# Quotient Category of a Multiring Category

Zhenbang Zuo, Gongxiang Liu  Department of Mathematics, Nanjing University,
Nanjing 210093, China<EMAIL_ADDRESS>Department of Mathematics,
Nanjing University, Nanjing 210093, China<EMAIL_ADDRESS>
###### Abstract.

The aim of this paper is to introduce a tensor structure for the quotient
category of an abelian monoidal category with biexact tensor product in order
to make the canonical funtor be a monoidal functor. In this tensor product, we
prove that the quotient category of a multiring category (resp. a multitensor
category) is still a multiring category (resp. a multitensor category).

###### Key words and phrases:

quotient category, Serre tensor-ideal, tensor category

##  1\. Introduction

Quotient categories attract many mathematicians, although there are various
definitions for a quotient category. Panyue Zhou researched the quotient
category of a triangulated category with a Serre functor and a non-zero
contravariantly finite rigid subcategory in [7]. He also researched mutation
pairs in an abelian category in order to make a certain quotient category
carry a triangulated structure in [8]. Steffen Koenig and Bin Zhu proved that
the quotient category of a triangulated category modulo a tilting subcategory
carries an induced abelian structure in [5]. Furthermore, Xiao-Wu Chen and
Henning Krause studied expansions of abelian categories in [2] in which
quotient categories and quotient functors played an important role.

This paper build upon the groundwork of [4], in which Gabriel introduced the
quotient category $\mathcal{A}/\mathcal{C}$ of an abelian category
$\mathcal{A}$ by a Serre subcategory $\mathcal{C}$. The process of obtaining
the quotient category of an abelian category is often called as the
localization of an abelian category. Gabriel proved that the quotient category
of an abelian category is still an abelian category, and the canonical functor
$T:\mathcal{A}\to\mathcal{A}/\mathcal{C}$ is an exact functor.

Our aim is to introduce a tensor product for the quotient category of a
multitensor category. To make the results seem more interesting, we divide our
results into two parts. The first part focuses on cases of abelian category,
in which we show that the quotient category of a locally finite abelian
category is a locally finite abelian category. In fact, the first part
provides necessary conditions for the second part.

The second part focuses on cases of monoidal category, in which we provide a
definition for tensor product in the quotient category of an abelian monoidal
category with biexact tensor product, and show that the quotient category of a
multiring category (resp. a multitensor category) is a multiring category
(resp. a multitensor category). Moreover, we show that a two-sided Serre
tensor-ideal of a multiring category with left duals must be a direct sum of
the component subcategories, and therefore the corresponding quotient category
is also a direct sum of the component subcategories.

In this paper, we present some basic definitions and results as preparation in
section 2. We discuss the locally finiteness of the quotient category of a
locally finite abelian category in section 3 as the first part of our main
results. We devote section 4 to give a definition for tensor product in the
quotient category of an abelian monoidal category with biexact tensor product,
to study the quotient category of a multiring category (resp. a multitensor
category), and to study the structure of a two-sided Serre tensor-ideal of a
multiring category with left duals as the second part of our main results.

##  2\. Preliminaries

Let $k$ be an algebraically closed field throughout this paper. Recall the
definition of Serre subcategory.

######  Definition 2.1 (Serre subcategory).

Let $\mathcal{A}$ be an abelian category. A full additive subcategory
$\mathcal{C}\subset\mathcal{A}$ is a Serre subcategory provided that
$\mathcal{C}$ is closed under taking subobjects, quotients and extensions.

Note that there is an equivalent definition for Serre subcategory.

######  Lemma 2.2.

A non-empty full subcategory $\mathcal{C}$ of $\mathcal{A}$ is a Serre
subcategory if and only if for any $X^{\prime}\to X\to X^{\prime\prime}$ exact
in $\mathcal{A}$ with $X^{\prime},X^{\prime\prime}\in\mathcal{C}$, then also
$X\in\mathcal{C}$.

Recall the definition of quotient category of an abelian category. One can
refer to [4] for more details about the localization of an abelian category.

######  Definition 2.3 (Quotient Category).

Let $\mathcal{A}$ be an abelian category, $\mathcal{C}$ be a Serre subcategory
of $\mathcal{A}$. The category $\mathcal{A}/\mathcal{C}$ is defined as
follows:

  1. (1)

the objects of $\mathcal{A}/\mathcal{C}$ coincide with the objects of
$\mathcal{A}$.

  2. (2)

the set of morphisms from $M$ to $N$ is defined by

$\mathrm{Hom}_{\mathcal{A}/\mathcal{C}}(M,N):=\varinjlim_{M^{\prime},N^{\prime}}\mathrm{Hom}_{\mathcal{A}}(M^{\prime},N/N^{\prime})$

in which $M^{\prime}$ and $N^{\prime}$ go through the subobjects of $M$ and
$N$ respectively such that $M/M^{\prime},N^{\prime}\in\mathcal{C}$.

Denote the canonical functor by $T:\mathcal{A}\to\mathcal{A}/\mathcal{C}$. In
fact, every morphism $\bar{f}$ in $\mathcal{A}/\mathcal{C}$ can be written as
$Tf$ for some $f$ in the direct system. One can observe this fact by either
understanding that the direct system in the definition of quotient category is
directed or referring to [4].

The following lemma, which is Lemma 2 in [4], describes how we can determine
whether $Tu$ is a zero morphism, a monomorphism, or an epimorphism in the
quotient category $\mathcal{A}/\mathcal{C}$ for a given morphism $u$ in
$\mathcal{A}$.

######  Lemma 2.4.

Let $\mathcal{A}$ be an abelian category, $\mathcal{C}$ be a Serre subcategory
of $\mathcal{A}$. Let $u:M\to N$ be a morphism of $\mathcal{A}$, then

  1. (1)

$Tu$ is zero if and only if $\mathrm{Im}\ u\in\mathcal{C}$;

  2. (2)

$Tu$ is a monomorphism if and only if $\mathrm{Ker}\ u\in\mathcal{C}$;

  3. (3)

$Tu$ is an epimorphism if and only if $\mathrm{Coker}\ u\in\mathcal{C}$;

Recall that an additive category $\mathcal{A}$ is said to be $k$-linear if for
any objects $X,Y\in\mathcal{A}$, $\mathrm{Hom}_{\mathcal{A}}(X,Y)$ is equipped
with a structure of a vector space over $k$, such that composition of
morphisms is $k$-linear. The following definitions refer to [3].

######  Definition 2.5 (Locally finite).

A $k$-linear abelian category $\mathcal{A}$ is said to be locally finite if
the following two conditions are satisfied:

  1. (1)

for any two objects $X,Y$ in $\mathcal{A}$, the vector space
$\mathrm{Hom}_{\mathcal{A}}(X,Y)$ is finite dimensional;

  2. (2)

every object in $\mathcal{A}$ has finite length.

Recall also the definitions for multiring category and multitensor category as
follows.

######  Definition 2.6.

  1. (1)

A multiring category $\mathcal{A}$ over $k$ is a locally finite $k$-linear
abelian monoidal category with bilinear and biexact tensor product. If in
addition $\mathrm{End}_{\mathcal{A}}(1)=k$, we will call $\mathcal{A}$ a ring
category.

  2. (2)

Let $\mathcal{A}$ be a locally finite $k$-linear abelian rigid monoidal
category. We will call $\mathcal{A}$ a multitensor category over $k$ if the
bifunctor $\otimes:\mathcal{A}\times\mathcal{A}\to\mathcal{A}$ is bilinear on
morphisms. If in addition $\mathrm{End}_{\mathcal{A}}(1)=k$, and $\mathcal{A}$
is indecomposable i.e. $\mathcal{A}$ is not equivalent to a direct sum of
nonzero multitensor categories, then we will call $\mathcal{A}$ a tensor
category.

Recall also Theorem 4.3.8 in [3] as the following:

######  Theorem 2.7.

  1. (1)

In a ring category with left duals, the unit object $1$ is simple.

  2. (2)

In a multiring category with left duals, the unit object $1$ is semisimple,
and is a direct sum of pairwise non-isomorphic simple objects $1_{i}$.

Note that $1_{i}$ is the image of $p_{i}$, where $\\{p_{i}\\}_{i\in I}$ is the
set of primitive idempotents of the algebra $\mathrm{End}(1)$. And
$1=\bigoplus\limits_{i\in I}1_{i}$.

We also write down the definition of two-sided Serre tensor-ideal. In some
literature, for example in [1], ’tensor-ideal’ is written as
’$\otimes$-ideal’, and it means one side absorption. In some other literature,
for example in [6], the name ’tensor-ideal’ is used.

######  Definition 2.8.

Let $\mathcal{A}$ be an abelian monoidal category. A Serre subcategory
$\mathcal{C}$ of $\mathcal{A}$ is called a two-sided Serre tensor-ideal of
$\mathcal{A}$ if for any $X\in\mathcal{A}$, $Y\in\mathcal{C}$, we have
$X\otimes Y\in\mathcal{C}$ and $Y\otimes X\in\mathcal{C}$.

##  3\. Localization preserves locally finiteness

Let $\mathcal{A}$ be an abelian category with a Serre subcategory
$\mathcal{C}$. As proved in [4], $\mathcal{A}/\mathcal{C}$ is an abelian
category. This conclusion allows us to discuss the length of an object in
$\mathcal{A}/\mathcal{C}$.

We have the following lemma about simple objects in $\mathcal{A}/\mathcal{C}$.

######  Lemma 3.1.

Every simple object in $\mathcal{A}$ has length $1$ or $0$ in
$\mathcal{A}/\mathcal{C}$.

###### Proof.

Suppose $X$ is a simple object in $\mathcal{A}$, consider its subobject $Y$ in
$\mathcal{A}/\mathcal{C}$, and denote the corresponding monomorphism by
$\bar{f}:Y\to X$.

We also view $Y$ as an object in $\mathcal{A}$. Because

$\mathrm{Hom}_{\mathcal{A}/\mathcal{C}}(Y,X)=\varinjlim_{Y^{\prime},X^{\prime}}\mathrm{Hom}_{\mathcal{A}}(Y^{\prime},X/X^{\prime}),$

where $Y/Y^{\prime},\ X^{\prime}\in\mathcal{C}$, there exists a subobject
$Y_{1}$ of $Y$ such that $Y/Y_{1}\in\mathcal{C}$ such that the following
diagram commute

$\textstyle{TY_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{Ti}$$\scriptstyle{Tf}$$\textstyle{TY\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\bar{f}}$$\textstyle{TX\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{TX}$

where $T$ is the canonical functor, $i$ is the injection, and $f:Y_{1}\to X$
is an epimorphism since $X$ is simple in $\mathcal{A}$. Thus, we have the
following exact sequence

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathrm{Ker}\
f\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{Y_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\textstyle{X\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0.}$

Because the canonical functor $T$ is exact, we have the following exact
sequence

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{T\mathrm{Ker}\
f\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{TY_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{Tf}$$\textstyle{TX\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0.}$

This means $Tf$ is an epimorphism in $\mathcal{A}/\mathcal{C}$. Note that $Ti$
is an isomorphism in $\mathcal{A}/\mathcal{C}$ and $\bar{f}$ is a
monomorphism, we know $Tf=\bar{f}\circ Ti$ is a monomorphism. Therefore, $Tf$
is an isomorphism. Consequently, $\bar{f}$ is an isomorphism. Thus, every
monomorphism in $\mathcal{A}/\mathcal{C}$ to $X$ is an isomorphism in
$\mathcal{A}/\mathcal{C}$. It shows that $X$ has length $1$ or $0$ in
$\mathcal{A}/\mathcal{C}$. Note that the case of length $0$ means $X$ is in
$\mathcal{C}$, and it is zero in $\mathcal{A}/\mathcal{C}$. ∎

Next, we prove that the localization preserves finite length.

######  Proposition 3.2.

Suppose every object in $\mathcal{A}$ has finite length, then every object in
$\mathcal{A}/\mathcal{C}$ has finite length.

###### Proof.

For an arbitrary object $X$ in $\mathcal{A}$, it has finite length in
$\mathcal{A}$ and one can assume its Jordan-Hölder series as following without
loss of generality.

$0=X_{0}\subset X_{1}\subset\cdots\subset X_{n-1}\subset X_{n}=X$

Since $X_{i+1}/X_{i}$ is a simple object in $\mathcal{A}$, it has length $1$
or $0$ in $\mathcal{A}/\mathcal{C}$ by the above Lemma. Because

$l(X_{i+1})=l(X_{i})+l(X_{i+1}/X_{i}),\ \text{ for all }\ 0\leq i\leq n-1,$

we know that

$\displaystyle l(X)$ $\displaystyle=l(X_{n})=l(X_{n-1})+l(X_{n}/X_{n-1})$
$\displaystyle=l(X_{n-2})+l(X_{n-1}/T_{n-2})+l(X_{n-1}/T_{n})$
$\displaystyle=\cdots$ $\displaystyle=\sum_{i=0}^{n-1}l(X_{i+1}/X_{i})$
$\displaystyle\leq n.$

This means $X$ has finite length in $\mathcal{A}/\mathcal{C}$, which is
actually smaller than its length in $\mathcal{A}$. ∎

In order to show that the quotient category of a locally finite abelian
category is locally finite, we are required to prove that the Hom-spaces of
the quotient category are finite dimensional.

######  Lemma 3.3.

Let $\mathcal{A}$ be a locally finite $k$-linear abelian category, then
$\mathrm{Hom}_{\mathcal{A}/\mathcal{C}}(M,N)$ is a finite dimensional vector
space, where $M,N$ are objects in $\mathcal{A}$.

###### Proof.

By definition, $\mathrm{Hom}_{\mathcal{A}/\mathcal{C}}(M,N)$ is obviously a
vector space. We now prove $\mathrm{Hom}_{\mathcal{A}/\mathcal{C}}(M,N)$ is
finite dimensional by induction on the lengths of $M,N$ in $\mathcal{A}$.
Firstly, suppose $l(M)=l(N)=1$. If $M\in\mathcal{C}$ or $N\in\mathcal{C}$,
then $\mathrm{Hom}_{\mathcal{A}/\mathcal{C}}(M,N)=0$. If
$M,N\notin\mathcal{C}$, it is clear that
$\mathrm{Hom}_{\mathcal{A}/\mathcal{C}}(M,N)=\mathrm{Hom}_{\mathcal{A}}(M,N)$,
which is finite dimensional.

Secondly, we want to show $\mathrm{Hom}_{\mathcal{A}/\mathcal{C}}(M,N)$ is a
finite dimensional vector space for $l(N)=1$. We prove it by induction on the
length of $M$, suppose $\mathrm{Hom}_{\mathcal{A}/\mathcal{C}}(M,N)$ is finite
dimensional for all $M,\ N$ such that $l(M)\leq m$ and $l(N)=1$. Now we
consider the case of $l(M)=m+1$ and $l(N)=1$. On one hand, if there is a
subobject $X$ of $M$ such that $0\leq l(X)\leq m$ and $M/X\in\mathcal{C}$, we
know that $Ti_{M}^{X}$ is an isomorphism in $\mathcal{A}/\mathcal{C}$, where
$i_{M}^{X}:X\to M$ is the monomorphism. This means that
$\mathrm{Hom}_{\mathcal{A}/\mathcal{C}}(M,N)\cong\mathrm{Hom}_{\mathcal{A}/\mathcal{C}}(X,N)$,
which is finite dimensional by the induction assumption. On the other hand, if
for any subobject $X$ of $M$ satisfying $0\leq l(X)\leq m$,
$M/X\notin\mathcal{C}$. It follows that
$\mathrm{Hom}_{\mathcal{A}/\mathcal{C}}(M,N)=\mathrm{Hom}_{\mathcal{A}}(M,N)$,
which is finite dimensional. Therefore,
$\mathrm{Hom}_{\mathcal{A}/\mathcal{C}}(M,N)$ is a finite dimensional vector
space for $l(N)=1$. Similarly, $\mathrm{Hom}_{\mathcal{A}/\mathcal{C}}(M,N)$
is a finite dimensional vector space for $l(M)=1$.

Thirdly, given two positive integers $m,n\geq 1$, suppose
$\mathrm{Hom}_{\mathcal{A}/\mathcal{C}}(M,N)$ is a finite dimensional vector
space for $l(M)\leq m$ and $l(N)\leq n+1$ and for $l(M)\leq m+1$ and $l(N)\leq
n$. We consider the case of $l(M)=m+1$ and $l(N)=n+1$.

If there is a subobject $X$ of $M$ such that $0\leq l(X)\leq m$ and
$M/X\in\mathcal{C}$, then
$\mathrm{Hom}_{\mathcal{A}/\mathcal{C}}(M,N)\cong\mathrm{Hom}_{\mathcal{A}/\mathcal{C}}(X,N)$
since $Ti_{M}^{X}$ is an isomorphism in $\mathcal{A}/\mathcal{C}$, where
$i_{M}^{X}:X\to M$ is the monomorphism. By the induction assumption,
$\mathrm{Hom}_{\mathcal{A}/\mathcal{C}}(M,N)\cong\mathrm{Hom}_{\mathcal{A}/\mathcal{C}}(X,N)$
is finite dimensional.

If there is a subobject $Y$ of $N$ such that $1\leq l(Y)\leq n+1$ and
$Y\in\mathcal{C}$, then
$\mathrm{Hom}_{\mathcal{A}/\mathcal{C}}(M,N)\cong\mathrm{Hom}_{\mathcal{A}/\mathcal{C}}(M,N/Y)$
since $Tp_{N/Y}^{N}$ is an isomorphism in $\mathcal{A}/\mathcal{C}$, where
$p_{N/Y}^{N}:N\to N/Y$ is the epimorphism. By the induction assumption,
$\mathrm{Hom}_{\mathcal{A}/\mathcal{C}}(M,N)\cong\mathrm{Hom}_{\mathcal{A}/\mathcal{C}}(M,N/Y)$
is finite dimensional.

If for any subobject $X$ of $M$ and any subobject $Y$ of $N$ satisfying $0\leq
l(X)\leq m$ and $1\leq l(Y)\leq n+1$, $M/X,Y\notin\mathcal{C}$. It follows
that
$\mathrm{Hom}_{\mathcal{A}/\mathcal{C}}(M,N)=\mathrm{Hom}_{\mathcal{A}}(M,N)$,
which is finite dimensional.

In summary, $\mathrm{Hom}_{\mathcal{A}/\mathcal{C}}(M,N)$ is a finite
dimensional vector space for $l(M)=m+1$ and $l(N)=n+1$. Consequently, we
obtain that $\mathrm{Hom}_{\mathcal{A}/\mathcal{C}}(M,N)$ is a finite
dimensional vector space for any objects $M,N$ in $\mathcal{A}$. ∎

It follows directly from Proposition 3.2 and Lemma 3.3 that
$\mathcal{A}/\mathcal{C}$ is locally finite.

######  Proposition 3.4.

The quotient category $\mathcal{A}/\mathcal{C}$ of a locally finite $k$-linear
abelian category $\mathcal{A}$ is a locally finite $k$-linear abelian
category.

##  4\. Tensor product in quotient category

In order to give a definition for tensor product in the quotient category of
an abelian monoidal category, we first introduce a lemma.

######  Lemma 4.1.

Let $\mathcal{A}$ be an abelian monoidal category, $\mathcal{C}$ be a two-
sided Serre tensor-ideal of $\mathcal{A}$. Suppose $f,g$ are morphisms in
$\mathcal{A}$ such that $\mathrm{ker}\ f$, $\mathrm{ker}\ g$ belong to
$\mathcal{C}$, then $\mathrm{ker}(f\otimes g)$ belongs to $\mathcal{C}$.
Similarly, if $\mathrm{coker}\ f$, $\mathrm{coker}\ g$ belong to
$\mathcal{C}$, then $\mathrm{coker}(f\otimes g)$ belongs to $\mathcal{C}$.

###### Proof.

Consider

$f:M\to N\ \text{and}\ g:X\to Y.$

Note that there is a composition

${f\otimes g:M\otimes X}$${N\otimes X}$${N\otimes Y,}$$\scriptstyle{f\otimes
id_{X}}$$\scriptstyle{id_{N}\otimes g}$

and there is a commutative diagram

${M\otimes X}$${N\otimes X}$${\mathrm{coker}(f\otimes
id)}$${0}$${0}$${N\otimes Y}$${N\otimes Y}$${0}$$\scriptstyle{f\otimes
id}$$\scriptstyle{f\otimes g}$$\scriptstyle{id\otimes g}$$\scriptstyle{id}$

By snake lemma, this implies an exact sequence

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathrm{ker}(f\otimes
id)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathrm{ker}(f\otimes
g)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathrm{ker}(id\otimes
g)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathrm{coker}(f\otimes
id)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathrm{coker}(f\otimes
g)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathrm{coker}(id\otimes
g)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$

which means

$\left\\{\begin{aligned} &\mathrm{ker}(f\otimes id)\to\mathrm{ker}(f\otimes
g)\to\mathrm{ker}(id\otimes g)\\\ &\mathrm{coker}(f\otimes
id)\to\mathrm{coker}(f\otimes g)\to\mathrm{coker}(id\otimes
g)\end{aligned}\right.$

exact. Note that $\mathrm{ker}(f\otimes id)=\mathrm{ker}\ f\otimes
X\in\mathcal{C}$, and $\mathrm{ker}(id\otimes g)=N\otimes\mathrm{ker}\
g\in\mathcal{C}$ provided $\mathrm{ker}\ f$, $\mathrm{ker}\ g$ belong to
$\mathcal{C}$. By Lemma 2.2, we know $\mathrm{ker}(f\otimes g)\in\mathcal{C}$.
Similarly, $\mathrm{coker}(f\otimes g)\in\mathcal{C}$ provided
$\mathrm{coker}\ f$, $\mathrm{coker}\ g$ belong to $\mathcal{C}$. ∎

The following two lemmas study the tensor product of two subobjects, and the
tensor product of two quotient objects respectively.

######  Lemma 4.2.

Let $\mathcal{A}$ be an abelian monoidal category with biexact tensor product.
Let $M$, $X$ be two objects in $\mathcal{A}$, $M^{\prime}$ be a subobject of
$M$, $X^{\prime}$ be a subobject of $X$, then $M^{\prime}\otimes X^{\prime}$
is a subobject of $M\otimes X$.

###### Proof.

Consider monomorphisms $i_{1}:M^{\prime}\to M$ and $i_{2}:X^{\prime}\to X$,
and exact sequences

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{M^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{i_{1}}$$\textstyle{M\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathrm{coker}\
i_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0;}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{i_{2}}$$\textstyle{X\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathrm{coker}\
i_{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0.}$

Because the tensor product is biexact, we have two following exact sequences

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{M^{\prime}\otimes
X^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{i_{1}\otimes
id}$$\textstyle{M\otimes
X^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathrm{coker}\
i_{1}\otimes
X^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0;}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{M\otimes
X^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{id\otimes
i_{2}}$$\textstyle{M\otimes
X\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{M\otimes\mathrm{coker}\
i_{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0.}$

This means

${i_{1}\otimes i_{2}:M^{\prime}\otimes X^{\prime}}$${M\otimes
X^{\prime}}$${M\otimes X}$$\scriptstyle{i_{1}\otimes
id}$$\scriptstyle{id\otimes i_{2}}$

is a monomorphism. Hence, $M^{\prime}\otimes X^{\prime}$ is a subobject of
$M\otimes X$. ∎

######  Lemma 4.3.

Let $\mathcal{A}$ be an abelian monoidal category with biexact tensor product.
Let $N$, $Y$ be two objects in $\mathcal{A}$, $N/N^{\prime}$ be a quotient
object of $N$, $Y/Y^{\prime}$ be a quotient object of $Y$, then
$N/N^{\prime}\otimes Y/Y^{\prime}$ is a quotient object of $N\otimes Y$.

###### Proof.

Consider epimorphisms $p_{1}:N\to N/N^{\prime}$ and $p_{2}:Y\to Y/Y^{\prime}$,
and exact sequences

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{N^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{N\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{p_{1}}$$\textstyle{N/N^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0;}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{Y^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{Y\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{p_{2}}$$\textstyle{Y/Y^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0;}$

Because the tensor product is biexact, we have two following exact sequences

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{N^{\prime}\otimes
Y/Y^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{N\otimes
Y/Y^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{p_{1}\otimes
id}$$\textstyle{N/N^{\prime}\otimes
Y/Y^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0;}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{N\otimes
Y^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{N\otimes
Y\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{id\otimes
p_{2}}$$\textstyle{N\otimes
Y/Y^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0.}$

This means

${p_{1}\otimes p_{2}:N\otimes Y}$${N\otimes
Y/Y^{\prime}}$${N/N^{\prime}\otimes Y/Y^{\prime}}$$\scriptstyle{id\otimes
p_{2}}$$\scriptstyle{p_{1}\otimes id}$

is an epimorphism. Hence, $N/N^{\prime}\otimes Y/Y^{\prime}$ is a quotient
object of $N\otimes Y$. ∎

In fact, we have an isomorphism

$N/N^{\prime}\otimes Y/Y^{\prime}\cong N\otimes Y/\mathrm{ker}(p_{1}\otimes
p_{2}).$

Now, we can prove that $T(f\otimes
g)\in\mathrm{Hom}_{\mathcal{A}/\mathcal{C}}(M\otimes X,N\otimes Y)$ for any
$\bar{f}=Tf:M\to N$, $\bar{g}=Tg:X\to Y$ in $\mathcal{A}/\mathcal{C}$.

######  Proposition 4.4.

Let $\mathcal{A}$ be an abelian monoidal category with biexact tensor product,
$\mathcal{C}$ be a two-sided Serre tensor-ideal of $\mathcal{A}$. Let
$\bar{f}:M\to N$, $\bar{g}:X\to Y$ be two morphisms in
$\mathcal{A}/\mathcal{C}$. Then $T(f\otimes g)$ is a morphism in
$\mathrm{Hom}_{\mathcal{A}/\mathcal{C}}(M\otimes X,N\otimes Y)$ where $f,g$
are in direct systems, and $T$ is the canonical functor.

###### Proof.

Suppose $f:M^{\prime}\to N/N^{\prime}$, $g:X^{\prime}\to Y/Y^{\prime}$ with
$M/M^{\prime}$, $N^{\prime}$, $X/X^{\prime}$, $Y^{\prime}$ are in
$\mathcal{C}$. It follows that

$f\otimes g:M^{\prime}\otimes X^{\prime}\to N/N^{\prime}\otimes Y/Y^{\prime}.$

Denote $i_{1}:M^{\prime}\to M$, $i_{2}:X^{\prime}\to X$ to be inclusions.
Since $Ti_{1}$, $Ti_{2}$ are isomorphisms in $\mathcal{A}/\mathcal{C}$, we
know that $\mathrm{coker}\ i_{1}$, $\mathrm{coker}\ i_{2}$ belong to
$\mathcal{C}$. By Lemma 4.1, $\mathrm{coker}(i_{1}\otimes
i_{2})\in\mathcal{C}$. This means $(M\otimes X)/(M^{\prime}\otimes
X^{\prime})\in\mathcal{C}$.

Denote $p_{1}:N\to N/N^{\prime}$, $p_{2}:Y\to Y/Y^{\prime}$. We have already
known that $N/N^{\prime}\otimes Y/Y^{\prime}\cong N\otimes
Y/\mathrm{ker}(p_{1}\otimes p_{2})$. Since $Tp_{1}$, $Tp_{2}$ are isomorphisms
in $\mathcal{A}/\mathcal{C}$, we know that $\mathrm{ker}\ p_{1}$,
$\mathrm{ker}\ p_{2}$ are in $\mathcal{C}$. By Lemma 4.1, it follows that
$\mathrm{ker}(p_{1}\otimes p_{2})\in\mathcal{C}$.

In summary, $f\otimes g$ is in the direct system. Therefore, $T(f\otimes
g)\in\mathrm{Hom}_{\mathcal{A}/\mathcal{C}}(M\otimes X,N\otimes Y)$. ∎

######  Proposition 4.5.

Let $\mathcal{A}$ be an abelian monoidal category with biexact tensor product,
$\mathcal{C}$ be a two-sided Serre tensor-ideal of $\mathcal{A}$. Let
$\bar{f}:M\to N$, $\bar{g}:X\to Y$ be two morphisms in
$\mathcal{A}/\mathcal{C}$. Suppose $\bar{f}=Tf=Tf_{1}$ and
$\bar{g}=Tg=Tg_{1}$, then $T(f\otimes g)=T(f_{1}\otimes g_{1})$.

###### Proof.

First of all, we claim that

$T(id\otimes g)=T(id\otimes g_{1}).$

Suppose $g:X^{\prime}\to Y/Y^{\prime}$, $g_{1}:X_{1}^{\prime}\to
Y/Y_{1}^{\prime}$. Since the direct system in the definition of quotient
categroty is directed, we can obtain a morphism $g_{2}:X_{2}^{\prime}\to
Y/Y_{2}^{\prime}$ such that the following diagrams commute:

${X_{2}^{\prime}}$${X^{\prime}}$${Y/Y_{2}^{\prime}}$${Y/Y^{\prime}}$$\scriptstyle{i^{X_{2}^{\prime}}_{X^{\prime}}}$$\scriptstyle{g_{2}}$$\scriptstyle{g}$$\scriptstyle{p^{Y/Y^{\prime}}_{Y/Y_{2}^{\prime}}}$

and

${X_{2}^{\prime}}$${X_{1}^{\prime}}$${Y/Y_{2}^{\prime}}$${Y/Y_{1}^{\prime}.}$$\scriptstyle{i^{X_{2}^{\prime}}_{X_{1}^{\prime}}}$$\scriptstyle{g_{2}}$$\scriptstyle{g_{1}}$$\scriptstyle{p^{Y/Y_{1}^{\prime}}_{Y/Y_{2}^{\prime}}}$

Thus, we obtain two commutative diagrams:

${M\otimes X_{2}^{\prime}}$${M\otimes X^{\prime}}$${M\otimes
Y/Y_{2}^{\prime}}$${M\otimes Y/Y^{\prime}}$$\scriptstyle{id\otimes
i^{X_{2}^{\prime}}_{X^{\prime}}}$$\scriptstyle{id\otimes
g_{2}}$$\scriptstyle{id\otimes g}$$\scriptstyle{id\otimes
p^{Y/Y^{\prime}}_{Y/Y_{2}^{\prime}}}$

and

${M\otimes X_{2}^{\prime}}$${M\otimes X_{1}^{\prime}}$${M\otimes
Y/Y_{2}^{\prime}}$${M\otimes Y/Y_{1}^{\prime}.}$$\scriptstyle{id\otimes
i^{X_{2}^{\prime}}_{X_{1}^{\prime}}}$$\scriptstyle{id\otimes
g_{2}}$$\scriptstyle{id\otimes g_{1}}$$\scriptstyle{id\otimes
p^{Y/Y_{1}^{\prime}}_{Y/Y_{2}^{\prime}}}$

Because $id\otimes i^{X_{2}^{\prime}}_{X^{\prime}}=i^{M\otimes
X_{2}^{\prime}}_{M\otimes X^{\prime}}$, $id\otimes
i^{X_{2}^{\prime}}_{X_{1}^{\prime}}=i^{M\otimes X_{2}^{\prime}}_{M\otimes
X_{1}^{\prime}}$, $id\otimes p^{Y/Y^{\prime}}_{Y/Y_{2}^{\prime}}=p^{M\otimes
Y/Y^{\prime}}_{M\otimes Y/Y_{2}^{\prime}}$ and $id\otimes
p^{Y/Y_{1}^{\prime}}_{Y/Y_{2}^{\prime}}=p^{M\otimes
Y/Y_{1}^{\prime}}_{M\otimes Y/Y_{2}^{\prime}}$, we obtain two commutative
diagrams:

${M\otimes X_{2}^{\prime}}$${M\otimes X^{\prime}}$${M\otimes
Y/Y_{2}^{\prime}}$${M\otimes Y/Y^{\prime}}$$\scriptstyle{i^{M\otimes
X_{2}^{\prime}}_{M\otimes X^{\prime}}}$$\scriptstyle{id\otimes
g_{2}}$$\scriptstyle{id\otimes g}$$\scriptstyle{p^{M\otimes
Y/Y^{\prime}}_{M\otimes Y/Y_{2}^{\prime}}}$

and

${M\otimes X_{2}^{\prime}}$${M\otimes X_{1}^{\prime}}$${M\otimes
Y/Y_{2}^{\prime}}$${M\otimes Y/Y_{1}^{\prime}.}$$\scriptstyle{i^{M\otimes
X_{2}^{\prime}}_{M\otimes X_{1}^{\prime}}}$$\scriptstyle{id\otimes
g_{2}}$$\scriptstyle{id\otimes g_{1}}$$\scriptstyle{p^{M\otimes
Y/Y_{1}^{\prime}}_{M\otimes Y/Y_{2}^{\prime}}}$

This means

$T(id\otimes g)=T(id\otimes g_{2})=T(id\otimes g_{1}).$

The claim has been proven, and similarly one can obtain that

$T(f\otimes id)=T(f_{1}\otimes id).$

Therefore,

$\displaystyle T(f\otimes g)$ $\displaystyle=T((f\otimes id)\circ(id\otimes
g))$ $\displaystyle=T(f\otimes id)\circ T(id\otimes g)$
$\displaystyle=T(f_{1}\otimes id)\circ T(id\otimes g_{1})$
$\displaystyle=T((f_{1}\otimes id)\circ(id\otimes g_{1}))$
$\displaystyle=T(f_{1}\otimes g_{1}).$

∎

Hence, we can give a definition of the tensor product of two morphisms in the
quotient category. The propositions above guarantee that the following
definition is well-defined.

######  Definition 4.6.

Let $\mathcal{A}$ be an abelian monoidal category with biexact tensor product,
$\mathcal{C}$ be a two-sided Serre tensor-ideal of $\mathcal{A}$. Define the
tensor product of objects in $\mathcal{A}/\mathcal{C}$ by the same tensor
product in $\mathcal{A}$, and define the tensor product of morphisms in
$\mathcal{A}/\mathcal{C}$ by

$\bar{f}\otimes\bar{g}:=T(f\otimes g)$

where $f,g$ are in direct systems, and $T$ is the canonical functor. It is
clear from the definition that

$Tf\otimes Tg=T(f\otimes g).$

Additionally, we define the associativity constraint in
$\mathcal{A}/\mathcal{C}$ by

$\bar{a}_{X,Y,Z}=Ta_{X,Y,Z}.$

By the way, it is also clear that the left and right unit isomorphisms in
$\mathcal{A}/\mathcal{C}$ are $Tl_{X}$ and $Tr_{X}$ where $l_{X}$ and $r_{X}$
are the left and right unit isomorphisms in $\mathcal{A}$.

Let’s consider an example of two-sided Serre tensor-ideal.

######  Example 4.7.

Let $I$ be a finite indexed set, $\mathcal{A}=\bigoplus\limits_{i\in
I}\mathrm{Mat}_{n_{i}}(\mathrm{Vec})$, where
$\mathrm{Mat}_{n_{i}}(\mathrm{Vec})$ is the category whose objects are
$n_{i}$-by-$n_{i}$ matrices of finite dimensional vector spaces. The tensor
product of two objects from distinct direct summands is defined to be zero.
One can observe that $\mathcal{A}$ is a multitensor category, and each
$\mathrm{Mat}_{n_{i}}(\mathrm{Vec})$ is a two-sided Serre tensor-ideal of
$\mathcal{A}$. In fact, let $\mathcal{C}=\mathrm{Mat}_{n_{i}}(\mathrm{Vec})$
to be such a two-sided Serre tensor-ideal, then
$\mathcal{A}/\mathcal{C}=\bigoplus\limits_{j\in
I\backslash\\{i\\}}\mathrm{Mat}_{n_{i}}(\mathrm{Vec})$

Next, we show that, with the above tensor product, the quotient category of a
monoidal category is also a monoidal category.

######  Proposition 4.8.

Let $\mathcal{A}$ be an abelian monoidal category with biexact tensor product,
$\mathcal{C}$ be a two-sided Serre tensor-ideal of $\mathcal{A}$, then
$\mathcal{A}/\mathcal{C}$ is a monoidal category.

###### Proof.

Consider the following pentagon axiom diagram of $\mathcal{A}$:

$\textstyle{((W\otimes X)\otimes Y)\otimes
Z\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{a_{W,X,Y}\otimes
id_{Z}}$$\scriptstyle{a_{W\otimes X,Y,Z}}$$\textstyle{(W\otimes(X\otimes
Y))\otimes
Z\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{a_{W,X\otimes
Y,Z}}$$\textstyle{(W\otimes X)\otimes(Y\otimes
Z)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{a_{W,X,Y\otimes
Z}}$$\textstyle{W\otimes((X\otimes Y)\otimes
Z)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{id_{W}\otimes
a_{X,Y,Z}}$$\textstyle{W\otimes(X\otimes(Y\otimes Z))}$

Applying the canonical functor $T$ gives that

$\textstyle{((W\otimes X)\otimes Y)\otimes
Z\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\bar{a}_{W,X,Y}\otimes
id_{Z}}$$\scriptstyle{\bar{a}_{W\otimes X,Y,Z}}$$\textstyle{(W\otimes(X\otimes
Y))\otimes
Z\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\bar{a}_{W,X\otimes
Y,Z}}$$\textstyle{(W\otimes X)\otimes(Y\otimes
Z)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\bar{a}_{W,X,Y\otimes
Z}}$$\textstyle{W\otimes((X\otimes Y)\otimes
Z)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{id_{W}\otimes\bar{a}_{X,Y,Z}}$$\textstyle{W\otimes(X\otimes(Y\otimes
Z)).}$

This is actually the pentagon axiom of $\mathcal{A}/\mathcal{C}$.

Consider the following triangle axiom diagram of $\mathcal{A}$:

$\textstyle{(X\otimes 1)\otimes
Y\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{a_{X,1,Y}}$$\scriptstyle{r_{X}\otimes
id_{Y}}$$\textstyle{X\otimes(1\otimes
Y)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{id_{X}\otimes
l_{Y}}$$\textstyle{X\otimes Y}$

Applying the canonical functor $T$ gives that

$\textstyle{(X\otimes 1)\otimes
Y\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\bar{a}_{X,1,Y}}$$\scriptstyle{Tr_{X}\otimes
id_{Y}}$$\textstyle{X\otimes(1\otimes
Y)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{id_{X}\otimes
Tl_{Y}}$$\textstyle{X\otimes Y}$.

Thus, $\mathcal{A}/\mathcal{C}$ is a monoidal category. ∎

Recall that $Tf\otimes Tg=T(f\otimes g)$, this implies that $T$ is a monoidal
functor. Now, we study the quotient category of a multiring category (resp. a
multitensor category) by a two-sided Serre tensor-ideal.

######  Proposition 4.9.

Let $\mathcal{A}$ be a multiring category, $\mathcal{C}$ be a two-sided Serre
tensor-ideal of $\mathcal{A}$. Then $\mathcal{A}/\mathcal{C}$ is a multiring
category.

###### Proof.

By Proposition 3.4 and Proposition 4.8, the quotient category of a locally
finite $k$-linear abelian monoidal category is a locally finite $k$-linear
abelian monoidal category. Therefore, it suffices to show that the tensor
product
$\otimes:\mathcal{A}/\mathcal{C}\times\mathcal{A}/\mathcal{C}\to\mathcal{A}/\mathcal{C}$
is bilinear and biexact.

Firstly, we show that the tensor product is bilinear. Note that the canonical
functor $T$ is linear, thus

$\displaystyle(\bar{f_{1}}+\bar{f_{2}})\otimes\bar{g}=(Tf_{1}+Tf_{2})\otimes
Tg=T(f_{1}+f_{2})\otimes Tg$ $\displaystyle=$ $\displaystyle
T((f_{1}+f_{2})\otimes g)=T(f_{1}\otimes g+f_{2}\otimes g)=T(f_{1}\otimes
g)+T(f_{2}\otimes g)$ $\displaystyle=$ $\displaystyle T(f_{1})\otimes
Tg+T(f_{2})\otimes Tg=\bar{f_{1}}\otimes\bar{g}+\bar{f_{2}}\otimes\bar{g}$

and similarly

$\displaystyle\bar{f}\otimes(\bar{g_{1}}+\bar{g_{2}})=Tf\otimes(Tg_{1}+Tg_{2})=Tf\otimes
T(g_{1}+g_{2})$ $\displaystyle=$ $\displaystyle
T(f\otimes(g_{1}+g_{2}))=T(f\otimes g_{1}+f\otimes g_{2})=T(f\otimes
g_{1})+T(f\otimes g_{2})$ $\displaystyle=$ $\displaystyle Tf\otimes
Tg_{1}+Tf\otimes Tg_{2}=\bar{f}\otimes\bar{g_{1}}+\bar{f}\otimes\bar{g_{2}}.$

Furthermore, for any $a\in k$, we have

$\displaystyle a\bar{f}\otimes\bar{g}=aTf\otimes Tg=T(af)\otimes
Tg=T(af\otimes g)$ $\displaystyle=$ $\displaystyle T(f\otimes ag)=Tf\otimes
T(ag)=Tf\otimes aTg=\bar{f}\otimes a\bar{g}.$

This implies
$\otimes:\mathcal{A}/\mathcal{C}\times\mathcal{A}/\mathcal{C}\to\mathcal{A}/\mathcal{C}$
is bilinear on morphisms.

Secondly, we show that the tensor product is biexact. Consider the following
exact sequence

$\textstyle{L\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\bar{f}}$$\textstyle{M\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\bar{g}}$$\textstyle{N.}$

Because we can write $\bar{g}=Tg$, $\bar{f}=Tf$, the exact sequence means that

$\mathrm{ker}(Tg)=\mathrm{Im}(Tf).$

By Lemma 3 in [4], it follows that

$T(\mathrm{ker}\ g)=T(\mathrm{Im}\ f).$

Because $T$ is an exact functor, the short exact sequence

${0}$${\mathrm{Im}f}$${\mathrm{ker}\ g}$${\mathrm{ker}\ g/\mathrm{Im}\
f}$${0}$$\scriptstyle{i}$$\scriptstyle{\pi}$

implies that

${0}$${T\mathrm{Im}f}$${T\mathrm{ker}\ g}$${T(\mathrm{ker}\ g/\mathrm{Im}\
f)}$${0}$$\scriptstyle{Ti}$$\scriptstyle{T\pi}$

is exact. Consequently,

$T(\mathrm{ker}(g)/\mathrm{Im}(f))\cong T(\mathrm{ker}\ g)/T(\mathrm{Im}\
f)=0\ \text{in}\ \mathcal{A}/\mathcal{C},$

which means $\mathrm{ker}(g)/\mathrm{Im}(f)\in\mathcal{C}$.

For any object $A$ in $\mathcal{A}$, since $A\otimes-$ is exact in
$\mathcal{A}$, we have the following short exact sequence

${0}$${A\otimes\mathrm{Im}f}$${A\otimes\mathrm{ker}\
g}$${A\otimes(\mathrm{ker}\ g/\mathrm{Im}f)}$${0.}$$\scriptstyle{id_{A}\otimes
i}$$\scriptstyle{id_{A}\otimes\pi}$

Note that there is a short exact sequence

${0}$${A\otimes\mathrm{Im}f}$${A\otimes\mathrm{ker}\
g}$${(A\otimes\mathrm{ker}\
g)/(A\otimes\mathrm{Im}f)}$${0.}$$\scriptstyle{id_{A}\otimes
i}$$\scriptstyle{\pi^{\prime}}$

Therefore,

$(A\otimes\mathrm{ker}\ g)/(A\otimes\mathrm{Im}f)\cong A\otimes(\mathrm{ker}\
g/\mathrm{Im}f)\in\mathcal{C}.$

Now, we claim that

$A\otimes\mathrm{ker}\ g=\mathrm{ker}(id_{A}\otimes\ g).$

Consider the following exact sequence

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathrm{ker}\
g\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{M\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{g}$$\textstyle{N,}$

applying $A\otimes-$ gives the following exact sequence

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{A\otimes\mathrm{ker}\
g\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{A\otimes
M\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{id_{A}\otimes
g}$$\textstyle{A\otimes N.}$

This indicates that

$A\otimes\mathrm{ker}\ g=\mathrm{ker}(id_{A}\otimes g).$

Besides, note that

$A\otimes\mathrm{Im}f=\mathrm{Im}(id_{A}\otimes f).$

It follows that

$\mathrm{ker}(id_{A}\otimes g)/\mathrm{Im}(id_{A}\otimes
f)=(A\otimes\mathrm{ker}\ g)/(A\otimes\mathrm{Im}f)\in\mathcal{C}.$

This implies that

$T(\mathrm{ker}(id_{A}\otimes g)/\mathrm{Im}(id_{A}\otimes f))=0$

i.e.

$T(\mathrm{ker}(id_{A}\otimes g))/T(\mathrm{Im}(id_{A}\otimes f))=0$

i.e.

$T(\mathrm{ker}(id_{A}\otimes g))=T(\mathrm{Im}(id_{A}\otimes f))$

By Lemma 3 in [4], this means

$\mathrm{ker}(T(id_{A}\otimes g))=\mathrm{Im}(T(id_{A}\otimes f))$

i.e.

$\mathrm{ker}(id_{A}\otimes Tg)=\mathrm{Im}(id_{A}\otimes Tf)$

i.e.

$\mathrm{ker}(id_{A}\otimes\bar{g})=\mathrm{Im}(id_{A}\otimes\bar{f}).$

This means $A\otimes-$ is exact in $\mathcal{A}/\mathcal{C}$. Similarly, one
can show $-\otimes A$ is exact in $\mathcal{A}/\mathcal{C}$. As a result, the
tensor product is biexact in $\mathcal{A}/\mathcal{C}$. Thus,
$\mathcal{A}/\mathcal{C}$ is a multiring category. ∎

######  Proposition 4.10.

Let $\mathcal{A}$ be a multitensor category, $\mathcal{C}$ be a two-sided
Serre tensor-ideal of $\mathcal{A}$, then $\mathcal{A}/\mathcal{C}$ is a
multitensor category.

###### Proof.

As proved in the above proposition, we know that $\mathcal{A}/\mathcal{C}$ is
a locally finite $k$-linear abelian monoidal category with bilinear tensor
product. Therefore, it suffices to show $\mathcal{A}/\mathcal{C}$ is rigid.

For any object $X$ in $\mathcal{A}/\mathcal{C}$, it has a left dual $X^{*}$,
which means there exist an evaluation $ev_{X}:X^{*}\otimes X\to 1$ and a
coevaluation $coev_{X}:1\to X\otimes X^{*}$ such that the compositions

${X}$${(X\otimes X^{*})\otimes X}$${X\otimes(X^{*}\otimes
X)}$${X,}$$\scriptstyle{coev_{X}\otimes
id_{X}}$$\scriptstyle{a_{X,X^{*},X}}$$\scriptstyle{id_{X}\otimes ev_{X}}$
${X^{*}}$${X^{*}\otimes(X\otimes X^{*})}$${(X^{*}\otimes X)\otimes
X^{*}}$${X^{*}}$$\scriptstyle{id_{X^{*}}\otimes
coev_{X}}$$\scriptstyle{a_{X^{*},X,X^{*}}^{-1}}$$\scriptstyle{ev_{X}\otimes
id_{X^{*}}}$

are the identity morphisms. Applying $T$ gives that the compositions

${X}$${(X\otimes X^{*})\otimes X}$${X\otimes(X^{*}\otimes
X)}$${X,}$$\scriptstyle{Tcoev_{X}\otimes
id_{X}}$$\scriptstyle{\bar{a}_{X,X^{*},X}}$$\scriptstyle{id_{X}\otimes
Tev_{X}}$ ${X^{*}}$${X^{*}\otimes(X\otimes X^{*})}$${(X^{*}\otimes X)\otimes
X^{*}}$${X^{*}}$$\scriptstyle{id_{X^{*}}\otimes
Tcoev_{X}}$$\scriptstyle{\bar{a}_{X^{*},X,X^{*}}^{-1}}$$\scriptstyle{Tev_{X}\otimes
id_{X^{*}}}$

are the identity morphisms. As a result, every object in
$\mathcal{A}/\mathcal{C}$ has a left dual. Similarly, one can show every
object in $\mathcal{A}/\mathcal{C}$ has a right dual. Thus,
$\mathcal{A}/\mathcal{C}$ is rigid. Consequently, $\mathcal{A}/\mathcal{C}$ is
a multitensor category. ∎

The following proposition shows that a two-sided Serre tensor-ideal of a
tensor category is always trivial.

######  Proposition 4.11.

Let $\mathcal{A}$ be a tensor category, $\mathcal{C}$ be a two-sided Serre
tensor-ideal of $\mathcal{A}$, then $\mathcal{C}$ is trivial.

###### Proof.

Suppose $\mathcal{C}$ is not zero, choose a non-zero object $B$ in
$\mathcal{C}$. Because $\mathcal{C}$ is a two-sided Serre tensor-ideal,
$B^{*}\otimes B\in\mathcal{C}$. Note that

$ev_{B}:B^{*}\otimes B\to 1$

is a non-zero epimorphism in $\mathcal{A}$, since $1$ is simple and the
composition $(id_{B}\otimes ev_{B})\circ a_{B,B^{*},B}\circ(coev_{B}\otimes
id_{B})$ is identity. Therefore, $1\in\mathcal{C}$. Consequently,
$\mathcal{A}=\mathcal{C}$. ∎

This proposition means that $\mathcal{A}/\mathcal{C}$ could only be zero or
$\mathcal{A}$ itself.

However, someone may be interested in other definitions on tensor structures
of $\mathcal{A}/\mathcal{C}$ in order to make $\mathcal{A}/\mathcal{C}$ be a
monoidal category. Interestingly, we will prove next that no matter how we
define the tensor structure, the canonical functor being monoidal implies the
Serre subcategory is trivial.

######  Proposition 4.12.

Let $\mathcal{A}$ be a tensor category, $\mathcal{C}$ be a Serre subcategory
of $\mathcal{A}$. Suppose the canonical functor $T$ is a monoidal functor,
then $\mathcal{C}$ is trivial.

###### Proof.

We prove it by contradiction. Assume $\mathcal{C}$ is non-trivial.

Choose a non-zero object $B$ in $\mathcal{C}$, we know that $TB=0$ in
$\mathcal{A}/\mathcal{C}$. This means $T(B^{*}\otimes B)\cong TB^{*}\otimes
TB=0$ in $\mathcal{A}/\mathcal{C}$. However,

$ev_{B}:B^{*}\otimes B\to 1$

is a non-zero epimorphism in $\mathcal{A}$ because $1$ is a simple object in
$\mathcal{A}$. Therefore,

$T(ev_{B}):T(B^{*}\otimes B)\cong TB^{*}\otimes TB=0\to T1$

is an epimorphism by Lemma 2.4. Since $T$ is a monoidal functor, $T1\not=0$ in
$\mathcal{A}/\mathcal{C}$. However, this contradicts $T(ev_{B})$ is an
epimorphism in $\mathcal{A}/\mathcal{C}$. ∎

Now, we consider the representation category of $k\mathbb{Z}_{2}$.

######  Example 4.13.

Consider $\mathbb{Z}_{2}=\\{1,g\\}$ and a field $k$ such that $\mathrm{char}\
k\nmid 2$, and denote the representation category of $k\mathbb{Z}_{2}$ by
$\mathcal{A}$. As is well known, $\mathcal{A}$ is semisimple and it has only
two irreducible representations $W_{1}=k(1-g)$ and $W_{2}=k(1+g)$. Define a
homomorphism by

$\displaystyle\varphi:$ $\displaystyle k(1-g)\otimes k(1-g)$
$\displaystyle\to$ $\displaystyle k(1+g)$ $\displaystyle a(1-g)\otimes(1-g)$
$\displaystyle\mapsto$ $\displaystyle a(1+g)$

where $a\in k$.

Let $\mathcal{C}$ be a Serre subcategory containing $W_{1}$. We now show that
if $\mathcal{A}/\mathcal{C}$ is a tensor category, then
$\mathcal{C}=\mathcal{A}$. Since $W_{1}\in\mathcal{C}$, we know that
$TW_{1}=0$ in $\mathcal{A}/\mathcal{C}$. It follows that $TW_{1}\otimes
TW_{1}\cong T(W_{1}\otimes W_{1})=T(k(1-g)\otimes k(1-g))=0$ in
$\mathcal{A}/\mathcal{C}$. This implies $W_{1}\otimes W_{1}\in\mathcal{C}$.
Consequently, $W_{2}=k(1+g)\in\mathcal{C}$ because $\varphi$ is a
$k\mathbb{Z}_{2}$-isomorphism. Thus, both $W_{1}$ and $W_{2}$ are in
$\mathcal{C}$, and $\mathcal{C}=\mathcal{A}$.

In order to make $\mathcal{A}/\mathcal{C}$ be a tensor category, it is clear
that $W_{2}=k(1+g)$ cannot be in $\mathcal{C}$, because $W_{2}=k(1+g)$ is the
unit object in $\mathcal{A}$.

We end our paper by showing that a two-sided Serre tensor-ideal of a multiring
category is actually a direct sum of some component subcategories. First of
all, recall that a multiring category $\mathcal{A}$ can be written as a direct
sum of its component subcategories $\mathcal{A}=\bigoplus\limits_{i,j\in
I}\mathcal{A}_{i,j}$, where $\mathcal{A}_{i,j}=1_{i}\otimes\mathcal{A}\otimes
1_{j}$ and $I$ is an indexed set such that $1=\bigoplus\limits_{i\in I}1_{i}$.

######  Lemma 4.14.

Let $\mathcal{A}$ be a multiring category with left duals, $\mathcal{C}$ be a
two-sided Serre tensor-ideal of $\mathcal{A}$, then
$\mathcal{C}\cap\mathcal{A}_{i,j}$ is either $\mathcal{A}_{i,j}$ or $0$, where
$\mathcal{A}_{i,j}$ is a component subcategory of $\mathcal{A}$.

###### Proof.

Suppose $\mathcal{C}\cap\mathcal{A}_{i,j}\not=0$. Choose $X\not=0$ in
$\mathcal{C}\cap\mathcal{A}_{i,j}$, we know that $X=1_{i}\otimes X\otimes
1_{j}$. Because $X^{*}=(1_{i}\otimes X\otimes 1_{j})^{*}=1_{j}^{*}\otimes
X^{*}\otimes 1_{i}^{*}=1_{j}\otimes X^{*}\otimes 1_{i}$, one can observe that

$(1_{i}\otimes X\otimes 1_{j})^{*}\otimes 1_{i}\otimes X\otimes
1_{j}\in\mathcal{C}\cap\mathcal{A}_{j,j}.$

We claim that $\mathrm{Im}(ev_{X})=1_{j}$. Assume $1_{k}$ is a direct summand
of $\mathrm{Im}(ev_{X})$, where $k\not=j$. Consider the exact sequence

${X^{*}\otimes X}$${\mathrm{Im}(ev_{X})}$${0.}$$\scriptstyle{ev_{X}}$

Tensoring this sequence with $1_{k}$ on the left, we obtain an exact sequence

${1_{k}\otimes X^{*}\otimes X}$${1_{k}}$${0.}$$\scriptstyle{id_{1_{k}}\otimes
ev_{X}}$

Since

$1_{k}\otimes X^{*}\otimes X=1_{k}\otimes 1_{j}\otimes X^{*}\otimes
1_{i}\otimes 1_{i}\otimes X\otimes 1_{j}=0,$

the above exact sequence means $1_{k}=0$, which is absurd. Therefore, $1_{k}$
is not a direct summand of $\mathrm{Im}(ev_{X})$. Because $ev_{X}:X^{*}\otimes
X\to 1$ is non-zero, we get that $\mathrm{Im}(ev_{X})=1_{j}$. Consequently,
$1_{j}\in\mathcal{C}$. This implies $\mathcal{A}_{i,j}\subset\mathcal{C}$, and
thus $\mathcal{C}\cap\mathcal{A}_{i,j}=\mathcal{A}_{i,j}$. ∎

In fact, if $\mathcal{A}_{i,j}\subset\mathcal{C}$, one can know
$1_{j}\in\mathcal{C}$ from the above proof. As a result,
$\mathcal{A}_{l,j}\subset\mathcal{C}$ and
$\mathcal{A}_{j,l}\subset\mathcal{C}$ for all $l\in I$. Similar to the process
of the above proof, if we consider

$coev_{X}:1\to X\otimes X^{*}=1_{i}\otimes X\otimes 1_{j}\otimes 1_{j}\otimes
X^{*}\otimes 1_{i},$

then we can obtain $1_{i}\in\mathcal{C}$. Consequently,
$\mathcal{A}_{l,i}\subset\mathcal{C}$ and
$\mathcal{A}_{i,l}\subset\mathcal{C}$ for all $l\in I$.

In particular, if $0\not=\mathcal{A}_{i,j}\subset\mathcal{C}$, then
$0\not=\mathcal{A}_{i,i}\subset\mathcal{C}$,
$0\not=\mathcal{A}_{j,j}\subset\mathcal{C}$, and
$0\not=\mathcal{A}_{j,i}\subset\mathcal{C}$. Besides,

$\displaystyle\mathcal{C}$ $\displaystyle=\mathcal{C}\cap\mathcal{A}$
$\displaystyle=\mathcal{C}\cap\bigoplus_{i,j\in I}\mathcal{A}_{i,j}$
$\displaystyle=\bigoplus_{i,j\in I}(\mathcal{C}\cap\mathcal{A}_{i,j}).$

It follows from the above Lemma that $\mathcal{C}$ is a direct sum of
$\mathcal{A}_{i,j}$’s.

The following proposition provides a deeper understanding for two-sided Serre
tensor-ideal.

######  Proposition 4.15.

Let $\mathcal{A}$ be a multiring category with left duals, $\mathcal{C}$ be a
two-sided Serre tensor-ideal of $\mathcal{A}$. Let $J=\\{i\in
I|1_{i}\in\mathcal{C}\\}$, then $\mathcal{A}_{i,k}=0$, $\mathcal{A}_{k,i}=0$
for all $k\notin J$, $i\in J$.

###### Proof.

Assume $\mathcal{A}_{i,k}\not=0$ for a given $i\in J$ and $k\notin J$. Since
$1_{i}\in\mathcal{C}$, we know that $\mathcal{A}_{i,k}\subset\mathcal{C}$. For
any $0\not=X\in\mathcal{A}_{i,k}$, we can obtain
$\mathrm{Im}(ev_{X})=1_{k}\in\mathcal{C}$ from a process similar to the proof
of Lemma 4.14. This contradicts $k\notin J$. Thus, $\mathcal{A}_{i,k}=0$.
Similarly, $\mathcal{A}_{k,i}=0$. ∎

Let $\mathcal{A}$ be a multiring category with left duals, the above
proposition implies that a two-sided Serre tensor-ideal $\mathcal{C}$ of
$\mathcal{A}$ can be written as $\mathcal{C}=\bigoplus\limits_{i,j\in
J}\mathcal{A}_{i,j}$, where $J=\\{i\in I|1_{i}\in\mathcal{C}\\}$. Furthermore,
for any $0\not=X\in\mathcal{A}$, the above proposition indicates that

$\displaystyle X$ $\displaystyle=\bigoplus_{i,j\in I}(1_{i}\otimes X\otimes
1_{j})$ $\displaystyle=\bigoplus_{i,j\notin J}(1_{i}\otimes X\otimes
1_{j})\bigoplus\bigoplus_{i,j\in J}(1_{i}\otimes X\otimes 1_{j}).$

Let

$X^{\prime}=\bigoplus_{i,j\notin J}(1_{i}\otimes X\otimes 1_{j})\text{ and
}X^{\prime\prime}=\bigoplus_{i,j\in J}(1_{i}\otimes X\otimes 1_{j}),$

it is clear that $X^{\prime\prime}\in\mathcal{C}$ and $X^{\prime}\cong X$ in
$\mathcal{A}/\mathcal{C}$. Therefore,
$\mathcal{A}/\mathcal{C}\cong\bigoplus\limits_{i,j\notin J}\mathcal{A}_{i,j}$.

Conversely, given a suitable subset $J$ of $I$, is $\bigoplus\limits_{i,j\in
J}\mathcal{A}_{i,j}$ a two-sided Serre tensor-ideal? We show next that the
answer is yes.

######  Proposition 4.16.

Let $\mathcal{A}$ be a multiring category with left duals, $J\subset I$
satisfying that $\mathcal{A}_{i,k}=0$, $\mathcal{A}_{k,i}=0$ for all $k\notin
J$, $i\in J$. Then $\mathcal{C}=\bigoplus\limits_{i,j\in J}\mathcal{A}_{i,j}$
is a two-sided Serre tensor-ideal of $\mathcal{A}$.

###### Proof.

It suffices to show $\mathcal{C}$ is a Serre subcategory. For any
$X\in\mathcal{C}$, let $Y$ be a subobject of $X$. We know that

${0}$${Y}$${X}$

is exact. For $p\notin J$ or $q\notin J$, tensoring $1_{p}$ and $1_{q}$ on the
left and right respectively gives that

${0}$${1_{p}\otimes Y\otimes 1_{q}}$${1_{p}\otimes X\otimes 1_{q}}$

is exact. Because $1_{p}\otimes X\otimes 1_{q}$ is $0$, we obtain
$1_{p}\otimes Y\otimes 1_{q}=0$. This means $Y\in\mathcal{C}$. Hence,
$\mathcal{C}$ is closed under taking subobjects. Similarly, $\mathcal{C}$ is
closed under taking quotient objects. Now, suppose there is an exact sequence

${0}$${X}$${Y}$${Z}$${0}$

where $X,Z\in\mathcal{C}$. For $p\notin J$ or $q\notin J$, tensoring $1_{p}$
and $1_{q}$ on the left and right respectively gives that

${0}$${1_{p}\otimes X\otimes 1_{q}}$${1_{p}\otimes Y\otimes
1_{q}}$${1_{p}\otimes Z\otimes 1_{q}}$${0}$

is exact. Because $1_{p}\otimes X\otimes 1_{q}=0=1_{p}\otimes Z\otimes 1_{q}$,
we obtain that $1_{p}\otimes Y\otimes 1_{q}=0$. This means $Y\in\mathcal{C}$.
Hence, $\mathcal{C}$ is closed under taking extensions. It follows that
$\mathcal{C}$ is a Serre subcategory. ∎

In summary, on one hand for $J\subset I$ satisfying that
$\mathcal{A}_{i,k}=0$, $\mathcal{A}_{k,i}=0$ for all $k\notin J$, $i\in J$,
$\mathcal{C}=\bigoplus\limits_{i,j\in J}\mathcal{A}_{i,j}$ is a two-sided
Serre tensor-ideal of $\mathcal{A}$. On the other hand, every two-sided Serre
tensor-ideal of $\mathcal{A}$ can be written as
$\mathcal{C}=\bigoplus\limits_{i,j\in J}\mathcal{A}_{i,j}$ for some $J\subset
I$.

Consider the two-sided Serre tensor-ideal
$\mathcal{C}=\bigoplus\limits_{i,j\in J}\mathcal{A}_{i,j}$ and the
corresponding quotient category
$\mathcal{A}/\mathcal{C}=\bigoplus\limits_{i,j\notin J}\mathcal{A}_{i,j}$. It
is easy to see that
$T(1_{\mathcal{A}})=T(\bigoplus_{i}1_{i})=\bigoplus\limits_{i\notin J}1_{i}$,
where $T$ is the canonical functor. Since $T$ is a monoidal functor,
$1_{\mathcal{A}/\mathcal{C}}=T(1_{\mathcal{A}})=\bigoplus\limits_{i\notin
J}1_{i}$. In fact, for $i,j\notin J$, $\mathcal{A}_{i,j}$ is a component
subcategory of $\mathcal{A}/\mathcal{C}$.

Now, we claim that the image of another two-sided Serre tensor-ideal
$\mathcal{C}^{\prime}$ is a two-sided Serre tensor-ideal of the quotient
category. Let $\mathcal{C}=\bigoplus\limits_{i,j\in J}\mathcal{A}_{i,j}$,
$\mathcal{C}^{\prime}=\bigoplus\limits_{i,j\in J^{\prime}}\mathcal{A}_{i,j}$
be two two-sided Serre tensor-ideals of $\mathcal{A}$. Because
$\mathcal{A}/\mathcal{C}=\bigoplus\limits_{i,j\notin J}\mathcal{A}_{i,j}$,
$T(\mathcal{C}^{\prime})\cong\bigoplus\limits_{i,j\in J^{\prime}\backslash
J}\mathcal{A}_{i,j}$ in $\mathcal{A}/\mathcal{C}$. We know that
$\mathcal{A}_{i,k}=0$, $\mathcal{A}_{k,i}=0$ for all $k\notin J^{\prime}$,
$i\in J^{\prime}$. Hence $\mathcal{A}_{i,k}=0$, $\mathcal{A}_{k,i}=0$ for all
$k\notin J^{\prime}\backslash J$, $i\in J^{\prime}\backslash J$. This means
the image of $\mathcal{C}^{\prime}$ is a two-sided Serre tensor-ideal of
$\mathcal{A}/\mathcal{C}$.

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