Definition 4.43.1. A triple $(\mathcal{C}, \otimes , \phi )$ where $\mathcal{C}$ is a category, $\otimes : \mathcal{C} \times \mathcal{C} \to \mathcal{C}$ is a functor, and $\phi $ is an associativity constraint is called a *monoidal category* if there exists a unit $\mathbf{1}$.

## 4.43 Monoidal categories

Let $\mathcal{C}$ be a category. Suppose we are given a functor

We often want to know whether $\otimes $ satisfies an associative rule and whether there is a unit for $\otimes $.

An *associativity constraint* for $(\mathcal{C}, \otimes )$ is a functorial isomorphism

such that for all objects $X, Y, Z, W$ the diagram

is commutative where every arrow is determined by a suitable application of $\phi $ and functoriality of $\otimes $. Given an associativity constraint there are well defined functors

for all $n \geq 1$.

Let $\phi $ be an associativity constraint. A *unit* for $(\mathcal{C}, \otimes , \phi )$ is an object $\mathbf{1}$ of $\mathcal{C}$ together with functorial isomorphisms

such that for all objects $X, Y$ the diagram

is commutative where the diagonal arrows are given by the isomorphisms introduced above.

An equivalent definition would be that a unit is a pair $(\mathbf{1}, 1)$ where $\mathbf{1}$ is an object of $\mathcal{C}$ and $1 : \mathbf{1} \otimes \mathbf{1} \to \mathbf{1}$ is an isomorphism such that the functors $L : X \mapsto \mathbf{1} \otimes X$ and $R : X \mapsto X \otimes \mathbf{1}$ are equivalences. Certainly, given a unit as above we get the isomorphism $1 : \mathbf{1} \otimes \mathbf{1} \to \mathbf{1}$ for free and $L$ and $R$ are equivalences as they are isomorphic to the identity functor. Conversely, given $(\mathbf{1}, 1)$ such that $L$ and $R$ are equivalences, we obtain functorial isomorphisms $l : \mathbf{1} \otimes X \to X$ and $r : X \otimes \mathbf{1} \to X$ characterized by $L(l) = 1 \otimes \text{id}_ X$ and $R(r) = \text{id}_ X \otimes 1$. Then we can use $r$ and $l$ in the notion of unit as above.

A unit is unique up to unique isomorphism if it exists (exercise).

We always write $\mathbf{1}$ to denote a unit of a monoidal category; as it is determined up to unique isomorphism there is no harm in choosing one. From now on we no longer write the brackets when taking tensor products in monoidal categories and we always identify $X \otimes \mathbf{1}$ and $\mathbf{1} \otimes X$ with $X$. Moreover, we will say “let $\mathcal{C}$ be a monoidal category” with $\otimes , \phi , \mathbf{1}$ understood.

Definition 4.43.2. Let $\mathcal{C}$ and $\mathcal{C}'$ be monoidal categories. A *functor of monoidal categories* $F : \mathcal{C} \to \mathcal{C}'$ is given by a functor $F$ as indicated and a isomorphism

functorial in $X$ and $Y$ such that for all objects $X$, $Y$, and $Z$ the diagram

commutes and such that $F(\mathbf{1})$ is a unit in $\mathcal{C}'$.

By our conventions about units, we may always assume $F(\mathbf{1}) = \mathbf{1}$ if $F$ is a functor of monoidal categories. As an example, if $A \to B$ is a ring homomorphism, then the functor $M \mapsto M \otimes _ A B$ is functor of monoidal categories from $\text{Mod}_ A$ to $\text{Mod}_ B$.

Lemma 4.43.3. Let $\mathcal{C}$ be a monoidal category. Let $X$ be an object of $\mathcal{C}$. The following are equivalent

the functor $L : Y \mapsto X \otimes Y$ is an equivalence,

the functor $R : Y \mapsto Y \otimes X$ is an equivalence,

there exists an object $X'$ such that $X \otimes X' \cong X' \otimes X \cong \mathbf{1}$.

**Proof.**
Assume (1). Choose $X'$ such that $L(X') = \mathbf{1}$, i.e., $X \otimes X' \cong \mathbf{1}$. Denote $L'$ and $R'$ the functors corresponding to $X'$. The equation $X \otimes X' \cong \mathbf{1}$ implies $L \circ L' \cong \text{id}$. Thus $L'$ must be the quasi-inverse to $L$ (which exists by assumption). Hence $L' \circ L \cong \text{id}$. Hence $X' \otimes X \cong \mathbf{1}$. Thus (3) holds.

The proof of (2) $\Rightarrow $ (3) is dual to what we just said.

Assume (3). Then it is clear that $L'$ and $L$ are quasi-inverse to each other and it is clear that $R'$ and $R$ are quasi-inverse to each other. Thus (1) and (2) hold. $\square$

Definition 4.43.4. Let $\mathcal{C}$ be a monoidal category. An object $X$ of $\mathcal{C}$ is called *invertible* if any (or all) of the equivalent conditions of Lemma 4.43.3 hold.

Observe that if $F : \mathcal{C} \to \mathcal{C}'$ is a functor of monoidal categories, then $F$ sends invertible objects to invertible objects.

Definition 4.43.5. Given a monoidal category $(\mathcal{C}, \otimes , \phi )$ and an object $X$ a *left dual* is an object $Y$ together with morphisms $\eta : \mathbf{1} \to X \otimes Y$ and $\epsilon : Y \otimes X \to \mathbf{1}$ such that the diagrams

commute. In this situation we say that $X$ is a *right dual* of $Y$.

Observe that if $F : \mathcal{C} \to \mathcal{C}'$ is a functor of monoidal categories, then $F(Y)$ is a left dual of $F(X)$ if $Y$ is a left dual of $X$.

Lemma 4.43.6. Let $\mathcal{C}$ be a monoidal category. If $Y$ is a left dual to $X$, then

functorially in $Z$ and $Z'$.

**Proof.**
Consider the maps

where we use $\eta $ in the second arrow and the sequence of maps

where we use $\epsilon $ in the second arrow. A straightforward calculation using the properties of $\eta $ and $\epsilon $ shows that the compositions of these are mutually inverse. Similarly for the other equality. $\square$

Remark 4.43.7. Lemma 4.43.6 says in particular that $Z \mapsto Z \otimes Y$ is the right adjoint of $Z' \mapsto Z' \otimes X$. In particular, uniqueness of adjoint functors guarantees that a left dual of $X$, if it exists, is unique up to unique isomorphism. Conversely, assume the functor $Z \mapsto Z \otimes Y$ is a right adjoint of the functor $Z' \mapsto Z' \otimes X$, i.e., we're given a bijection

functorial in both $Z$ and $Z'$. The unit of the adjunction produces maps

functorial in $Z$ and the counit of the adjoint produces maps

functorial in $Z'$. In particular, we find $\eta = \eta _\mathbf {1} : \mathbf{1} \to X \otimes Y$ and $\epsilon = \epsilon _\mathbf {1} : Y \otimes X \to \mathbf{1}$. As an exercise in the relationship between units, counits, and the adjunction isomorphism, the reader can show that we have

However, this isn't enough to show that $(\epsilon \otimes \text{id}_ Y) \circ (\text{id}_ Y \otimes \eta ) = \text{id}_ Y$ and $(\text{id}_ X \otimes \epsilon ) \circ (\eta \otimes \text{id}_ X) = \text{id}_ X$, because we don't know in general that $\eta _ Y = \text{id}_ Y \otimes \eta $ and we don't know that $\epsilon _ X = \epsilon \otimes \text{id}_ X$. For this it would suffice to know that our adjunction isomorphism has the following property: for every $W, Z, Z'$ the diagram

If this holds, we will say *the adjunction is compatible with the given tensor structure*. Thus the requirement that $Z \mapsto Z \otimes Y$ be the right adjoint of $Z' \mapsto Z' \otimes X$ compatible with the given tensor structure is an equivalent formulation of the property of being a left dual.

Lemma 4.43.8. Let $\mathcal{C}$ be a monoidal category. If $Y_ i$, $i = 1, 2$ are left duals of $X_ i$, $i = 1, 2$, then $Y_2 \otimes Y_1$ is a left dual of $X_1 \otimes X_2$.

**Proof.**
Follows from uniqueness of adjoints and Remark 4.43.7.
$\square$

A *commutativity constraint* for $(\mathcal{C}, \otimes )$ is a functorial isomorphism

such that the composition

is the identity. We say $\psi $ is *compatible* with a given associativity constraint $\phi $ if for all objects $X, Y, Z$ the diagram

commutes.

Definition 4.43.9. A quadruple $(\mathcal{C}, \otimes , \phi , \psi )$ where $\mathcal{C}$ is a category, $\otimes : \mathcal{C} \otimes \mathcal{C} \to \mathcal{C}$ is a functor, $\phi $ is an associativity constraint, and $\psi $ is a commutativity constraint compatible with $\phi $ is called a *symmetric monoidal category* if there exists a unit.

To be sure, if $(\mathcal{C}, \otimes , \phi , \psi )$ is a symmetric monoidal category, then $(\mathcal{C}, \otimes , \phi )$ is a monoidal category.

Lemma 4.43.10. Let $(\mathcal{C}, \otimes , \phi , \psi )$ be a symmetric monoidal category. Let $X$ be an object of $\mathcal{C}$ and let $Y$, $\eta : \mathbf{1} \to X \otimes Y$, and $\epsilon : Y \otimes X \to \mathbf{1}$ be a left dual of $X$ as in Definition 4.43.5. Then $\eta ' = \psi \circ \eta : \mathbf{1} \to Y \otimes X$ and $\epsilon ' = \epsilon \circ \psi : X \otimes Y \to \mathbf{1}$ makes $X$ into a left dual of $Y$.

**Proof.**
Omitted. Hint: pleasant exercise in the definitions.
$\square$

Definition 4.43.11. Let $\mathcal{C}$ and $\mathcal{C}'$ be symmetric monoidal categories. A *functor of symmetric monoidal categories* $F : \mathcal{C} \to \mathcal{C}'$ is given by a functor $F$ as indicated and an isomorphism

functorial in $X$ and $Y$ such that $F$ is a functor of monoidal categories and such that for all objects $X$ and $Y$ the diagram

commutes.

Remark 4.43.12. Let $\mathcal{C}$ be a monoidal category. We say $\mathcal{C}$ has an *internal hom* if for every pair of objects $X, Y$ of $\mathcal{C}$ there is an object $hom(X, Y)$ of $\mathcal{C}$ such that we have

functorially in $X, Y, Z$. By the Yoneda lemma the bifunctor $(X, Y) \mapsto hom(X, Y)$ is determined up to unique isomorphism if it exists. Given an internal hom we obtain canonical maps

$hom(X, Y) \otimes X \to Y$,

$hom(Y, Z) \otimes hom(X, Y) \to hom(X, Z)$,

$Z \otimes hom(X, Y) \to hom(X, Z \otimes Y)$,

$Y \to hom(X, Y \otimes X)$, and

$hom(Y, Z) \otimes X \to hom(hom(X, Y), Z)$ in case $\mathcal{C}$ is symmetric monoidal.

Namely, the map in (1) is the image of $\text{id}_{hom(X, Y)}$ by $\mathop{\mathrm{Mor}}\nolimits (hom(X, Y), hom(X, Y)) \to \mathop{\mathrm{Mor}}\nolimits (hom(X, Y) \otimes X, Y)$. To construct the map in (2) by the defining property of $hom(X, Z)$ we need to construct a map

and such a map exists since by (1) we have maps $hom(X, Y) \otimes X \to Y$ and $hom(Y, Z) \otimes Y \to Z$. To construct the map in (3) by the defining property of $hom(X, Z \otimes Y)$ we need to construct a map

for which we use $\text{id}_ Z \otimes a$ where $a$ is the map in (1). To construct the map in (4) we note that we already have the map $Y \otimes hom(X, X) \to hom(X, Y \otimes X)$ by (3). Thus it suffices to construct a map $\mathbf{1} \to hom(X, X)$ and for this we take the element in $\mathop{\mathrm{Mor}}\nolimits (\mathbf{1}, hom(X, X))$ corresponding to the canonical isomorphism $\mathbf{1} \otimes X \to X$ in $\mathop{\mathrm{Mor}}\nolimits (\mathbf{1} \otimes X, X)$. Finally, we come to (5). By the universal property of $hom(hom(X, Y), Z)$ it suffices to construct a map

We do this by swapping the last two tensor products using the commutativity constraint and then using the maps $hom(X, Y) \otimes X \to Y$ and $hom(Y, Z) \otimes Y \to Z$.

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