# On Structures in Arrow Categories

Paulina L. A. Goedicke Institute for Theoretical Physics University of Cologne
<EMAIL_ADDRESS>Jamie Vicary Department of Computer Science
University of Cambridge<EMAIL_ADDRESS>
###### Abstract

In this article we investigate which categorical structures of a category
$\mathcal{C}$ are inherited by its arrow category. In particular, we show that
a monoidal equivalence between two categories gives rise to a monoidal
equivalence between their arrow categories. Moreover, we examine under which
circumstances an arrow category is rigid and pivotal. Finally, we derive what
the (co)algebra, bialgebra and Hopf algebra objects are in an arrow category.

###### Contents

  1. 1 Introduction
  2. 2 Structures in Arrow Categories
    1. 2.1 Functors in Arrow Categories
    2. 2.2 Monoidal Products and Braidings in Arrow Categories
    3. 2.3 Duals and Pivots in Arrow Categories
    4. 2.4 (Co)monoids, Bialgebras, Frobenius Structures and Hopf Algebras in Arrow Categories
  3. A Proofs

##  1 Introduction

Given a category $\mathcal{C}$, the category of arrows of $\mathcal{C}$ is a
very fundamental concept in category theory [3]. This article is a revision of
a chapter from the first author’s master’s thesis111The thesis was submitted
in October 2021 at the Department of Mathematics at University of Hamburg.,
investigating which categorical structures an arrow category inherits from its
underlying category $\mathcal{C}$. We start by discussing functors and natural
transformations in Section 2.1 and show that an equivalence between two
categories gives rise to an equivalence between their arrow categories. In
Section 2.2 we then extend this to braided monoidal categories and functors.
In Section 2.3 we prove that the arrow category of a rigid monoidal category
restricted to objects that are isomorphisms is also rigid. We further show
that in that case, if the underlying category is in addition pivotal, its
arrow category is also pivotal. In fact, it is a ribbon category. In Section
2.4 we then discuss (co)monoids, bialgebras, Frobenius structures and Hopf
algebras in arrow categories.

There already have been definitions of monoidal products in arrow categories
[4] and the rest of our results also seem to be quite fundamental, however we
still believe them to have some novelty.

We assume familiarity with the basic concepts of category theory, the
graphical calculus and quantum groups and refer to [5], [3] and [2] for an
introduction. Here we will only briefly review the definition of an arrow
category:

######  Definition 1.1.

([3], p. 24-25) Let $\mathcal{C}$ be a category. The arrow category of
$\mathcal{C}$ Arr($\mathcal{C}$) is defined as follows:

  * •

objects are triples ($A$, $B$, $h$) where $A$, $B$ $\in$ obj($\mathcal{C}$)
and $h:A\longrightarrow B$.

  * •

morphisms $\phi:(A,B,h)\longrightarrow(A^{\prime},B^{\prime},h^{\prime})$ are
pairs ($\phi_{A}$, $\phi_{B}$) of morphims $\phi_{A}:A\longrightarrow
A^{\prime}$ and $\phi_{B}:B\longrightarrow B^{\prime}$ in $\mathcal{C}$ such
that the following diagram commutes:

${A}$${A^{\prime}}$${B}$${B^{\prime}}$$\scriptstyle{h}$$\scriptstyle{\phi_{A}}$$\scriptstyle{h^{\prime}}$$\scriptstyle{\phi_{B}}$

######  Example 1.2.

The arrow category of Mat($\mathbb{N}$) has $b\times v$-matrices $\chi$ as
objects, where $b$ and $v$ are natural numbers. A morphism $S:M\longrightarrow
N$ between two matrices $M:v\longrightarrow b$ and
$N:v^{\prime}\longrightarrow b^{\prime}$ is given by a pair of matrices
$(S_{b},S_{v}):(v,b)\longrightarrow(v^{\prime},b^{\prime})$ such that the
following diagram commutes:

${b}$${b^{\prime}}$${v}$${v^{\prime}}$$\scriptstyle{S_{b}}$$\scriptstyle{M}$$\scriptstyle{N}$$\scriptstyle{S_{v}}$

##  2 Structures in Arrow Categories

We will now discuss what kind of categorical structures one can define in
arrow categories.

###  2.1 Functors in Arrow Categories

In the following we will show that a functor between two categories gives rise
to a functor between their arrow categories and prove a similar statement for
natural transformations.

######  Proposition 2.1.

Let $\mathcal{C}$ and $\mathcal{D}$ be two categories and let
$\mathcal{F}:\mathcal{C}\longrightarrow\mathcal{D}$ be a covariant functor
between those categories. Then $\mathcal{F}$ gives rise to a covariant functor
$\tilde{\mathcal{F}}$ between Arr($\mathcal{C}$) and Arr($\mathcal{D}$).

###### Proof.

Given a functor $\mathcal{F}:\mathcal{C}\longrightarrow\mathcal{D}$ that
assigns to every object $A$ in $\mathcal{C}$ an object $\mathcal{F}(A)$ in
$\mathcal{D}$ and to every morphism $f:A\longrightarrow B$ in $\mathcal{C}$ a
morphism $\mathcal{F}(f):\mathcal{F}(A)\longrightarrow\mathcal{F}(B)$ in
$\mathcal{D}$, we can define a functor $\tilde{\mathcal{F}}$ that maps every
object $f:A\longrightarrow B$ in Arr($\mathcal{C}$) to an object
$\tilde{\mathcal{F}}(f)=\mathcal{F}(f):\mathcal{F}(A)\longrightarrow\mathcal{F}(B)$
in Arr($\mathcal{D}$) and every morphism $(\phi,\psi):f\longrightarrow
f^{\prime}$ in Arr($\mathcal{C}$) to a morphism
$\tilde{\mathcal{F}}(\phi,\psi)=(\mathcal{F}(\phi),\mathcal{F}(\psi)):\mathcal{F}(f)\longrightarrow\mathcal{F}(f^{\prime})$
in Arr($\mathcal{D}$). This is valid because the diagram

${\mathcal{F}(A)}$${\mathcal{F}(A^{\prime})}$${\mathcal{F}(B)}$${\mathcal{F}(B^{\prime})}$$\scriptstyle{\mathcal{F}(\phi)}$$\scriptstyle{\mathcal{F}(f)}$$\scriptstyle{\mathcal{F}(f^{\prime})}$$\scriptstyle{\mathcal{F}(\psi)}$

commutes due to functoriality of $\mathcal{F}$. It is easy to show that
$\tilde{\mathcal{F}}$ preserves composition and the identity morphism in
Arr($\mathcal{C}$). ∎

Similarly, one can prove the following statement:

######  Proposition 2.2.

A contravariant functor $\mathcal{F}:\mathcal{C}\longrightarrow\mathcal{D}$
gives rise to a contravariant functor
$\tilde{\mathcal{F}}:\text{Arr}(\mathcal{C})\longrightarrow\text{Arr}(\mathcal{D})$.

######  Example 2.3.

The contravariant functor
$\mathcal{T}:\mathbf{Mat}(\mathbb{N})\longrightarrow\mathbf{Mat}(\mathbb{N})$
which maps each set $v$ to itself and each matrix $Mi:v\longrightarrow b$ to
its transpose $M^{T}:b\longrightarrow v$ gives rise to a contravariant functor
$\tilde{\mathcal{T}}:\mathrm{Arr}(\mathbf{Mat}(\mathbb{N}))\longrightarrow\mathrm{Arr}(\mathbf{Mat}(\mathbb{N}))$
which maps each matrix $M:v\longrightarrow b$ to its transpose
$M^{T}:b\longrightarrow v$ and each morphism $(S_{v},S_{b}):M\longrightarrow
N$ to its transpose $(S_{v}^{T},S_{b}^{T}):N^{T}\longrightarrow M^{T}$ such
that the following diagram commutes:

${b^{\prime}}$${b}$${v^{\prime}}$${v}$$\scriptstyle{S_{v}^{T}}$$\scriptstyle{N^{T}}$$\scriptstyle{M^{T}}$$\scriptstyle{S^{T}_{v}}$

######  Proposition 2.4.

Let $\mathcal{C}$ and $\mathcal{D}$ be two categories and let
$\mathcal{F},\mathcal{G}:\mathcal{C}\longrightarrow\mathcal{D}$ be two
covariant functors between those categories that induce the functors
$\tilde{\mathcal{F}},\tilde{\mathcal{G}}:\mathrm{Arr}(\mathcal{C})\longrightarrow\mathrm{Arr}(\mathcal{D})$.
Consider a natural transformation $\eta:\mathcal{F}\Longrightarrow\mathcal{G}$
between $\mathcal{F}$ and $\mathcal{G}$. Then $\eta$ induces a natural
transformation
$\tilde{\eta}:\tilde{\mathcal{F}}\Longrightarrow\tilde{\mathcal{G}}$.

###### Proof.

Let $\eta:\mathcal{F}\Longrightarrow\mathcal{G}$ be a natural transformation
that assigns to every object $A$ in $\mathcal{C}$ a morphism
$\eta_{A}:\mathcal{F}(A)\longrightarrow\mathcal{G}(A)$ such that for any
morphism $f:A\longrightarrow B$ in $\mathcal{C}$ the following diagram
(naturality condition) commutes:

${\mathcal{F}(A)}$${\mathcal{G}(A)}$${\mathcal{F}(B)}$${\mathcal{G}(B)}$$\scriptstyle{\eta_{A}}$$\scriptstyle{\mathcal{F}(f)}$$\scriptstyle{\mathcal{G}(f)}$$\scriptstyle{\eta_{B}}$

Using this, once can assign to every object $f:A\longrightarrow B$ in
Arr($\mathcal{C}$) a morphism
$\tilde{\eta}_{f}=(\eta_{A},\eta_{B}):\tilde{\mathcal{F}}(f)\longrightarrow\tilde{\mathcal{G}}(f)$,
such that for any morphism
$(\phi,\psi):f\longrightarrow f^{\prime}$ in Arr($\mathcal{C}$), where
$f^{\prime}:A^{\prime}\longrightarrow B^{\prime}$, the following diagram
(naturality condition in the arrow category) commutes:

${\mathcal{F}(A)}$${\mathcal{G}(A)}$${\mathcal{F}(A^{\prime})}$${\mathcal{G}(A^{\prime})}$${\mathcal{F}(B)}$${\mathcal{G}(B)}$${\mathcal{F}(B^{\prime})}$${\mathcal{G}(B^{\prime})}$$\scriptstyle{\mathcal{F}(\phi)}$$\scriptstyle{\eta_{A}}$$\scriptstyle{\mathcal{F}(f)}$$\scriptstyle{\mathcal{G}(\phi)}$$\scriptstyle{\mathcal{G}(f)}$$\scriptstyle{\eta_{A^{\prime}}}$$\scriptstyle{\mathcal{F}(f^{\prime})}$$\scriptstyle{\mathcal{G}(f^{\prime})}$$\scriptstyle{\mathcal{F}(\psi)}$$\scriptstyle{\eta_{B}}$$\scriptstyle{\mathcal{G}(\psi)}$$\scriptstyle{\eta_{B^{\prime}}}$

Here the the top, the back, the front and the bottom face commute due to
naturality of $\eta$ and the two side faces commute by definition. Hence the
whole diagram commutes and we have defined a natural transformation
$\tilde{\eta}:\tilde{\mathcal{F}}\Longrightarrow\tilde{\mathcal{G}}$. ∎

It is straightforward to verify that the following has to hold:

######  Proposition 2.5.

If $\eta:\mathcal{F}\Longrightarrow\mathcal{G}$ is a natural isomorphism, so
is $\tilde{\eta}:\tilde{\mathcal{F}}\Longrightarrow\tilde{\mathcal{G}}$.

######  Theorem 2.6.

Let $\mathcal{C}$ and $\mathcal{D}$ be two equivalent categories, i. e. there
exist a pair of functors $\mathcal{F}:\mathcal{C}\longrightarrow\mathcal{D}$
and $\mathcal{G}:\mathcal{D}\longrightarrow\mathcal{C}$ and natural
isomorphisms $\mathcal{F}\circ\mathcal{G}\cong\mathrm{id}_{\mathcal{D}}$ and
$\mathcal{G}\circ\mathcal{F}\cong\mathrm{id}_{\mathcal{C}}$. Then
Arr($\mathcal{C}$) and Arr($\mathcal{D}$) are also equivalent.

###### Proof.

By Proposition 2.1 the functors
$\mathcal{F}:\mathcal{C}\longrightarrow\mathcal{D}$ and
$\mathcal{G}:\mathcal{D}\longrightarrow\mathcal{C}$ give rise to functors
$\tilde{\mathcal{F}}:\mathrm{Arr}(\mathcal{C})\longrightarrow\mathrm{Arr}(\mathcal{D})$
and
$\tilde{\mathcal{G}}:\mathrm{Arr}(\mathcal{D})\longrightarrow\mathrm{Arr}(\mathcal{C})$.
From Proposition 2.5 we know that the natural isomorphisms
$\mathcal{F}\circ\mathcal{G}\cong\mathrm{id}_{\mathcal{D}}$ and
$\mathcal{G}\circ\mathcal{F}\cong\mathrm{id}_{\mathcal{C}}$ give rise to
natural isomorphisms
$\tilde{\mathcal{F}}\circ\tilde{\mathcal{G}}\cong\mathrm{id}_{\mathrm{Arr}(\mathcal{D})}$
and
$\tilde{\mathcal{G}}\circ\tilde{\mathcal{F}}\cong\mathrm{id}_{\mathrm{Arr}(\mathcal{C})}$.
Hence we have an equivalence. ∎

With that, we can define a dagger structure on Arr($\mathcal{C}$), using the
dagger structure in $\mathcal{C}$:

######  Proposition 2.7.

Let $\mathcal{C}$ be a dagger category and let $\mathcal{C}_{uni}$ be the
subcategory where all morphisms are unitary, i. e. all
$f\in\mathrm{Hom}_{\mathcal{C}}$ are invertible with $f^{-1}=f^{\dagger}$.
Then $\mathrm{Arr}(\mathcal{C}_{uni})$ is a dagger category.

###### Proof.

Let $\mathcal{C}$ be a dagger category, i. e. there exists an involutive
contravariant functor $\dagger:\mathcal{C}\longrightarrow\mathcal{C}$ such
that

$\displaystyle(g\circ f)^{\dagger}=f^{\dagger}\circ g^{\dagger}$
$\displaystyle\mathrm{id}^{\dagger}=\mathrm{id}$
$\displaystyle(f^{\dagger})^{\dagger}=f.$

Restricting to the subcategory where all morphisms are unitary, i. e. we have
for each morphism $\psi:A\longrightarrow A^{\prime}$ in $\mathcal{C}$:

$\displaystyle\psi\circ\psi^{\dagger}=\mathrm{id}_{A^{\prime}}\ \ \
\text{and}$ (1) $\displaystyle\psi^{\dagger}\circ\psi=\mathrm{id}_{A}$ (2)

and taking the arrow category of $\mathcal{C}$, we can construct a functor
$\mathrm{Arr}^{\dagger}:\mathrm{Arr}(\mathcal{C}_{uni})\longrightarrow\mathrm{Arr}(\mathcal{C}_{uni})$
which sends each object $f:A\longrightarrow B$ in
$\mathrm{Arr}(\mathcal{C}_{uni})$ to itself and each morphism

${A}$${A^{\prime}}$${B}$${B^{\prime}}$$\scriptstyle{\psi}$$\scriptstyle{f}$$\scriptstyle{g}$$\scriptstyle{\psi^{\prime}}$

to its adjoint
$(\psi,\psi^{\prime})^{\dagger}=(\psi^{\dagger},\psi^{\prime\dagger})$:

${A^{\prime}}$${A}$${B^{\prime}}$${B}$$\scriptstyle{\psi^{\dagger}}$$\scriptstyle{f}$$\scriptstyle{g}$$\scriptstyle{\psi^{\prime\dagger}}$

The above diagram commutes because we have $\psi^{\prime}\circ f=g\circ\psi$
per definition and $\psi\circ\psi^{\dagger}=\mathrm{id}_{B^{\prime}}$. It is
easy to verify that this construction fulfils the requirements for a dagger
functor. ∎

###  2.2 Monoidal Products and Braidings in Arrow Categories

In this section we will define a braiding in arrow categories and demonstrate
that a monoidal functor between two categories induces a monoidal functor
between their arrow categories. Moreover, we will show that, given a symmetric
monoidal category $\mathcal{C}$, its arrow category is also symmetric.

The following statement can be found in a similar manner in [4]. We will still
give its proof.

######  Proposition 2.8.

If $\mathcal{C}$ is a monoidal category, we can use the monoidal product in
$\mathcal{C}$ to define a pointwise monoidal product in Arr($\mathcal{C}$).

###### Proof.

Let
$(\mathcal{C},\otimes_{\mathcal{C}},\mathbb{I}_{\mathcal{C}},\lambda_{\mathcal{C}},\rho_{\mathcal{C}})$
be a monoidal category and Arr($\mathcal{C}$) its arrow category. A monoidal
structure in Arr($\mathcal{C}$) can be defined as follows:

  * •

On objects we have

$\displaystyle(f:A_{1}\longrightarrow
B_{1})\otimes_{\mathrm{Arr}(\mathcal{C})}(g:A_{2}\longrightarrow
B_{2}):=f\otimes_{\mathcal{C}}g:A_{1}\otimes_{\mathcal{C}}A_{2}\longrightarrow
B_{1}\otimes_{\mathcal{C}}B_{2}$

where $f:A_{1}\longrightarrow B_{1}$ and $g:A_{2}\longrightarrow B_{2}$ are
objects in Arr($\mathcal{C}$).

  * •

On morphisms $(\phi_{A_{1}},\phi_{B_{1}})$ and $(\phi_{A_{2}},\phi_{B_{2}})$,
where $\phi_{A_{i}}:A_{i}\longrightarrow A_{i}^{\prime}$ and
$\phi_{B_{i}}:B_{i}\longrightarrow B_{i}^{\prime}$ for $i=1,2$, we have

$\displaystyle(\phi_{A_{1}},\phi_{B_{1}})\otimes_{\mathrm{Arr}(\mathcal{C})}(\phi_{A_{2}},\phi_{B_{2}}):=(\phi_{A_{1}}\otimes_{\mathcal{C}}\phi_{A_{2}},\phi_{B_{1}}\otimes_{\mathcal{C}}\phi_{B_{2}})$
(3)

such that the following diagram commutes:

${A_{1}\otimes_{\mathcal{C}}A_{2}}$${B_{1}\otimes_{\mathcal{C}}B_{2}}$${A_{1}^{\prime}\otimes_{\mathcal{C}}A_{2}^{\prime}}$${B_{1}^{\prime}\otimes_{\mathcal{C}}B_{2}^{\prime}}$$\scriptstyle{f\otimes_{\mathcal{C}}g}$$\scriptstyle{\phi_{A_{1}}\otimes_{\mathcal{C}}\phi_{A_{2}}}$$\scriptstyle{\phi_{B_{1}^{\prime}}\otimes_{\mathcal{C}}\phi_{B_{2}^{\prime}}}$$\scriptstyle{f^{\prime}\otimes_{\mathcal{C}}g^{\prime}}$

  * •

The unit object is given by the identity morphism on the monoidal unit in
$\mathcal{C}$:

$\displaystyle\mathrm{id}_{\mathbb{I}_{\mathcal{C}}}:\mathbb{I}_{\mathcal{C}}\longrightarrow\mathbb{I}_{\mathcal{C}}.$
(4)

  * •

Since left- and right-unitors are natural isomorphisms one can use Propisition
2.5 to define

$\displaystyle\lambda_{f}:\mathrm{id}_{\mathbb{I}_{\mathcal{C}}}\otimes_{\mathcal{C}}f\longrightarrow
f$ (5)

and

$\displaystyle\rho_{f}:f\otimes_{\mathcal{C}}\mathrm{id}_{\mathbf{I}_{\mathcal{C}}}\longrightarrow
f$ (6)

where $f:A\longrightarrow B$, i. e. the following diagrams commute:

${A\otimes_{\mathcal{C}}\mathbb{I}_{\mathcal{C}}}$${A}$${B\otimes_{\mathcal{C}}\mathbb{I}_{\mathcal{C}}}$${B}$$\scriptstyle{\rho_{A}}$$\scriptstyle{f\otimes_{\mathcal{C}}\mathrm{id}_{\mathbb{I}_{\mathcal{C}}}}$$\scriptstyle{f}$$\scriptstyle{\rho_{B}}$

${\mathbb{I}_{\mathcal{C}}\otimes_{\mathcal{C}}A}$${A}$${\mathbb{I}_{\mathcal{C}}\otimes_{\mathcal{C}}B}$${B}$$\scriptstyle{\lambda_{A}}$$\scriptstyle{\mathrm{id}_{\mathbb{I}_{\mathcal{C}}}\otimes_{\mathcal{C}}f}$$\scriptstyle{f}$$\scriptstyle{\lambda_{B}}$

  * •

Finally, the associator is a natural isomorphism

$\displaystyle\alpha:(f_{1}\otimes_{\mathcal{C}}f_{2})\otimes_{\mathcal{C}}f_{3}\longrightarrow
f_{1}\otimes_{\mathcal{C}}(f_{2}\otimes_{\mathcal{C}}f_{3})$ (7)

where $f_{i}:A_{i}\longrightarrow B_{i}$ for $i=1,2,3$, such that the diagram

${(A_{1}\otimes_{\mathcal{C}}A_{2})\otimes_{\mathcal{C}}A_{3}}$${A_{1}\otimes_{\mathcal{C}}(A_{2}\otimes_{\mathcal{C}}A_{3})}$${(B_{1}\otimes_{\mathcal{C}}B_{2})\otimes_{\mathcal{C}}B_{3}}$${B_{1}\otimes_{\mathcal{C}}(B_{2}\otimes_{\mathcal{C}}B_{3})}$$\scriptstyle{\alpha_{A_{1},A_{2},A_{3}}}$$\scriptstyle{(f_{1}\otimes_{\mathcal{C}}f_{2})\otimes_{\mathcal{C}}f_{3}}$$\scriptstyle{f_{1}\otimes_{\mathcal{C}}(f_{2}\otimes_{\mathcal{C}}f_{3})}$$\scriptstyle{\alpha_{B_{1},B_{2},B_{3}}}$

commutes.

We still have to show that the pentagon and the triangle axiom are satisfied.
The proof for this can be found in the appendix. ∎

######  Example 2.9.

The monoidal product in Arr(Mat($\mathbb{N}$)) is defined on objects
$M:v_{1}\longrightarrow b_{1}$, $N:v_{2}\longrightarrow b_{2}$ via the
Kronecker product of matrices: $M\otimes N:v_{1}\cdot v_{2}\to b_{1}\cdot
b_{2}$. On morphisms the monoidal product is defined as follows:
$(S_{v_{1}},S_{b_{1}})\otimes(S_{v_{2}},S_{b_{2}})=(S_{v_{1}}\otimes
S_{v_{2}},S_{b_{1}}\otimes S_{b_{2}})$, i. e. we have the commutative diagram:

${v_{1}\cdot v_{2}}$${v_{1}^{\prime}\cdot v_{2}^{\prime}}$${b_{1}\cdot
b_{2}}$${b_{1}^{\prime}\cdot b_{2}^{\prime}}$$\scriptstyle{S_{v_{1}}\otimes
S_{v_{2}}}$$\scriptstyle{M\otimes N}$$\scriptstyle{M^{\prime}\otimes
N^{\prime}}$$\scriptstyle{S_{b_{1}}\otimes S_{b_{2}}}$

The tensor unit is given by the $1\times 1$-matrix:

$\displaystyle\mathbb{I}:1\longrightarrow 1.$ (8)

In a similar way one can define a braiding in Arr($\mathcal{C}$) using the
braiding in $\mathcal{C}$.

######  Proposition 2.10.

If $\mathcal{C}$ is a braided monoidal category with braiding
$\sigma_{A,B}:A\otimes_{\mathcal{C}}B\longrightarrow B\otimes_{\mathcal{C}}A$,
then Arr($\mathcal{C})$ has a braiding given by
$\sigma_{f,g}=(\sigma_{A,C},\sigma_{B,D}):f\otimes_{\mathcal{C}}g\longrightarrow
g\otimes_{\mathcal{C}}f$ for $f:A\longrightarrow B$ and $g:C\longrightarrow
D$, i. e. we have the following commutative diagram:

${A\otimes_{\mathcal{C}}C}$${C\otimes_{\mathcal{C}}A}$${B\otimes_{\mathcal{C}}D}$${D\otimes_{\mathcal{C}}B}$$\scriptstyle{\sigma_{A,C}}$$\scriptstyle{f\otimes_{\mathcal{C}}g}$$\scriptstyle{g\otimes_{\mathcal{C}}f}$$\scriptstyle{\sigma_{B,D}}$

###### Proof.

We can use Prop. 2.5 to define a natural isomorphism
$\sigma_{f,g}=(\sigma_{A,C},\sigma_{B,D})$ on Arr($\mathcal{C})$ using the
natural isomorphisms $\sigma_{A,C}$ and $\sigma_{B,D}$. In order to define a
braiding, $\sigma_{f,g}=(\sigma_{A,C},\sigma_{B,D})$ has to satisfy the
hexagon identities. Here we will only discuss one of them as the other one can
be proved analogously. In the diagram below the top and the bottom face
commute because $\sigma_{A,C}$ resp. $\sigma_{B,D}$ is a braiding in
$\mathcal{C}$. The left side face and the right square in the front commute,
because the braiding in $\mathcal{C}$ is natural and because of the definition
of the monoidal product in Arr($\mathcal{C}$). The left square in the front
commutes because the associator is natural in $\mathcal{C}$. Similarly, the
back face commutes because of naturality of the associator in $\mathcal{C}$.
Finally, the right side face commutes due to naturality of the braiding in
$\mathcal{C}$ and hence the whole diagram commutes.222We have again
abbreviated the labels of the monoidal products and the associator in order to
make the diagram more readable.

${(A\otimes D)\otimes X}$${A\otimes(D\otimes X)}$${(D\otimes A)\otimes
X}$${D\otimes(A\otimes X)}$${D\otimes(X\otimes A)}$${(D\otimes X)\otimes
A}$${(B\otimes C)\otimes Y}$${B\otimes(C\otimes Y)}$${(C\otimes B)\otimes
Y}$${C\otimes(B\otimes Y)}$${C\otimes(Y\otimes B)}$${(C\otimes Y)\otimes
B}$$\scriptstyle{\sigma_{A,D}\otimes\mathrm{id}_{X}}$$\scriptstyle{\alpha}$$\scriptstyle{(f\otimes
g)\otimes h}$$\scriptstyle{\sigma_{A,D,X}}$$\scriptstyle{f\otimes(g\otimes
h)}$$\scriptstyle{\alpha}$$\scriptstyle{(g\otimes f)\otimes
h}$$\scriptstyle{g\otimes(f\otimes
h)}$$\scriptstyle{\mathrm{id}_{D}\otimes\sigma_{A,X}}$$\scriptstyle{g\otimes(h\otimes
f)}$$\scriptstyle{\alpha}$$\scriptstyle{(g\otimes h)\otimes
f}$$\scriptstyle{\sigma_{B,C}\otimes\mathrm{id}_{Y}}$$\scriptstyle{\alpha}$$\scriptstyle{\sigma_{B,C,Y}}$$\scriptstyle{\alpha}$$\scriptstyle{\mathrm{id}_{C}\otimes\sigma_{B,Y}}$$\scriptstyle{\alpha}$

∎

######  Example 2.11.

On Arr(Mat($\mathbb{N}$)) one can define a braiding via

$\displaystyle\sigma_{M,N}=(\sigma_{v_{1},v_{2}},\sigma_{b_{1},b_{2}}):M\otimes
N\longrightarrow N\otimes M$ (9)

where $M$ is a $v_{1}\cdot v_{2}\times b_{1}\cdot b_{2}$-matrix and $N$ is a
$v_{2}\cdot v_{1}\times b_{2}\cdot b_{1}$-matrix.

######  Theorem 2.12.

If $\mathcal{C}$ is a symmetric monoidal category, then so is
Arr($\mathcal{C}$).

###### Proof.

If $\mathcal{C}$ is symmetric the braiding fulfils the following condition for
arbitrary objects $A$, $C$ in $\mathcal{C}$

$\displaystyle\sigma_{A,C}\circ\sigma_{C,A}=\mathrm{id}_{A\otimes_{\mathcal{C}}C}.$
(10)

In Arr($\mathcal{C}$) symmetry would require that

$\displaystyle\sigma_{f,g}\circ\sigma_{g,f}=\mathrm{id}_{f\otimes_{Arr(\mathcal{C})}g}$

where $f$ and $g$ are morphisms in $\mathcal{C}$. In others words, the
following diagram has to commute

${A\otimes_{\mathcal{C}}D}$${A\otimes_{\mathcal{C}}D}$${D\otimes_{\mathcal{C}}A}$${B\otimes_{\mathcal{C}}C}$${B\otimes_{\mathcal{C}}C}$${C\otimes_{\mathcal{C}}B}$$\scriptstyle{f\otimes_{\mathcal{C}}g}$$\scriptstyle{\sigma_{A,D}}$$\scriptstyle{\mathrm{id}_{A\otimes_{\mathcal{C}}D}}$$\scriptstyle{f\otimes_{\mathcal{C}}g}$$\scriptstyle{\sigma_{A,D}}$$\scriptstyle{g\otimes_{\mathcal{C}}f}$$\scriptstyle{\sigma_{B,C}}$$\scriptstyle{\mathrm{id}_{B\otimes_{\mathcal{C}}C}}$$\scriptstyle{\sigma_{B,C}}$

The top and the bottom face commute because $\mathcal{C}$ is symmetric and the
two side faces commute as the braiding in $\mathcal{C}$ is natural. The back
face commutes due to the definition of the identity morphism in
Arr($\mathcal{C}$) and hence the whole diagram commutes which proves the
assumption. ∎

As it is possible to define a monoidal product in arrow categories, one can
ask if a monoidal functor between two categories gives rise to a monoidal
functor between their arrow categories. This is indeed the case, as the
following proposition shows:

######  Proposition 2.13.

Let $(\mathcal{C},\otimes_{\mathcal{C}},\mathbb{I}_{\mathcal{C}})$ and
$(\mathcal{D},\otimes_{\mathcal{D}},\mathbb{I}_{\mathcal{D}})$ be two monoidal
categories and $(\mathcal{F},\mathcal{F}_{2},\mathcal{F}_{0})$ be a monoidal
functor between them. Then $(\mathcal{F},\mathcal{F}_{2},\mathcal{F}_{0})$
gives rise to a monoidal functor
$(\tilde{\mathcal{F}},\tilde{\mathcal{F}}_{2},\tilde{\mathcal{F}}_{0})$
between Arr($\mathcal{C}$) and Arr($\mathcal{D}$).

###### Proof.

We have already shown that a functor $\mathcal{F}$ between two categories
$\mathcal{C}$ and $\mathcal{D}$ induces a functor $\tilde{\mathcal{F}}$
between their arrow categories and that a natural transformation
$\eta:\mathcal{F}\Longrightarrow\mathcal{G}$ gives rise to a natural
transformation
$\tilde{\eta}:\tilde{\mathcal{F}}\Longrightarrow\tilde{\mathcal{G}}$. Hence we
can use the natural transformations $\mathcal{F}_{2}$ and $\mathcal{F}_{0}$ to
define natural transformations

$\displaystyle\tilde{\mathcal{F}}_{2}=(\mathcal{F}_{2,A,C},\mathcal{F}_{2,B,D}):\tilde{\mathcal{F}}(f)\otimes_{\mathcal{D}}\tilde{\mathcal{F}}(g)\longrightarrow\tilde{\mathcal{F}}(f\otimes_{\mathcal{C}}g)$
(11)

for $f:A\longrightarrow B$ and $g:C\longrightarrow D$ and

$\displaystyle\tilde{\mathcal{F}}_{0}=(\mathcal{F}_{0},\mathcal{F}_{0}):\mathrm{id}_{\mathbb{I}_{\mathcal{D}}}\longrightarrow\tilde{\mathcal{F}}(\mathrm{id}_{\mathbb{I}_{\mathcal{C}}}).$
(12)

In diagrams we have

${\mathcal{F}(A)\otimes_{\mathcal{D}}\mathcal{F}(C)}$${\mathcal{F}(A\otimes_{\mathcal{C}}C)}$${\mathcal{F}(B)\otimes_{\mathcal{D}}\mathcal{F}(D)}$${\mathcal{F}(B\otimes_{\mathcal{C}}D)}$$\scriptstyle{\mathcal{F}_{2,A,C}}$$\scriptstyle{\mathcal{F}(f)\otimes_{\mathcal{D}}\mathcal{F}(g)}$$\scriptstyle{\mathcal{F}(f\otimes_{\mathcal{C}}g)}$$\scriptstyle{\mathcal{F}_{2,B,D}}$

and

${\mathbb{I}_{\mathcal{D}}}$${\mathcal{F}(\mathbb{I}_{\mathcal{C}})}$${\mathbb{I}_{\mathcal{D}}}$${\mathcal{F}(\mathbb{I}_{\mathcal{C}})}$$\scriptstyle{\mathcal{F}_{0}}$$\scriptstyle{\mathrm{id}_{\mathbb{I}_{\mathcal{D}}}}$$\scriptstyle{\mathcal{F}(\mathrm{id}_{\mathbb{I}_{\mathcal{C}}}}$$\scriptstyle{\mathcal{F}_{0}}$

It is left to show that if $\mathcal{F}:\mathcal{C}\longrightarrow\mathcal{D}$
is monoidal,
$\tilde{\mathcal{F}}:\text{Arr}(\mathcal{C})\longrightarrow\text{Arr}(\mathcal{D})$
preserves the monoidal product, i. e. $\tilde{\mathcal{F}}_{2}$ and
$\tilde{\mathcal{F}_{0}}$ fulfil the associativity and the unitality
conditions, respectively. The associativity condition in Arr($\mathcal{C}$) is
given by the diagram on the next page. Here the top and the bottom face
commute because $\mathcal{F}_{2}$ satisfies the associativity condition. The
left face and the right square in the front commute because $\mathcal{F}_{2}$
is natural and because of the monoidal product in Arr($\mathcal{D}$). The
right square in the front commutes because the associator in $\mathcal{C}$ is
natural. Moreover, the left and the right face in the back commute as
$\mathcal{F}_{2}$ is natural and the middle face in the back commutes due to
naturality of the associator in $\mathcal{C}$.333Again we have abbreviated the
labels of the monoidal products, associators and natural transformations
$\mathcal{F}_{2}$ to make the diagram more readable.

${(\mathcal{F}(A)\otimes\mathcal{F}(B))\otimes\mathcal{F}(C)}$${\mathcal{F}(A)\otimes(\mathcal{F}(B)\otimes\mathcal{F}(C))}$${\mathcal{F}(A\otimes
B)\otimes\mathcal{F}(C)}$${\mathcal{F}((A\otimes B)\otimes
C)}$${\mathcal{F}(A\otimes(B\otimes
C)}$${(\mathcal{F}(V)\otimes\mathcal{F}(W))\otimes\mathcal{F}(U)}$${\mathcal{F}(V)\otimes(\mathcal{F}(W)\otimes\mathcal{F}(U))}$${\mathcal{F}(V\otimes
W)\otimes\mathcal{F}(U)}$${\mathcal{F}((V\otimes W)\otimes
U)}$${\mathcal{F}(V\otimes(W\otimes
U)}$$\scriptstyle{\mathcal{F}_{2}}$$\scriptstyle{\alpha}$$\scriptstyle{(\mathcal{F}(f)\otimes\mathcal{F}(g))\otimes\mathcal{F}(h)}$$\scriptstyle{\mathrm{id}_{\mathcal{F}(A)}\otimes\phi_{B,C}}$$\scriptstyle{\mathcal{F}(f)\otimes(\mathcal{F}(g)\otimes\mathcal{F}(h))}$$\scriptstyle{\mathcal{F}_{2}}$$\scriptstyle{\mathcal{F}(f\otimes
g)\otimes\mathcal{F}(h)}$$\scriptstyle{\mathcal{F}((f\otimes g)\otimes
h)}$$\scriptstyle{\mathcal{F}(\alpha)}$$\scriptstyle{\mathcal{F}(f\otimes(g\otimes
h))}$$\scriptstyle{\mathcal{F}_{2}}$$\scriptstyle{\alpha}$$\scriptstyle{\mathrm{id}_{\mathcal{F}(V)}\otimes\phi_{W,U}}$$\scriptstyle{\mathcal{F}_{2}}$$\scriptstyle{\mathcal{F}(\alpha)}$

The unitality condition is given by the following diagram:

${\mathcal{F}(A)\otimes_{\mathcal{D}}\mathbb{I}_{\mathcal{D}}}$${\mathcal{F}(A)\otimes_{\mathcal{D}}\mathcal{F}(\mathbb{I}_{\mathcal{C}})}$${\mathcal{F}(A)}$${\mathcal{F}(A\otimes_{\mathcal{C}}\mathbb{I}_{\mathcal{C}})}$${\mathcal{F}(B)\otimes_{\mathcal{D}}\mathbb{I}_{\mathcal{D}}}$${\mathcal{F}(B)\otimes_{\mathcal{D}}\mathcal{F}(\mathbb{I}_{\mathcal{C}})}$${\mathcal{F}(B)}$${\mathcal{F}(B\otimes_{\mathcal{C}}\mathbb{I}_{\mathcal{C}})}$$\scriptstyle{\rho_{\mathcal{F}(A)}}$$\scriptstyle{\mathrm{id}_{\mathcal{F}(A)}\otimes_{\mathcal{D}}\mathcal{F}_{0}}$$\scriptstyle{\mathcal{F}(f)\otimes_{\mathcal{D}}\mathrm{id}_{\mathbb{I}}}$$\scriptstyle{\mathcal{F}_{2,A\mathbb{I}}}$$\scriptstyle{\mathcal{F}(f)\otimes_{\mathcal{D}}\mathcal{F}(\mathrm{id}_{\mathbb{I}})}$$\scriptstyle{\mathcal{F}(\rho_{A})}$$\scriptstyle{\mathcal{F}(f)}$$\scriptstyle{\mathcal{F}(f\otimes_{\mathcal{C}}\mathrm{id}_{\mathbb{I}_{\mathcal{C}}})}$$\scriptstyle{\rho_{\mathcal{F}(B)}}$$\scriptstyle{\mathrm{id}_{\mathcal{F}(B)}\otimes_{\mathcal{D}}\mathcal{F}_{0}}$$\scriptstyle{\mathcal{F}_{2,B,\mathbb{I}}}$$\scriptstyle{\mathcal{F}(\rho_{B})}$

The top and the bottom face commute because $\mathcal{F}_{0}$ satisfies the
unitality condition. The back face commutes as $\mathcal{F}_{0}$ is natural
and due to the definition of the monoidal product in Arr($\mathcal{D}$).
Moreover, the left face commutes because of the definition of the unitor in
$\mathcal{C}$ and the right face commutes since $\mathcal{F}_{2}$ is natural.
Finally, the front face also commutes due to the definition of the unitor in
$\mathcal{C}$. ∎

Similarly, a _braided_ monoidal functor
$\mathcal{F}:\mathcal{C}\longrightarrow\mathcal{D}$ gives rise to a braided
monoidal functor between Arr($\mathcal{C}$) and Arr($\mathcal{D}$):

######  Proposition 2.14.

Let
$(\mathcal{C},\otimes_{\mathcal{C}},\mathbb{I}_{\mathcal{C}},\sigma_{\mathcal{C}})$
and
$(\mathcal{D},\otimes_{\mathcal{D}},\mathbb{I}_{\mathcal{D}},\sigma_{\mathcal{D}})$
be two braided monoidal categories and
$(\mathcal{F},\mathcal{F}_{2},\mathcal{F}_{0})$ be a braided monoidal functor
between them. Then $(\mathcal{F},\mathcal{F}_{2},\mathcal{F}_{0})$ gives rise
to a braided monoidal functor
$(\tilde{\mathcal{F}},\tilde{\mathcal{F}}_{2},\tilde{\mathcal{F}}_{0})$
between Arr($\mathcal{C}$) and Arr($\mathcal{D}$).

###### Proof.

We only need to show that the following diagram commutes:

${\mathcal{F}(V)\otimes_{\mathcal{D}}\mathcal{F}(V^{\prime})}$${\mathcal{F}(V^{\prime})\otimes_{\mathcal{D}}\mathcal{F}(V)}$${\mathcal{F}(V\otimes_{\mathcal{C}}V^{\prime})}$${\mathcal{F}(V^{\prime}\otimes_{\mathcal{C}}V)}$${\mathcal{F}(W)\otimes_{\mathcal{D}}\mathcal{F}(W^{\prime})}$${\mathcal{F}(W^{\prime})\otimes_{\mathcal{D}}\mathcal{F}(W)}$${\mathcal{F}(W\otimes_{\mathcal{C}}W^{\prime})}$${\mathcal{F}(W^{\prime}\otimes_{\mathcal{C}}W)}$$\scriptstyle{\mathcal{F}_{2,V,V^{\prime}}}$$\scriptstyle{\sigma_{\mathcal{F}(V),\mathcal{F}(V^{\prime})}}$$\scriptstyle{\mathcal{F}(f)\otimes_{\mathcal{D}}\mathcal{F}(f^{\prime})}$$\scriptstyle{\mathcal{F}_{2,V,V^{\prime}}}$$\scriptstyle{\mathcal{F}(f^{\prime})\otimes_{\mathcal{D}}\mathcal{F}(f)}$$\scriptstyle{\mathcal{F}(\sigma_{V,V^{\prime}})}$$\scriptstyle{\mathcal{F}(f\otimes_{\mathcal{C}}f^{\prime})}$$\scriptstyle{\mathcal{F}(f^{\prime}\otimes_{\mathcal{C}}f)}$$\scriptstyle{\mathcal{F}_{2,W,W^{\prime}}}$$\scriptstyle{\sigma_{\mathcal{F}(W),\mathcal{F}(W^{\prime})}}$$\scriptstyle{\mathcal{F}_{2,W^{\prime},W}}$$\scriptstyle{\mathcal{F}(\sigma_{W,W^{\prime}})}$

The top and the bottom face commute because $\mathcal{F}$ is a braided
monoidal functor and the two side faces commute due to naturality of
$\mathcal{F}_{2}$. The front face commutes due to naturality of the braiding
in $\mathcal{C}$. Finally, the back face commutes per definition of the
braiding and hence the whole diagram commutes. ∎

######  Proposition 2.15.

If the functor $\mathcal{F}:\mathcal{C}\longrightarrow\mathcal{D}$ in the
above construction is symmetric, i. e. if $\mathcal{C}$ is symmetric, then so
is
$\tilde{\mathcal{F}}:\text{Arr}(\mathcal{C})\longrightarrow\text{Arr}(\mathcal{D})$.

It is straightforward to verify that the following statement is true:

######  Example 2.16.

Let $\mathbb{K}$ be a field. An $n$-dimensional topological quantum field
theory is given by a symmteric monoidal functor from the cobordism category to
the category of vector spaces (see [1]):

$\displaystyle\mathcal{Z}:\mathbf{Cob}_{n}\longrightarrow\mathbf{Vect}_{\mathbb{K}}$
(13)

This functor assigns to each closed oriented $(n-1)$-dimensional manifold $M$
a $\mathbb{K}$-vector space $\mathcal{Z}(M)$ and to each oriented bordism $B$
from an $(n-1)$-dimensional manifold $M$ to another $(n-1)$-dimensional
manifold $N$ a $\mathbb{K}$-linear map
$\mathcal{Z}(B):\mathcal{Z}(M)\longrightarrow\mathcal{Z}(N)$.

Applying Prop. 2.15, we can construct a symmetric monoidal functor

$\displaystyle\tilde{\mathcal{Z}}:\mathrm{Arr}(\mathbf{Cob}_{n})\longrightarrow\mathrm{Arr}(\mathbf{Vect}_{\mathbb{K}})$
(14)

which assigns to each oriented bordism $B$ from an $(n-1)$-dimensional
manifold $M$ to another $(n-1)$-dimensional manifold $N$ a $\mathbb{K}$-linear
map $\mathcal{Z}(B):\mathcal{Z}(M)\longrightarrow\mathcal{Z}(N)$ and to each
morphism $(\beta_{M},\beta_{N}):B\longrightarrow B^{\prime}$

${M}$${M^{\prime}}$${N}$${N^{\prime}}$$\scriptstyle{\beta_{M}}$$\scriptstyle{B}$$\scriptstyle{B^{\prime}}$$\scriptstyle{\beta_{N}}$

where $\beta_{M}:M\longrightarrow M^{\prime}$ and $\beta_{N}:N\longrightarrow
N^{\prime}$ are bordisms, a morphism
$(\mathcal{Z}(\beta_{M}),\mathcal{Z}(\beta_{N})):\mathcal{Z}(B)\longrightarrow\mathcal{Z}(B^{\prime})$

${\mathcal{Z}(M)}$${\mathcal{Z}(M^{\prime})}$${\mathcal{Z}(N)}$${\mathcal{Z}(N^{\prime})}$$\scriptstyle{\mathcal{Z}(\beta_{M})}$$\scriptstyle{\mathcal{Z}(B)}$$\scriptstyle{\mathcal{Z}(B^{\prime})}$$\scriptstyle{\mathcal{Z}(\beta_{N})}$

where
$\mathcal{Z}(\beta_{M}):\mathcal{Z}(M)\longrightarrow\mathcal{Z}(M^{\prime})$
and
$\mathcal{Z}(\beta_{N}):\mathcal{Z}(N)\longrightarrow\mathcal{Z}(N^{\prime})$
are $\mathbb{K}$-linear maps.

######  Proposition 2.17.

Let $(\mathcal{C},\otimes_{\mathcal{C}},\mathbb{I}_{\mathcal{C}})$ and
$(\mathcal{D},\otimes_{\mathcal{D}},\mathbb{I}_{\mathcal{D}})$ be two monoidal
categories and $(\mathcal{F},\mathcal{F}_{2},\mathcal{F}_{0})$ and
$(\mathcal{G},\mathcal{G}_{2},\mathcal{G}_{0})$ be two monoidal functors
between those categories. A monoidal natural transformation
$\eta:\mathcal{F}\Longrightarrow\mathcal{G}$ gives rise to a monoidal natural
transformation
$\tilde{\eta}:\tilde{\mathcal{F}}\Longrightarrow\tilde{\mathcal{G}}$, where
$\tilde{\mathcal{F}}$ and $\tilde{\mathcal{G}}$ are the functors defined on
the arrow categories.

###### Proof.

We only need to show that $\tilde{\eta}$ preserves the monoidal product and
the monoidal unit in Arr($\mathcal{C}$), if $\eta$ preserves the monoidal
product and the monoidal unit in $\mathcal{C}$. For this, the following
diagram has to commute:

${\mathcal{F}(A)\otimes_{\mathcal{D}}\mathcal{F}(A)}$${\mathcal{G}(A)\otimes_{\mathcal{D}}\mathcal{G}(C)}$${\mathcal{F}(A\otimes_{\mathcal{C}}C)}$${\mathcal{G}(A\otimes_{\mathcal{C}}C)}$${\mathcal{F}(B)\otimes_{\mathcal{D}}\mathcal{F}(D)}$${\mathcal{G}(B)\otimes_{\mathcal{D}}\mathcal{G}(D)}$${\mathcal{F}(B\otimes_{\mathcal{C}}D)}$${\mathcal{G}(B\otimes_{\mathcal{C}}D)}$$\scriptstyle{\mathcal{F}_{2,A,C}}$$\scriptstyle{\eta_{A}\otimes_{\mathcal{D}}\eta_{B}}$$\scriptstyle{\mathcal{F}(f)\otimes_{\mathcal{D}}\mathcal{F}(g)}$$\scriptstyle{\mathcal{G}_{2,A,C}}$$\scriptstyle{\mathcal{G}(f)\otimes_{\mathcal{D}}\mathcal{G}(g)}$$\scriptstyle{\eta_{A\otimes_{\mathcal{C}}C}}$$\scriptstyle{\mathcal{F}(f\otimes_{\mathcal{C}}g)}$$\scriptstyle{\mathcal{G}(f\otimes_{\mathcal{C}}g)}$$\scriptstyle{\mathcal{F}_{2,B,D}}$$\scriptstyle{\eta_{B}\otimes_{\mathcal{D}}\eta_{D}}$$\scriptstyle{\mathcal{G}_{2,B,D}}$$\scriptstyle{\eta_{B\otimes_{\mathcal{C}}D}}$

The top and the bottom face commute because $\eta$ is a monoidal natural
transformation and the two side faces commute due to naturality of
$\mathcal{F}_{2}$ resp. $\mathcal{G}_{2}$. Finally, the front and the back
face commute due to naturality of $\eta$ and thus the whole diagram commutes.
Moreover, the following diagram has to commute as well:

${\mathcal{F}(\mathbb{I}_{\mathcal{C}})}$${\mathcal{G}(\mathbb{I}_{\mathcal{C}})}$${\mathbb{I}_{\mathcal{D}}}$${\mathcal{F}(\mathbb{I}_{\mathcal{C}})}$${\mathcal{G}(\mathbb{I}_{\mathcal{C}})}$${\mathbb{I}_{\mathcal{D}}}$$\scriptstyle{\mathcal{F}(\mathrm{id}_{\mathbb{I}_{\mathcal{C}}})}$$\scriptstyle{\mathcal{F}_{0}}$$\scriptstyle{\eta_{\mathbb{I}_{\mathcal{C}}}}$$\scriptstyle{\mathcal{G}(\mathrm{id}_{\mathbb{I}_{\mathcal{C}}})}$$\scriptstyle{\mathcal{G}_{0}}$$\scriptstyle{\mathrm{id}_{\mathbb{I}_{\mathcal{D}}}}$$\scriptstyle{\mathcal{F}_{0}}$$\scriptstyle{\eta_{\mathbb{I}_{\mathcal{C}}}}$$\scriptstyle{\mathcal{G}_{0}}$

Here the top and the bottom face commute because $\eta$ is a monoidal natural
transformation and the two side faces commute due to the naturality of
$\mathcal{F}_{0}$ resp. $\mathcal{G}_{0}$. The back face also commutes as
$\eta$ is a natural and thus the whole diagram commutes. ∎

###  2.3 Duals and Pivots in Arrow Categories

In this section we will investigate under which circumstances an arrow
category is rigid. As it turns out, we can, similarly to the constructions
made earlier, use the structure of $\mathcal{C}$ to define a evaluation and
coevaluation map on the subcategory of Arr($\mathcal{C}$) where all objects
are isomorphisms, i.e. where all objects are morphisms in the _core_ of
$\mathcal{C}$. Furthermore, we will show that a pivot in a pivotal category
$\mathcal{C}$ induces a pivot in its arrow category and that a twist in a
ribbon category gives rise to a twist in its arrow category.

######  Theorem 2.18.

Let $(\mathcal{C},\otimes_{\mathcal{C}},\mathbb{I}_{\mathcal{C}})$ be a rigid
monoidal category. An object $f:A\longrightarrow B$ in Arr($\mathcal{C}$) has
a dual if and only if it is an isomorphism in $\mathcal{C}$.

###### Proof.

Let $d_{A}:A^{*}\otimes_{\mathcal{C}}A\longrightarrow\mathbb{I}$ and
$b_{A}:\mathbb{I}\longrightarrow A\otimes_{\mathcal{C}}A^{*}$, where $A,A^{*}$
$\in$ obj($\mathcal{C}$) respectively, define the evaluation and coevaluation
map in $\mathcal{C}$. For a morphism $f:A\longrightarrow B$ $\in$
obj(Arr($\mathcal{C}$)) and its dual $f^{*}:A^{*}\longrightarrow B^{*}$ we can
now define maps in Arr($\mathcal{C}$) :

$\displaystyle
d_{f}:f^{*}\otimes_{\mathrm{Arr}(\mathcal{C})}f\longrightarrow\mathrm{id}_{\mathbb{I}_{\mathcal{C}}}$
(15)

and

$\displaystyle b_{f}:\mathrm{id}_{\mathbb{I}_{\mathcal{C}}}\longrightarrow
f^{*}\otimes_{\mathrm{Arr}(\mathcal{C})}f,$ (16)

such that

${A^{*}\otimes_{\mathcal{C}}A}$${\mathbb{I}_{\mathcal{C}}}$${B^{*}\otimes_{\mathcal{C}}B}$${\mathbb{I}_{\mathcal{C}}}$$\scriptstyle{d_{A}}$$\scriptstyle{f^{*}\otimes_{\mathcal{C}}f}$$\scriptstyle{\mathrm{id}_{\mathbb{I}_{\mathcal{C}}}}$$\scriptstyle{d_{B}}$

and

${\mathbb{I}_{\mathcal{C}}}$${A\otimes_{\mathcal{C}}A^{*}}$${\mathbb{I}_{\mathcal{C}}}$${B\otimes_{\mathcal{C}}B^{*}}$$\scriptstyle{b_{A}}$$\scriptstyle{\mathrm{id}_{\mathbb{I}_{\mathcal{C}}}}$$\scriptstyle{f\otimes_{\mathcal{C}}f^{*}}$$\scriptstyle{b_{B}}$

commute. The first snake identity in Arr($\mathcal{C}$) is given by:

${A}$${A}$${A\otimes_{\mathcal{C}}(A\otimes_{\mathcal{C}}A^{*})}$${B}$${B}$${B\otimes_{\mathcal{C}}(B\otimes_{\mathcal{C}}B^{*})}$$\scriptstyle{\mathrm{id}_{A}}$$\scriptstyle{f}$$\scriptstyle{\mathrm{id}_{A}\otimes_{\mathcal{C}}b_{A}}$$\scriptstyle{f}$$\scriptstyle{\mathrm{id}_{A}\otimes_{\mathcal{C}}d_{A}}$$\scriptstyle{f\otimes_{\mathcal{C}}(f\otimes_{\mathcal{C}}f^{*})}$$\scriptstyle{\mathrm{id}_{B}\otimes_{\mathcal{C}}b_{B}}$$\scriptstyle{\mathrm{id}_{B}}$$\scriptstyle{\mathrm{id}_{B}\otimes_{\mathcal{C}}d_{B}}$

The top and the bottom face commute because $b$ and $d$ fulfil the snake
identities in $\mathcal{C}$ for the dualities $A\dashv A^{*}$ and $B\dashv
B^{*}$ respectively. The two side faces commute due to the definition of the
monoidal product and the (co)evaluation map in Arr($\mathcal{C}$). Finally, it
is easy to see that the back face also commutes and thus the whole diagram
commutes. Analogously, one can show that the second snake identity is
satisfied in Arr($\mathcal{C}$).

In terms of the graphical calculus, commutativity of the diagram corresponding
to Eq. 15 means that the following equation has to hold:

Inserting this into the snake identity in $\mathcal{C}$, yields:

For the other snake identity we get a similar expression. From this we can
conclude that $f$ has to be an isomorphism with inverse:

On the other hand, if we assume that $f$ is invertible with inverse given by
the morphism in the diagram above, it is easy to verify that the diagrams
corresponding to Eq. 16 and Eq. 15 and hence also the snake identities in
Arr($\mathcal{C}$) commute. From that one can conclude that an object $f$ in
Arr($\mathcal{C}$) has a dual if and only if it is invertible. ∎

######  Example 2.19.

In Mat($\mathbb{N}$), only invertible matrices have duals.

In the following we will denote the core of an arbitrary category
$\mathcal{C}$ with $\mathcal{C}_{core}$.

######  Theorem 2.20.

Let $\mathcal{C}$ be a pivotal category with pivot $\pi_{A}:A\longrightarrow
A^{**}$ for $A$ $\in$ obj($\mathcal{C}$). Then Arr($\mathcal{C}_{core}$) is
also pivotal with pivot $\tilde{\pi}_{f}=(\pi_{A},\pi_{B}):f\longrightarrow
f^{**}$ for $f:A\longrightarrow B$ $\in$ obj(Arr($\mathcal{C}$)).

###### Proof.

Since the pivot is a monoidal natural transformation between the identity
functor and the functor that sends each object to its double dual, we can
apply Proposition 2.13 and Proposition 2.4 to define a monoidal natural
transformation between the identity functor on Arr($\mathcal{C}_{core}$) and
the functor that sends each object in Arr($\mathcal{C}_{core}$) to its double
dual. ∎

######  Example 2.21.

The arrow category of the subcategory of $\mathbf{Mat}(\mathbb{N})$ where all
matrices are invertible is a pivotal symmetric monoidal category.

######  Proposition 2.22.

Let $(\mathcal{C},\otimes_{\mathcal{C}},\mathbb{I}_{\mathcal{C}})$ be a ribbon
category, i. e. a rigid monoidal category with a twist
$\theta_{A}:A\longrightarrow A$. Then $Arr(\mathcal{C}_{core})$ is a ribbon
category with twist $(\theta_{A},\theta_{B}):f\longrightarrow f$, i. e. we
have:

${A}$${A}$${B}$${B}$$\scriptstyle{\theta_{A}}$$\scriptstyle{f}$$\scriptstyle{f}$$\scriptstyle{\theta_{B}}$

###### Proof.

In order to define a twist $(\theta_{A},\theta_{B}):f\longrightarrow f$, the
following diagram has to commute:

${C\otimes_{\mathcal{C}}A}$${A\otimes_{\mathcal{C}}C}$${A\otimes_{\mathcal{C}}C}$${A\otimes_{\mathcal{C}}C}$${D\otimes_{\mathcal{C}}B}$${B\otimes_{\mathcal{C}}D}$${B\otimes_{\mathcal{C}}D}$${B\otimes_{\mathcal{C}}D}$$\scriptstyle{\sigma_{A,C}^{-1}}$$\scriptstyle{\sigma_{C,A}}$$\scriptstyle{g\otimes_{\mathcal{C}}f}$$\scriptstyle{\theta_{A}\otimes_{\mathcal{C}}\theta_{C}}$$\scriptstyle{f\otimes_{\mathcal{C}}g}$$\scriptstyle{\theta_{A\otimes
C}}$$\scriptstyle{f\otimes_{\mathcal{C}}g}$$\scriptstyle{f\otimes_{\mathcal{C}}g}$$\scriptstyle{\sigma_{B,D}^{-1}}$$\scriptstyle{\sigma_{D,B}}$$\scriptstyle{\theta_{B}\otimes_{\mathcal{C}}\theta_{D}}$$\scriptstyle{\theta_{B\otimes
D}}$

Here the bottom and the top face commute because $\theta$ is a twist in
$\mathcal{C}$. Moreover, the front face commutes due to naturality of $\theta$
in $\mathcal{C}$. The left and the back face commute due to naturality of the
braiding and its inverse and the right face commutes because of the definition
of the monoidal product and naturality of the braiding.

The second identity that has to be fulfilled is given by

$\displaystyle(\theta_{\mathbb{I}_{\mathcal{C}}},\theta_{\mathbb{I}_{\mathcal{C}}})=(\mathrm{id}_{\mathbb{I}_{\mathcal{C}}},\mathrm{id}_{\mathbb{I}_{\mathcal{C}}})$

This equation holds because $\theta$ is a twist in $\mathcal{C}$ and hence:
$\theta_{\mathbb{I}_{\mathcal{C}}}=\mathrm{id}_{\mathbb{I}_{\mathcal{C}}}$.
The last condition that has to be satisfied is given by:

$\displaystyle(\theta_{B^{*}},\theta_{A^{*}})=(\theta_{A},\theta_{B})^{*}=(\theta_{B}^{*},\theta_{A}^{*})$

which holds since $\theta$ is a twist and hence we have:
$(\theta_{B}^{*},\theta_{A}^{*})=(\theta_{B^{*}},\theta_{A^{*}})$. ∎

######  Example 2.23.

Consider again the arrow category of the subcategory of
$\mathbf{Mat}(\mathbb{N})$ where all matrices are invertible. The pivot
induces a twist on this category via $\theta_{A}=\mathrm{id}_{A}$ and hence
this category is a ribbon category. In particular, it is a compact category.

###  2.4 (Co)monoids, Bialgebras, Frobenius Structures and Hopf Algebras in
Arrow Categories

In this section we will show that the (co)monoids in arrow categories are
given by (co)monoid-morphisms. With that, we can then derive a notion of
bialgebras, Frobenius algebras and Hopf algebras in arrow categories. In the
following we will use $\eta$ to denote the unit of a monoid in a category.
This should not be confused with a natural transformation.

######  Theorem 2.24.

Let $\mathcal{C}$ be a monoidal category and ($A$, $\mu_{A}$, $\eta_{A}$) and
($B$, $\mu_{B}$, $\eta_{B}$) be monoids in $\mathcal{C}$. Then a morphism of
monoids $f:A\longrightarrow B$ in $\mathcal{C}$ is a monoid in
Arr($\mathcal{C}$). If the monoids $A$ and $B$ are commutative, so is the
monoid $f:A\longrightarrow B$ in Arr($\mathcal{C}$).

###### Proof.

Let $\mathcal{C}$ be a monoidal category and ($A$, $\mu_{A}$, $\eta_{A}$) and
($B$, $\mu_{B}$, $\eta_{B})$ be monoids in $\mathcal{C}$. We can then define a
monoid object $f:A\longrightarrow B$ in Arr($\mathcal{C}$), where $f$ is a
morphism of monoids, via the following construction:

  * •

the multiplication is given by a morphism $\tilde{\mu}=(\mu_{A},\mu_{B})$ such
that

${A\otimes_{\mathcal{C}}A}$${A}$${B\otimes_{\mathcal{C}}B}$${B}$$\scriptstyle{\mu_{A}}$$\scriptstyle{f\otimes_{\mathcal{C}}f}$$\scriptstyle{f}$$\scriptstyle{\mu_{B}}$

commutes.

  * •

the unitor is a morphism $\tilde{\eta}=(\eta_{A},\eta_{B})$ such that

${\mathbb{I}_{\mathcal{C}}}$${A}$${\mathbb{I}_{\mathcal{C}}}$${B}$$\scriptstyle{\eta_{A}}$$\scriptstyle{\mathrm{id}_{\mathbb{I}_{\mathcal{C}}}}$$\scriptstyle{f}$$\scriptstyle{\eta_{B}}$

commutes.

We still need to show that the pentagon axiom and the unitor diagrams commute.
This will be done in Appendix 2.4.

∎

Analogously, one can show that the following statement is true:

######  Theorem 2.25.

Given a category $\mathcal{C}$, then a comonoid morphism in $\mathcal{C}$ is a
comonoid in Arr($\mathcal{C}$). If the comonoid in $\mathcal{C}$ is
cocommutative, then this also holds for the comonoid in Arr($\mathcal{C}$).

Having defined both monoid and comonoid objects in a Arr($\mathcal{C}$), we
can now define bialgebra objects in Arr($\mathcal{C}$). In the following,
$\Delta$ denotes the comultiplication and $\epsilon$ is the counit of a
comonoid in $\mathcal{C}$.

######  Theorem 2.26.

Let $(\mathcal{C},\otimes_{\mathcal{C}},\mathbb{I}_{\mathcal{C}})$ be a
monoidal category and ($A$, $\mu_{A}$, $\eta_{A}$, $\Delta_{A}$,
$\epsilon_{A}$) and ($B$, $\mu_{B}$, $\eta_{B}$, $\Delta_{B}$, $\epsilon_{B}$)
be both a monoid and a comonoid in $\mathcal{C}$ such that the bialgebra axiom
is satisfied. Then a morphim $f:A\longrightarrow B$ in $\mathcal{C}$, which is
both a monoid and comonoid morphism, is a bialgebra object in
Arr($\mathcal{C}$).

###### Proof.

Let ($A$, $\mu_{A}$, $\eta_{A}$, $\Delta_{A}$, $\epsilon_{A}$) and ($B$,
$\mu_{B}$, $\eta_{B}$, $\Delta_{B}$, $\epsilon_{B}$) be two objects which are
both a monoid and a comonoid and both satisfy the bialgebra axiom. We then
have that each $f:A\longrightarrow B$ which is both a monoid and a comonoid
morphism is a monoid and a comonoid in Arr($\mathcal{C}$). The morphism $f$ is
a bialgebra object in Arr($\mathcal{C}$), if it satisfies the two bialgebra
axioms in Arr($\mathcal{C}$). The first axiom requires that the following
diagram commutes:

${A\otimes A}$${A}$${(A\otimes A)\otimes(A\otimes A)}$${(A\otimes
A)\otimes(A\otimes A)}$${A\otimes A}$${B\otimes B}$${B}$${(B\otimes
B)\otimes(B\otimes B)}$${(B\otimes B)\otimes(B\otimes B)}$${B\otimes
B}$$\scriptstyle{\Delta_{A}\otimes\Delta_{A}}$$\scriptstyle{\mu_{A}}$$\scriptstyle{f\otimes
f}$$\scriptstyle{\Delta_{A}}$$\scriptstyle{f}$$\scriptstyle{\mathrm{id}_{A}\otimes\sigma_{A,A}\otimes\mathrm{id}_{A}}$$\scriptstyle{(f\otimes
f)\otimes(f\otimes f)}$$\scriptstyle{(f\otimes f)\otimes(f\otimes
f)}$$\scriptstyle{\mu_{A}\otimes\mu_{A}}$$\scriptstyle{f\otimes
f}$$\scriptstyle{\Delta_{B}\otimes\Delta_{B}}$$\scriptstyle{\mu_{B}}$$\scriptstyle{\Delta_{B}}$$\scriptstyle{\mathrm{id}_{B}\otimes\sigma_{B,B}\otimes\mathrm{id}_{B}}$$\scriptstyle{\mu_{B}\otimes\mu_{B}}$

The left face commutes because $f$ is a morphism of comonoids and because of
the definition of the monoidal product in Arr($\mathcal{C}$). Analogously, the
right face commutes since $f$ is a morphism of comonoids. Because $A$ and $B$
are bialgebra objects in $\mathcal{C}$, the top and the bottom face commute.
The back face and the right square in the front commute because $f$ is a
monoid morphism and because of the definition of the monoidal product in
Arr($\mathcal{C}$). Finally, the left square in the front also commutes
because of the definition of the braiding in $\mathcal{C}$ and the monoidal
product in Arr($\mathcal{C}$) and thus the whole diagram commutes.

The second bialgebra axiom requires that the following diagram commutes:

${A\otimes A}$${\mathbb{I}\otimes\mathbb{I}}$${A}$${\mathbb{I}}$${B\otimes
B}$${\mathbb{I}\otimes\mathbb{I}}$${B}$${\mathbb{I}}$$\scriptstyle{\mu_{A}}$$\scriptstyle{\epsilon_{A}\otimes\epsilon_{A}}$$\scriptstyle{f\otimes
f}$$\scriptstyle{\tilde{=}}$$\scriptstyle{\mathrm{id}_{\mathbb{I}}\otimes\mathrm{id}_{\mathbb{I}}}$$\scriptstyle{\epsilon_{A}}$$\scriptstyle{f}$$\scriptstyle{\mathrm{id}_{\mathbb{I}}}$$\scriptstyle{\mu_{B}}$$\scriptstyle{\epsilon_{B}\otimes\epsilon_{B}}$$\scriptstyle{\tilde{=}}$$\scriptstyle{\epsilon_{B}}$

Here the back face commutes because of the definition of the monoidal product
in Arr($\mathcal{C}$) and because $f$ is a morphism of comonoids. The left and
the front face also commute, because $f$ is a morphism of monoids. Finally,
the top and bottom face commute as $A$ and $B$ are bialgebra objects in
$\mathcal{C}$ and thus the whole diagram commutes. ∎

######  Proposition 2.27.

Let $(\mathcal{C},\otimes_{\mathcal{C}},\mathbb{I}_{\mathcal{C}})$ be a
monoidal category and $A$, $B$ be two Frobenius structures in $\mathcal{C}$
(in particular, $A$ and $B$ are both monoids and comonoids). Then an object
$f:A\longrightarrow B$ in Arr($\mathcal{C}$), which is both a monoid and a
comonoid morphism is a Frobenius structure in Arr($\mathcal{C}$). If $A$ and
$B$ are special Frobenius algebras in $\mathcal{C}$, then $f:A\longrightarrow
B$ is a special Frobenius structure in Arr($\mathcal{C}$).

###### Proof.

The Frobenius law in Arr($\mathcal{C}$) is given by the following diagram:

${A\otimes A}$${A\otimes A\otimes A}$${A\otimes A\otimes A}$${A\otimes
A}$${B\otimes B}$${B\otimes B\otimes B}$${B\otimes B\otimes B}$${B\otimes
B}$$\scriptstyle{\mathrm{id}_{A}\otimes\Delta_{A}}$$\scriptstyle{\Delta_{A}\otimes\mathrm{id}_{A}}$$\scriptstyle{f\otimes
f}$$\scriptstyle{\mathrm{id}_{A}\otimes\mu_{A}}$$\scriptstyle{f\otimes
f\otimes
f}$$\scriptstyle{\mu_{A}\otimes\mathrm{id}_{A}}$$\scriptstyle{f\otimes
f\otimes f}$$\scriptstyle{f\otimes
f}$$\scriptstyle{\mathrm{id}_{B}\otimes\Delta_{B}}$$\scriptstyle{\Delta_{B}\otimes\mathrm{id}_{B}}$$\scriptstyle{\mathrm{id}_{B}\otimes\mu_{B}}$$\scriptstyle{\mu_{B}\otimes\mathrm{id}_{B}}$

The top and the bottom face of the diagram commutes because $A$ and $B$ are
Frobenius algebras in $\mathcal{C}$. The left face commutes because of the
definition of the monoidal product in Arr($\mathcal{C}$) and because $f$ is a
morphism of comonoids. Analogously, the back face commutes. Finally, the front
and the right face commute because of the definition of the monoidal product
in $\mathcal{C}$ and because $f$ is a morphism of monoids and thus the whole
diagram commutes.

If $A$ and $B$ are both special, then the following diagram also commutes as
$f$ is both a morphism of monoids and comonoids:

${A}$${A}$${A\otimes A}$${B}$${B}$${B\otimes
B}$$\scriptstyle{\Delta_{A}}$$\scriptstyle{\mathrm{id}_{A}}$$\scriptstyle{f}$$\scriptstyle{f}$$\scriptstyle{\mu_{A}}$$\scriptstyle{f\otimes
f}$$\scriptstyle{\Delta_{B}}$$\scriptstyle{\mathrm{id}_{B}}$$\scriptstyle{\mu_{B}}$

and hence $f$ is also special. ∎

######  Proposition 2.28.

Let $(\mathcal{C},\otimes_{\mathcal{C}},\mathbb{I}_{\mathcal{C}})$ be a
monoidal dagger category and $A$, $B$ be two objects in $\mathcal{C}$ which
are dagger Frobenius algebras in $\mathcal{C}$. Then a unitary morphism of
both monoids and comonoids $f:A\longrightarrow B$ is a dagger Frobenius
structure in Arr($\mathcal{C}$).

###### Proof.

Let $f:A\longrightarrow B$ be a morphism of bialgebras, i. e. a moprhism of
monoids and comonoids that is also unitary, i. e.: $f^{\dagger}=f^{-1}$. We
can define a multiplication and a comultiplication in Arr($\mathcal{C}$) via

$\displaystyle\mu_{f}$
$\displaystyle=(\mu_{A},\mu_{B}):f\otimes_{\mathcal{C}}f\longrightarrow f\ \ \
\ \text{and}$ $\displaystyle\Delta_{f}$
$\displaystyle=(\Delta_{A},\Delta_{B}):f\longrightarrow
f\otimes_{\mathcal{C}}f$

respectively. Applying the dagger functor from Prop. 2.7 on the
comultiplication then gives

$\displaystyle\Delta_{f}^{\dagger}=(\Delta_{A},\Delta_{B})^{\dagger}=(\Delta_{A}^{\dagger},\Delta_{B}^{\dagger})=(\mu_{A},\mu_{B})=\mu_{f}$

where we used the fact that $A$ and $B$ are dagger Frobenius algebras.
Similarly, we get that

$\displaystyle\epsilon_{f}^{\dagger}=(\epsilon_{A}^{\dagger},\epsilon_{B}^{\dagger})=(\eta_{A},\eta_{B})=\eta_{f}$

and hence $f:A\longrightarrow B$ is a dagger Frobenius structure in
Arr($\mathcal{C}$). ∎

This leads us to the following theorem:

######  Theorem 2.29.

Let $(\mathcal{C},\otimes_{\mathcal{C}},\mathbb{I}_{\mathcal{C}})$ be a
braided monoidal dagger category and let $A$ and $B$ be two objects in
$\mathcal{C}$ which are commutative special dagger Frobenius algebras in
$\mathcal{C}$. Then a unitary morphism of Frobenius algebras
$f:A\longrightarrow B$ is a commutative special dagger Frobenius structure in
Arr($\mathcal{C}$).

###### Proof.

We only need to apply Theorem 2.8, Proposition 2.27 and Proposition 2.28. ∎

######  Theorem 2.30.

Let $\mathcal{C}$ be a monoidal category and ($A$, $\mu_{A}$, $\eta_{A}$,
$\Delta_{A}$, $\epsilon_{A}$, $S_{A}$) and ($B$, $\mu_{B}$, $\eta_{B}$,
$\Delta_{B}$, $\epsilon_{B}$, $S_{B}$) be Hopf algebra objects in
$\mathcal{C}$ where $S_{A}:A\longrightarrow A$ and $S_{B}:B\longrightarrow B$
are antipodes in $\mathcal{C}$. Then morphisms of Hopf algebras are Hopf
algebra objects in Arr($\mathcal{C}$).

###### Proof.

Let $\mathcal{C}$ be a monoidal category and ($A$, $\mu_{A}$, $\eta_{A}$,
$\Delta_{A}$, $\epsilon_{A}$, $S_{A}$) and ($B$, $\mu_{B}$, $\eta_{B}$,
$\Delta_{B}$, $\epsilon_{B}$, $S_{B}$) be Hopf algebra objects in
$\mathcal{C}$, where $S_{A}:A\longrightarrow A$ and $S_{B}:B\longrightarrow B$
are antipodes in $\mathcal{C}$. We can then define an antipode
$S=(S_{A},S_{B})$

${A}$${A}$${B}$${B}$$\scriptstyle{S_{A}}$$\scriptstyle{f}$$\scriptstyle{f}$$\scriptstyle{S_{B}}$

where $f$ is a morphism of antipodes and thus a Hopf algebra object in
Arr($\mathcal{C}$).

We now have to prove that the Hopf algebra axiom is satisfied if $f$ is a
morphism of Hopf algebras. In Arr($\mathcal{C}$) this means that the following
diagram has to commute:

${A}$${\mathbb{I}}$${A\otimes A}$${A\otimes
A}$${A}$${B}$${\mathbb{I}}$${B\otimes B}$${B\otimes
B}$${B}$$\scriptstyle{\Delta_{A}}$$\scriptstyle{\epsilon_{A}}$$\scriptstyle{f}$$\scriptstyle{\eta_{A}}$$\scriptstyle{\mathrm{id}_{\mathbb{I}}}$$\scriptstyle{\mathrm{id}_{A}\otimes
S_{A}}$$\scriptstyle{f\otimes f}$$\scriptstyle{f\otimes
f}$$\scriptstyle{\mu_{A}}$$\scriptstyle{f}$$\scriptstyle{\Delta_{B}}$$\scriptstyle{\epsilon_{B}}$$\scriptstyle{\eta_{B}}$$\scriptstyle{\mathrm{id}_{B}\otimes
S_{B}}$$\scriptstyle{\mu_{B}}$

Because $f$ is a morphism of comonoids, the left and the right side face
commute. Analogously, the second square of the front face and the back face
commute, since $f$ is a morphism of monoids. The first square of the front
face commutes because of the definition of the monoidal product in
Arr($\mathcal{C}$) and the definition of the antipode in Arr($\mathcal{C}$).
Finally, the top and the bottom face commute as $A$ and $B$ are Hopf algebras
in $\mathcal{C}$ and hence the whole diagram commutes. ∎

##  Appendix A Proofs

#### Proof of 2.8 continued:

###### Proof.

We want to show that the pentagon and the triangle axiom are satisfied for the
monoidal product defined in the beginning of this proof. The pentagon axiom in
Arr($\mathcal{C}$) is given by the diagram on the next page where the front
and the back face commute because $\alpha$ satisfies the pentagon axiom
($\alpha$ is the associator in $\mathcal{C}$). The two side faces commute due
to the definition of the monoidal product and naturality of the associator in
$\mathcal{C}$. The two top faces and the bottom face also commute due to
naturality of the associator in $\mathcal{C}$ and hence the whole diagram
commutes.444We have abbreviated the labels of the monoidal products and the
associators trying to make the diagram more readable.

${A_{1}\otimes(A_{2}\otimes(A_{3}\otimes A_{4}))}$${(A_{1}\otimes
A_{2})\otimes(A_{3}\otimes A_{4})}$${((A_{1}\otimes A_{2})\otimes
A_{3})\otimes A_{4}}$${B_{1}\otimes(B_{2}\otimes(B_{3}\otimes
B_{4}))}$${(B_{1}\otimes B_{2})\otimes(B_{3}\otimes B_{4})}$${((B_{1}\otimes
B_{2})\otimes B_{3})\otimes B_{4}}$${A_{1}\otimes((A_{2}\otimes A_{3})\otimes
A_{4})}$${(A_{1}\otimes(A_{2}\otimes A_{3}))\otimes
A_{4}}$${B_{1}\otimes((B_{2}\otimes B_{3})\otimes
B_{4})}$${(B_{1}\otimes(B_{2}\otimes B_{3}))\otimes
B_{4}}$$\scriptstyle{f_{1}\otimes(f_{2}\otimes(f_{3}\otimes
f_{4}))}$$\scriptstyle{\alpha}$$\scriptstyle{\mathrm{id}_{A_{1}}\otimes\alpha}$$\scriptstyle{(f_{1}\otimes
f_{2})\otimes(f_{3}\otimes
f_{4})}$$\scriptstyle{\alpha}$$\scriptstyle{((f_{1}\otimes f_{2})\otimes
f_{3})\otimes
f_{4}}$$\scriptstyle{\alpha\otimes\mathrm{id}_{A_{4}}}$$\scriptstyle{\alpha}$$\scriptstyle{\mathrm{id}_{B_{1}}\otimes\alpha}$$\scriptstyle{\alpha}$$\scriptstyle{\alpha}$$\scriptstyle{f_{1}\otimes((f_{2}\otimes
f_{3})\otimes
f_{4})}$$\scriptstyle{\alpha}$$\scriptstyle{(f_{1}\otimes(f_{2}\otimes
f_{3}))\otimes f_{4}}$$\scriptstyle{\alpha}$

The triangle axiom is given by the diagram:

${(A\otimes\mathbb{I})\otimes A^{\prime}}$${A\otimes(\mathbb{I}\otimes
A^{\prime})}$${A\otimes A^{\prime}}$${(B\otimes\mathbb{I})\otimes
B^{\prime}}$${B\otimes(\mathbb{I}\otimes B^{\prime})}$${B\otimes
B^{\prime}}$$\scriptstyle{(f\otimes\mathrm{id}_{\mathbb{I}})\otimes
f^{\prime}}$$\scriptstyle{\rho\otimes\mathrm{id}_{A^{\prime}}}$$\scriptstyle{\alpha}$$\scriptstyle{f\otimes(\mathrm{id}_{\mathbb{I}}\otimes
f^{\prime})}$$\scriptstyle{\mathrm{id}_{A}\otimes\lambda}$$\scriptstyle{f\otimes
f^{\prime}}$$\scriptstyle{\rho\otimes\mathrm{id}_{B^{\prime}}}$$\scriptstyle{\alpha}$$\scriptstyle{\mathrm{id}_{B}\otimes\lambda}$

Here the top and the bottom face commute because $\rho$ and $\alpha$ satisfy
the triangle identity in $\mathcal{C}$ and the two side faces commute due to
naturality of the left and right unitors in $\mathcal{C}$. Finally, the back
face commutes because of the naturality of the associator in $\mathcal{C}$. ∎

#### Proof of Theorem 2.24 continued:

###### Proof.

We will first show that the multiplication in the above construction satisfies
the pentagon axiom. For an arbitrary morphism $f:A\longrightarrow B$ between
two monoid objects $A$, $B$ $\in$ obj($\mathcal{C}$) the pentagon axiom in
$\mathcal{C}$ gives rise to the following diagram:

${(A\otimes A)\otimes A}$${A\otimes(A\otimes A)}$${A\otimes A}$${(B\otimes
B)\otimes B}$${B\otimes(B\otimes B)}$${B\otimes B}$${A\otimes
A}$${A}$${B\otimes B}$${B}$$\scriptstyle{(f\otimes f)\otimes
f}$$\scriptstyle{\alpha}$$\scriptstyle{\mu_{A}\otimes\mathrm{id}_{A}}$$\scriptstyle{f\otimes(f\otimes
f)}$$\scriptstyle{\mathrm{id}_{A}\otimes\mu_{A}}$$\scriptstyle{f\otimes
f}$$\scriptstyle{\mu_{A}}$$\scriptstyle{\alpha}$$\scriptstyle{\mu_{B}\otimes\mathrm{id}_{B}}$$\scriptstyle{\mathrm{id}_{B}\otimes\mu_{B}}$$\scriptstyle{\mu_{B}}$$\scriptstyle{f\otimes
f}$$\scriptstyle{\mu_{A}}$$\scriptstyle{f}$$\scriptstyle{\mu_{B}}$

The front and the back face commute because $A$ and $B$ are monoids in
$\mathcal{C}$. Moreover, the bottom and the right face commute because $f$ is
a morphism of monoids. The left face commutes because of the definition of the
monoidal product in Arr($\mathcal{C}$) and because $f$ is a morphism of
monoids. The left square in the top commutes because of naturality the
associator in $\mathcal{C}$ and the right square in the top commutes because
of the definition of the monoidal product in Arr($\mathcal{C}$) and because
$f$ is a morphism of monoids. Hence the whole diagram commutes and the
pentagon axiom is satisfied.

Moreover, we need to show that the unitor diagram commutes:

${\mathbb{I}\otimes_{\mathcal{C}}A}$${A\otimes_{\mathcal{C}}A}$${A\otimes_{\mathcal{C}}\mathbb{I}}$${A}$${\mathbb{I}\otimes_{\mathcal{C}}B}$${B\otimes_{\mathcal{C}}B}$${B\otimes_{\mathcal{C}}\mathbb{I}}$${B}$$\scriptstyle{\mathrm{id}_{\mathbb{I}}\otimes_{\mathcal{C}}f}$$\scriptstyle{\lambda_{A}}$$\scriptstyle{\eta_{A}\otimes_{\mathcal{C}}\mathrm{id}_{A}}$$\scriptstyle{f\otimes_{\mathcal{C}}f}$$\scriptstyle{\mu_{A}}$$\scriptstyle{\mathrm{id}_{A}\otimes_{\mathcal{C}}\eta_{A}}$$\scriptstyle{f\otimes_{\mathcal{C}}\mathrm{id}_{\mathbb{I}}}$$\scriptstyle{\rho_{A}}$$\scriptstyle{f}$$\scriptstyle{\lambda_{B}}$$\scriptstyle{\eta_{B}\otimes_{\mathcal{C}}\mathrm{id}_{B}}$$\scriptstyle{\mu_{B}}$$\scriptstyle{\mathrm{id}_{B}\otimes_{\mathcal{C}}\eta_{B}}$$\scriptstyle{\rho_{B}}$

The top and the bottom face commute because $A$ and $B$ are monoids in
$\mathcal{C}$. The left and the right face commute due to the definition of
left and right unitors in $\mathcal{C}$ and the definition of the monoidal
product in Arr($\mathcal{C}$). Because $f$ is a morphism of monoids, the inner
face also commutes. The two squares in the back commute as well, because of
the definition of the monoidal product in Arr($\mathcal{C}$) and because $f$
is a monoid morphism. Hence the whole diagram commutes.

Finally, commutativity means that the following diagram has to commute

${A\otimes_{\mathcal{C}}A}$${A\otimes_{\mathcal{C}}A}$${A}$${B\otimes_{\mathcal{C}}B}$${B\otimes_{\mathcal{C}}B}$${B}$$\scriptstyle{f\otimes_{\mathcal{C}}f}$$\scriptstyle{\mu_{A}}$$\scriptstyle{\sigma_{A,A}}$$\scriptstyle{f\otimes_{\mathcal{C}}f}$$\scriptstyle{\mu_{A}}$$\scriptstyle{f}$$\scriptstyle{\mu_{B}}$$\scriptstyle{\sigma_{B,B}}$$\scriptstyle{\mu_{B}}$

which holds if both $A$ and $B$ are commutative monoids in $\mathcal{C}$. ∎

## References

  * [1] John C. Baez and James Dolan.  Higher-dimensional algebra and topological quantum field theory.  Journal of Mathematical Physics, 36(11):6073–6105, nov 1995.
  * [2] C. Heunen J. Vicary.  Categories for Quantum Theory.  Oxford University Press, 2020.
  * [3] Steven Roman.  An Introduction to the Language of Category Theory.  Birkhäuser, 2017.
  * [4] David White and Donald Yau.  Arrow categories of monoidal model categories, 2018.
  * [5] Christopher J. Wood, Jacob D. Biamonte, and David G. Cory.  Tensor networks and graphical calculus for open quantum systems.  2011\.