$ is a right $B$-comodule
map and $F(A_{s})=\mathbb{M}^{\varphi}(A_{s})$, the following diagram
commutes.

$M\otimes N$$M\otimes
N$$M\otimes_{A_{s}}A$$M\otimes_{A_{s}}A$$M\otimes_{B_{s}}N$$\bar{\omega}^{F}$$\bar{\omega}^{F}\otimes\text{id}$$\Delta_{A}|_{A_{s}}$$\text{id}\otimes\varphi$$\Delta_{B}|_{B_{s}}$

Thus
$\Delta_{B}\bigl{(}\bar{\omega}^{F}(y)\bigr{)}=\bar{\omega}^{F}(y_{(1)})\otimes\varphi(y_{(2)})$
for all $y\in A_{s}$. Applying $\varepsilon_{B}\otimes\text{id}$ to both
sides, we have

$\displaystyle\bar{\omega}^{F}(y)$
$\displaystyle=\varepsilon_{B}\bigl{(}\bar{\omega}^{F}(y_{(1)})\bigr{)}\varphi(y_{(2)})$
$\displaystyle=\varepsilon_{A}(y_{(1)})\varphi(y_{(2)})$
$\displaystyle=\varphi(y).$

This implies that $\bar{\omega}^{F}=\varphi|_{A_{s}}:A_{s}\longrightarrow
B_{s}$, and hence $\varphi(1)=\bar{\omega}^{F}(1)=1$.

Next we show that $\bar{\omega}^{F}$ preserves products. Since
$F=(F,\bar{\phi}^{F},\bar{\omega}^{F})$ is a comonoidal functor, the following
diagram commutes for all $M\in\mathbb{M}^{A}$:

$F(A_{s}\otimes_{A_{s}}M)$$F(M)$$F(A_{s})\otimes_{B_{s}}F(M)$$B_{s}\otimes_{B_{s}}F(M)$$F(l_{M}^{A})$$\bar{\omega}^{F}\otimes_{B_{s}}\text{id}$$\bar{\phi}^{F}_{A_{s},M}$$l^{B}_{M}$

In particular, the following diagram commutes:

$F(A_{s}\otimes_{A_{s}}A_{s})$$F(A_{s})\otimes_{B_{s}}F(A_{s})$$F(A_{s})$$B_{s}\otimes_{B_{s}}F(A_{s})$$B_{s}$$B_{s}\otimes_{B_{s}}B_{s}$$\bar{\phi}_{A_{s},A_{s}}^{F}$$l_{F(A_{s})}^{B}$$l_{B_{s}}^{B}$$F(l_{A_{s}}^{A})$$\bar{\omega}^{F}\otimes_{B_{s}}\text{id}$$\bar{\omega}^{F}$$\text{id}\otimes_{B_{s}}\bar{\omega}^{F}$

It follows that

$\bar{\omega}^{F}(y_{1}y_{2})=\bar{\omega}^{F}(y_{1})\bar{\omega}^{F}(y_{2})$

for all $y_{1},y_{2}\in A_{s}$ since $\bar{\phi}_{A_{s},A_{s}}^{F}$ is induced
from the identity map $\text{id}_{A_{s}\otimes A_{s}}$. Therefore, it is shown
that $\bar{\omega}^{F}=\varphi|_{A_{s}}$ is an algebra isomorphism.

Next let us show that $\varphi$ is a weak bialgebra map. For this it is enough
to show that $\varphi$ preserves products. For two subspaces $P,P^{\prime}$ of
$A$, let $PP^{\prime}$ denote the subspace of $A$ spanned by the set $\\{\
pp^{\prime}\ |\ p\in P,\ p^{\prime}\in P^{\prime}\ \\}$. Let
$\mu_{P,P^{\prime}}:P\otimes P^{\prime}\longrightarrow PP^{\prime}$ be the
restriction of the product $\mu_{A}$ of $A$. Then $\mu_{P,P^{\prime}}$ induces
a $\boldsymbol{k}$-linear map
$\bar{\mu}_{P,P^{\prime}}:P\otimes_{A_{s}}P^{\prime}\longrightarrow
PP^{\prime}$ since the equation $(p\cdot y)p^{\prime}=p(y\cdot p^{\prime})$
holds for $p\in P,\ y\in A_{s},\ p^{\prime}\in P^{\prime}$.

Let $a,a^{\prime}\in A$, and let $P,P^{\prime}$ be finite-dimensional
subcoalgebras of $A$ such that $a\in P$, $a^{\prime}\in P^{\prime}$. Then
$PP^{\prime}$ is also a finite-dimensional subcoalgebra of $A$ containing
$aa^{\prime}$, and the map $\bar{\mu}_{P,P^{\prime}}$ is a right $A$-comodule
map. Let $\varphi_{P},\varphi_{P^{\prime}}$ be the coalgebra maps defined in
the proof of Theorem 3.1. We will show that the following diagram commutes.

(3.1)

Here, $\bar{\mu}_{B}$ is the induced map from the product $\mu_{B}$ of $B$.
Since $\varphi_{P},\varphi_{P^{\prime}}$ are right $B$-comodule maps by Remark
3.2, the map $\varphi_{P}\otimes_{B_{s}}\varphi_{P^{\prime}}$ is well-defined.
Since $F=\text{\boldmath{$\mathsf{M}$}}^{\varphi}$ and
$\hat{U}^{B}\circ\text{\boldmath{$\mathsf{M}$}}^{\varphi}={}_{\varphi_{s}^{-1}}\text{\boldmath{$\mathsf{M}$}}_{\varphi_{s}^{-1}}\circ\hat{U}^{A}$
as $\boldsymbol{k}$-linear monoidal functors, one can verify that
$\bar{\phi}_{P,P^{\prime}}^{F}:F(P\otimes_{A_{s}}P^{\prime})\longrightarrow
F(P)\otimes_{B_{s}}F(P^{\prime})$ is a $\boldsymbol{k}$-linear map given by

$\bar{\phi}_{P,P^{\prime}}^{F}(p\otimes_{A_{s}}p^{\prime})=p\otimes_{B_{s}}p^{\prime}\qquad(p\in
F(P)=P,\ p^{\prime}\in F(P^{\prime})=P^{\prime}).$ (3.2)

Let $(F(P)\otimes_{B_{s}}F(P^{\prime}),\ \rho_{P,P^{\prime}}^{F})$ be the
tensor product in $\text{\boldmath{$\mathsf{M}$}}^{B}$ of right comodules
$(F(P),\rho_{P}^{F})$ and $(F(P^{\prime}),\rho_{P^{\prime}}^{F})$. Here,
$\rho_{P,P^{\prime}}^{F}$ is the right coaction given by

$\rho_{P,P^{\prime}}^{F}(m\otimes_{B_{s}}n)=(m_{(0)}\otimes_{B_{s}}n_{(0)})\otimes
m_{(1)}n_{(1)}\qquad(m\in F(P),\ n\in F(P^{\prime})),$

and $\rho_{P}^{F}(m)=m_{(0)}\otimes m_{(1)},\
\rho_{P^{\prime}}^{F}(n)=n_{(0)}\otimes n_{(1)}$. Since
$\bar{\phi}_{P,P^{\prime}}^{F}$ is a right $B$-comodule map from
$(F(P\otimes_{A_{s}}P^{\prime}),\ \rho_{P\otimes_{A_{s}}P^{\prime}}^{F})$ to
$(F(P)\otimes_{B_{s}}F(P^{\prime}),\ \rho_{P,P^{\prime}}^{F})$, the following
diagram commutes.

$F(P\otimes_{A_{s}}P^{\prime})$$F(P\otimes_{A_{s}}P^{\prime})\otimes
B$$F(P)\otimes_{B_{s}}F(P^{\prime})$$\bigl{(}F(P)\otimes_{B_{s}}F(P^{\prime})\bigr{)}\otimes
B$$\rho_{P\otimes_{A_{s}}P^{\prime}}^{F}$$\rho_{P,P^{\prime}}^{F}$$\bar{\phi}_{P,P^{\prime}}^{F}$$\bar{\phi}_{P,P^{\prime}}^{F}\otimes\text{id}_{B}$

On the other hand, there is a $(B_{s},B_{s})$-bimodule map
$\chi_{P,P^{\prime}}:(P\otimes B)\otimes_{B_{s}}(P^{\prime}\otimes
B)\longrightarrow(P\otimes_{B_{s}}P^{\prime})\otimes B$, which is defined by

$\chi_{P,P^{\prime}}\bigl{(}(p\otimes b)\otimes_{B_{s}}(p^{\prime}\otimes
b^{\prime})\bigr{)}=(p\otimes_{B_{s}}p^{\prime})\otimes bb^{\prime}\qquad(p\in
P,\ p^{\prime}\in P^{\prime},\ b,b^{\prime}\in B).$

We also define a $\boldsymbol{k}$-linear map
$\varepsilon_{P,P^{\prime}}:P\otimes_{A_{s}}P^{\prime}\longrightarrow\boldsymbol{k}$
by

$\varepsilon_{P,P^{\prime}}(p\otimes_{A_{s}}p^{\prime})=\varepsilon(p1_{(1)})\varepsilon(1_{(2)}p^{\prime})\qquad(p\in
P,\ p^{\prime}\in P^{\prime}).$

Then it can be shown that the following diagram commutes.

$F(P\otimes_{A_{s}}P^{\prime})$$F(P)\otimes_{B_{s}}F(P^{\prime})$$B\otimes_{B_{s}}B$$F(P)\otimes_{B_{s}}F(P^{\prime})\otimes
B$$(F(P)\otimes B)\otimes_{B_{s}}(F(P^{\prime})\otimes
B)$$F(P\otimes_{A_{s}}P^{\prime})\otimes B$$\boldsymbol{k}\otimes
B$$F(PP^{\prime})\otimes
B$$F(PP^{\prime})$$B$($\ast$)$\bar{\phi}_{P,P^{\prime}}^{F}$$\varphi_{P}\otimes_{B_{s}}\varphi_{P^{\prime}}$$\rho_{P\otimes_{A_{s}}P^{\prime}}^{F}$$\rho_{P,P^{\prime}}^{F}$$\bar{\phi}_{P,P^{\prime}}^{F}\otimes\text{id}$$\chi_{P,P^{\prime}}$$\rho_{P}^{F}\otimes_{B_{s}}\rho_{P^{\prime}}^{F}$$(F(\varepsilon_{P})\otimes\text{id})_{B_{s}}(F(\varepsilon_{P^{\prime}})\otimes\text{id})$$F(\varepsilon_{P,P^{\prime}})\otimes\text{id}$$\cong$$F(\bar{\mu}_{P,P^{\prime}})\otimes\text{id}$$F(\varepsilon_{PP^{\prime}})\otimes\text{id}$$\rho_{PP^{\prime}}^{F}$$F(\bar{\mu}_{P,P^{\prime}})$$\varphi_{PP^{\prime}}$$\bar{\mu}_{B}$

In fact, the commutativity of ($\ast$) in the above diagram comes from the
following three facts:

  1. (i)

$\bar{\phi}^{F}_{M,N}:F(M\otimes_{A_{s}}N)\longrightarrow
F(M)\otimes_{B_{s}}F(N)$ is an isomorphism,

  2. (ii)

the equation (3.2) holds,

  3. (iii)

for $p\in P,\ p^{\prime}\in P^{\prime}$,

    1. $\bullet$

$p\otimes_{A_{s}}p^{\prime}=(p\cdot 1)\otimes_{A_{s}}p^{\prime}=\bigl{(}p\cdot
1_{(1)}\varepsilon_{s}(1_{(2)})\bigr{)}\otimes_{A_{s}}p^{\prime}=p1_{(1)}\otimes_{A_{s}}\varepsilon_{s}(1_{(2)})p^{\prime}$,

    2. $\bullet$

$\varepsilon\bigl{(}\varepsilon_{s}(1_{(2)})p^{\prime}\bigr{)}=\varepsilon(1_{(2)}p^{\prime})$
by Lemma 2.2(2).

Thus, the diagram (3.1) commutes, and therefore $\varphi$ preserves products.

To complete the proof we need to verify that $F=\mathbb{M}^{\varphi}$ as
$\boldsymbol{k}$-linear comonoidal functors. This is an easy consequence of
the proof of Lemma 2.7(2) since $\bar{\phi}^{F}_{M,N}$ is induced from the
identity map $\text{id}_{M\otimes N}$ for all $M,N\in\mathbb{M}^{A}$. ∎

By Corollary 3.3 and Theorem 3.5 we have:

######  Corollary 3.6.

Let $A,B$ be weak bialgebras over $\boldsymbol{k}$, and
$F=(F,\bar{\phi}^{F},\bar{\omega}^{F}):\mathbb{M}^{A}\longrightarrow\mathbb{M}^{B}$
be a strong $\boldsymbol{k}$-linear comonoidal functor satisfying $U^{B}\circ
F=U^{A}$ as $\boldsymbol{k}$-linear monoidal categories. If $F$ is a
$\boldsymbol{k}$-linear monoidal equivalence, then the weak bialgebra map
$\varphi:A\longrightarrow B$ determined in Theorem 3.5 is an isomorphism of
weak bialgebras.

##  4 Categorical aspects of indecomposable weak bialgebras

Let $A=(A,\Delta_{A},\varepsilon_{A})$ and $B=(B,\Delta_{B},\varepsilon_{B})$
be two weak bialgebras over $\boldsymbol{k}$, and set $H=A\oplus B$. Then, $H$
has a weak bialgebra structure such that the algebra structure is the product
and the coalgebra structure is the direct sum of $A$ and $B$. The target and
source counital maps $\varepsilon_{t}$ and $\varepsilon_{s}$ are given by

$\displaystyle\varepsilon_{t}(x)$
$\displaystyle=(\varepsilon_{A})_{t}(a)+(\varepsilon_{B})_{t}(b),$
$\displaystyle\varepsilon_{s}(x)$
$\displaystyle=(\varepsilon_{A})_{s}(a)+(\varepsilon_{B})_{s}(b)$

for all $x=a+b\in H,\ a\in A,\ b\in B$, where
$(\varepsilon_{A})_{t},(\varepsilon_{A})_{s}$ are the target and source
counital maps of $A$, and $(\varepsilon_{B})_{t},(\varepsilon_{B})_{s}$ are
that of $B$, respectively. Moreover, the target and source subalgebras are
given as follows:

$\displaystyle H_{t}$
$\displaystyle=\varepsilon_{t}(H)=(\varepsilon_{A})_{t}(A)+(\varepsilon_{B})_{t}(B)=A_{t}+B_{t},$
$\displaystyle H_{s}$
$\displaystyle=\varepsilon_{s}(H)=(\varepsilon_{A})_{s}(A)+(\varepsilon_{B})_{s}(B)=A_{s}+B_{s}.$

We call the above weak bialgebra $H$ the direct sum of $A$ and $B$.

A weak bialgebra $H$ is called indecomposable if there are no weak bialgebras
$A$ and $B$ such that $H\cong A\oplus B$. Any finite-dimensional weak
bialgebra can be decomposed into the finitely many direct sum of
indecomposable ones. More precisely, we have:

######  Theorem 4.1.

Let $H$ be a finite-dimensional weak bialgebra over $\boldsymbol{k}$. Then

  1. $(1)$

there are finitely many indecomposable weak bialgebras $H_{i}\ (i=1,\ldots,k)$
such that $H=H_{1}\oplus\cdots\oplus H_{k}$.

  2. $(2)$

If indecomposable weak bialgebras $H_{1},\ldots,H_{k}$ and
$H_{1}^{\prime},\ldots,H_{l}^{\prime}$ satisfy

$H_{1}\oplus\cdots\oplus H_{k}=H=H_{1}^{\prime}\oplus\cdots\oplus
H_{l}^{\prime},$

then $k=l$, and there is a permutation $\sigma\in\mathfrak{S}_{k}$ such that
$H_{i}^{\prime}=H_{\sigma(i)}$ for all $i=1,\ldots,k$.

The above theorem can be proved by decomposability and uniqueness of finite-
dimensional algebras into indecomposable ones (see [21] for detail).

A $\boldsymbol{k}$-linear monoidal category $\mathcal{C}$ is called
indecomposable if there are no $\boldsymbol{k}$-linear monoidal categories
$\mathcal{C}_{1},\mathcal{C}_{2}$ such that
$\mathcal{C}\simeq\mathcal{C}_{1}\times\mathcal{C}_{2}$ as
$\boldsymbol{k}$-linear monoidal categories.

######  Proposition 4.2.

Let $H$ be a weak bialgebra over $\boldsymbol{k}$. If $H$ is decomposable,
then

  1. $(1)$

the left $H$-module category ${}_{H}\text{\boldmath{$\mathsf{M}$}}$ and the
finite-dimensional left $H$-module category ${}_{H}\mathbb{M}$ are
decomposable.

  2. $(2)$

the right $H$-comodule category $\text{\boldmath{$\mathsf{M}$}}^{H}$ and the
finite-dimensional right $H$-comodule category $\mathbb{M}^{H}$ are
decomposable.

###### Proof..

Suppose that $H=A\oplus B$ for some weak bialgebras $A,B$ over
$\boldsymbol{k}$.

(1) The left $H$-module category ${}_{H}\text{\boldmath{$\mathsf{M}$}}$ is
equivalent to the Cartesian product
${}_{A}\text{\boldmath{$\mathsf{M}$}}\times{}_{B}\text{\boldmath{$\mathsf{M}$}}$
as $\boldsymbol{k}$-linear monoidal categories. An equivalence is given by the
following monoidal functors, which are quasi-inverse each other:

$\displaystyle(F,\phi^{F},\omega^{F}):$
$\displaystyle{}_{H}\text{\boldmath{$\mathsf{M}$}}\longrightarrow{}_{A}\text{\boldmath{$\mathsf{M}$}}\times{}_{B}\text{\boldmath{$\mathsf{M}$}},$
$\displaystyle F(X)$ $\displaystyle=(1_{A}\cdot X,\ 1_{B}\cdot X),$
$\displaystyle\phi^{F}_{X,Y}$ $\displaystyle=\mathrm{id}_{F(X)\circledast
F(Y)}:F(X)\circledast F(Y)\longrightarrow F(X\circledast Y),$
$\displaystyle\omega^{F}$
$\displaystyle=\mathrm{id}_{(A_{t},B_{t})}:(A_{t},B_{t})\longrightarrow
F(H_{t}),$ $\displaystyle(G,\phi^{G},\omega^{G}):$
$\displaystyle{}_{A}\text{\boldmath{$\mathsf{M}$}}\times{}_{B}\text{\boldmath{$\mathsf{M}$}}\longrightarrow{}_{H}\text{\boldmath{$\mathsf{M}$}},$
$\displaystyle G(U,V)$ $\displaystyle=U\times V,$
$\displaystyle\phi^{G}_{(U_{1},V_{1}),(U_{2},V_{2})}$
$\displaystyle=\mathrm{id}_{G(U_{1},V_{1})\circledast
G(U_{2},V_{2})}:G(U_{1},V_{1})\circledast G(U_{2},V_{2})\longrightarrow
G\bigl{(}(U_{1},V_{1})\circledast(U_{2},V_{2})\bigr{)},$
$\displaystyle\omega^{G}$ $\displaystyle:H_{t}\longrightarrow
G(A_{t},B_{t})=A_{t}\times B_{t},\ \ \omega^{G}(z)=(1_{A}\cdot z,\ 1_{B}\cdot
z)\qquad(z\in H_{t}).$

By restricting the above equivalence
${}_{H}\text{\boldmath{$\mathsf{M}$}}\simeq{}_{A}\text{\boldmath{$\mathsf{M}$}}\times{}_{B}\text{\boldmath{$\mathsf{M}$}}$
to finite dimension a $\boldsymbol{k}$-linear monoidal equivalence
${}_{H}\mathbb{M}\simeq{}_{A}\mathbb{M}\times{}_{B}\mathbb{M}$ is obtained.

(2) The right $H$-comodule category $\text{\boldmath{$\mathsf{M}$}}^{H}$ is
equivalent to the Cartesian product
$\text{\boldmath{$\mathsf{M}$}}^{A}\times\text{\boldmath{$\mathsf{M}$}}^{B}$
as $\boldsymbol{k}$-linear monoidal categories. An equivalence is given by the
following monoidal functors, which are quasi-inverse each other:

$\displaystyle(F,\phi^{F},\omega^{F}):$
$\displaystyle\text{\boldmath{$\mathsf{M}$}}^{H}\longrightarrow\text{\boldmath{$\mathsf{M}$}}^{A}\times\text{\boldmath{$\mathsf{M}$}}^{B},$
$\displaystyle F\bigl{(}(X,\rho)\bigr{)}$
$\displaystyle=\Bigl{(}\bigl{(}(\varepsilon_{A}\circ\pi_{A})\cdot X,\
(\mathrm{id}\otimes\pi_{A})\circ\rho|_{(\varepsilon_{A}\circ\pi_{A})\cdot
X}\bigr{)},\bigl{(}(\varepsilon_{B}\circ\pi_{B})\cdot X,\
(\mathrm{id}\otimes\pi_{B})\circ\rho|_{(\varepsilon_{B}\circ\pi_{B})\cdot
X}\bigr{)}\Bigr{)}$ $\displaystyle\phi^{F}_{X,Y}$
$\displaystyle=\mathrm{id}_{F(X)\circledast F(Y)}:F(X)\circledast
F(Y)\longrightarrow F(X\circledast Y),$ $\displaystyle\omega^{F}$
$\displaystyle=\mathrm{id}_{(A_{s},B_{s})}:(A_{s},B_{s})\longrightarrow
F(H_{s}),$ $\displaystyle(G,\phi^{G},\omega^{G}):$
$\displaystyle\text{\boldmath{$\mathsf{M}$}}^{A}\times\text{\boldmath{$\mathsf{M}$}}^{B}\longrightarrow\text{\boldmath{$\mathsf{M}$}}^{H},$
$\displaystyle G(U,V)$ $\displaystyle=U\times V,$
$\displaystyle\phi^{G}_{(U_{1},V_{1}),(U_{2},V_{2})}$
$\displaystyle=\mathrm{id}_{G(U_{1},V_{1})\circledast
G(U_{2},V_{2})}:G(U_{1},V_{1})\circledast G(U_{2},V_{2})\longrightarrow
G\bigl{(}(U_{1},V_{1})\circledast(U_{2},V_{2})\bigr{)},$
$\displaystyle\omega^{G}$ $\displaystyle:H_{s}\longrightarrow
G(A_{s},B_{s})=A_{s}\times B_{s},\ \ \omega^{G}(y)=\bigl{(}\pi_{A}(y),\
\pi_{B}(y)\bigr{)}\qquad(y\in H_{s}),$

where $\pi_{A}$ and $\pi_{B}$ are natural surjections associated with the
direct sum decomposition $H=A\oplus B$.

By restricting the above equivalence
$\text{\boldmath{$\mathsf{M}$}}^{H}\simeq\text{\boldmath{$\mathsf{M}$}}^{A}\times\text{\boldmath{$\mathsf{M}$}}^{B}$
to finite dimension a $\boldsymbol{k}$-linear monoidal equivalence
$\mathbb{M}^{H}\simeq\mathbb{M}^{A}\times\mathbb{M}^{B}$ is obtained. ∎

The converse of the above proposition is true. To prove it we need the
following reconstruction theorem of bialgebras.

######  Theorem 4.3 (Ulbrich [19], Schauenburg [17, Theorem 5.4]).

Let $\mathcal{C}$ be a $\boldsymbol{k}$-linear monoidal category, and
$\omega:\mathcal{C}\longrightarrow\text{Vect}_{\boldsymbol{k}}^{\text{f.d.}}$
be a faithful and exact $\boldsymbol{k}$-linear monoidal functor. Then, there
are a bialgebra $B$ and a monoidal equivalence
$F:\mathcal{C}\longrightarrow\mathbb{M}^{B}$ such that $U^{B}\circ F=\omega$,
where
$U^{B}:\mathbb{M}^{B}\longrightarrow\text{Vect}_{\boldsymbol{k}}^{\text{f.d.}}$
is the forgetful functor. ∎

Combining Theorems 3.5 and 4.3 one can show the following.

######  Theorem 4.4.

Let $H$ be a finite-dimensional weak bialgebra over $\boldsymbol{k}$. Then $H$
is indecomposable as a weak bialgebra if and only if the finite-dimensional
left $H$-module category ${}_{H}\mathbb{M}$ is indecomposable as a
$\boldsymbol{k}$-linear monoidal category.

###### Proof..

By the contraposition of Proposition 4.2 “if part” holds. We will show that
“only if part”. Suppose that $H$ is indecomposable whereas ${}_{H}\mathbb{M}$
is not. Then, there are $\boldsymbol{k}$-linear monoidal categories
$\mathcal{C}_{1},\mathcal{C}_{2}$ such that
${}_{H}\mathbb{M}\simeq\mathcal{C}_{1}\times\mathcal{C}_{2}$ as
$\boldsymbol{k}$-linear monoidal categories. Let
${}_{H}U:{}_{H}\mathbb{M}\longrightarrow\text{Vect}_{\boldsymbol{k}}^{\text{f.d.}}$
be the forgetful functor, and
$F:\mathcal{C}_{1}\times\mathcal{C}_{2}\longrightarrow{}_{H}\mathbb{M}$ be a
$\boldsymbol{k}$-linear monoidal category equivalence. Since the two
$\boldsymbol{k}$-linear monoidal functors

$\displaystyle\omega_{1}:$ $\displaystyle\
\mathcal{C}_{1}\cong\mathcal{C}_{1}\times 0\xrightarrow{\ \ F\ \
}{}_{H}\mathbb{M}\xrightarrow{\ \ {}_{H}U\ \
}\text{Vect}_{\boldsymbol{k}}^{\text{f.d.}},$ $\displaystyle\omega_{2}:$
$\displaystyle\ \mathcal{C}_{2}\cong 0\times\mathcal{C}_{2}\xrightarrow{\ \ F\
\ }{}_{H}\mathbb{M}\xrightarrow{\ \ {}_{H}U\ \
}\text{Vect}_{\boldsymbol{k}}^{\text{f.d.}}$

are faithful and exact, there are bialgebras $A,B$ and $\boldsymbol{k}$-linear
monoidal equivalences $G_{1}:\mathcal{C}_{1}\simeq\mathbb{M}^{A},\
G_{2}:\mathcal{C}_{2}\simeq\mathbb{M}^{B}$ such that $U^{A}\circ
G_{1}=\omega_{1},\ U^{B}\circ G_{2}=\omega_{2}$ by Theorem 4.3. Then

$\mathcal{C}_{1}\times\mathcal{C}_{2}\simeq\mathbb{M}^{A}\times\mathbb{M}^{B}\cong\mathbb{M}^{A\oplus
B}$

as $\boldsymbol{k}$-linear monoidal categories. Thus we have
$\boldsymbol{k}$-linear monoidal equivalence
$G:\mathbb{M}^{H^{\ast}}\longrightarrow\mathbb{M}^{A\oplus B}$. This
equivalence satisfies $U^{A\oplus B}\circ G=U^{H^{\ast}}$. So, by Corollary
3.6, there is a weak bialgebra isomorphism $g:A\oplus B\longrightarrow
H^{\ast}$. Therefore, $H\cong H^{\ast\ast}\cong(A\oplus B)^{\ast}\cong
A^{\ast}\oplus B^{\ast}$. This contradicts the fact that $H$ is indecomposable
as a weak bialgebra. ∎

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