Lemma 4.43.1. Let $(\mathcal{C}, \otimes , \phi )$ be as above. There is a 1-to-1 correspondence between units $(\mathbf{1}, l, r)$ in $\mathcal{C}$ and pairs $(\mathbf{1}, 1)$ 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.
4.43 Monoidal categories
Let $\mathcal{C}$ be a category. Suppose we are given a functor
\[ \otimes : \mathcal{C} \times \mathcal{C} \longrightarrow \mathcal{C} \]
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
\[ \phi _{X, Y, Z} : X \otimes (Y \otimes Z) \to (X \otimes Y) \otimes Z \]
such that for all objects $X, Y, Z, W$ the diagram
4.43.0.1
\begin{equation} \label{categories-equation-pentagon} \vcenter { \xymatrix{ X \otimes (Y \otimes ( Z \otimes W)) \ar[r] \ar[d] & X \otimes ((Y \otimes Z) \otimes W) \ar[d] \\ (X \otimes Y) \otimes (Z \otimes W) \ar[d] & (X \otimes (Y \otimes Z)) \otimes W \ar[ld] \\ ((X \otimes Y) \otimes Z) \otimes W } } \end{equation}
is commutative where every arrow is determined by a suitable application of $\phi $ and functoriality of $\otimes $.
In [Theorem 3.1, associativity] it is explained how a triple $(\mathcal{C}, \otimes , \phi )$ as a above is coherent. A reformulation is that such a triple gives rise to a system of well defined functors
\[ \mathcal{C} \times \ldots \times \mathcal{C} \longrightarrow \mathcal{C}, \quad (X_1, \ldots , X_ n) \longmapsto X_1 \otimes \ldots \otimes X_ n \]
for all $n \geq 1$ and for $n, m \geq 1$ functorial isomorphisms
\[ (X_1 \otimes \ldots \otimes X_ n) \otimes (Y_1 \otimes \ldots \otimes Y_ m) \to X_1 \otimes \ldots \otimes X_ n \otimes Y_1 \otimes \ldots \otimes Y_ m \]
such that all possible diagrams formed from these isomorphisms commute (the pentagram diagram above is an example). For $n = 5$ this means that the expressions
\[ \begin{matrix} (((X_1 \otimes X_2) \otimes X_3) \otimes X_4) \otimes X_5
& ((X_1 \otimes (X_2 \otimes X_3)) \otimes X_4) \otimes X_5
\\ ((X_1 \otimes X_2) \otimes (X_3 \otimes X_4)) \otimes X_5
& (X_1 \otimes ((X_2 \otimes X_3) \otimes X_4)) \otimes X_5
\\ (X_1 \otimes (X_2 \otimes (X_3 \otimes X_4))) \otimes X_5
& ((X_1 \otimes X_2) \otimes X_3) \otimes (X_4 \otimes X_5)
\\ (X_1 \otimes (X_2 \otimes X_3)) \otimes (X_4 \otimes X_5)
& (X_1 \otimes X_2) \otimes ((X_3 \otimes X_4) \otimes X_5)
\\ (X_1 \otimes X_2) \otimes (X_3 \otimes (X_4 \otimes X_5))
& X_1 \otimes (((X_2 \otimes X_3) \otimes X_4) \otimes X_5)
\\ X_1 \otimes ((X_2 \otimes (X_3 \otimes X_4)) \otimes X_5)
& X_1 \otimes ((X_2 \otimes X_3) \otimes (X_4 \otimes X_5))
\\ X_1 \otimes (X_2 \otimes ((X_3 \otimes X_4) \otimes X_5))
& X_1 \otimes (X_2 \otimes (X_3 \otimes (X_4 \otimes X_5)))
\end{matrix} \]
are all functorially isomorphic to $X_1 \otimes \ldots \otimes X_5$ and that these isomorphisms are all compatible with all possible isomorphisms one gets between these functors using $\phi $ in 3 consecutive spots (when possible). We will use this without further mention in the following and we will no longer write parentheses when writing iterations of $\otimes $.
A unit for a triple $(\mathcal{C}, \otimes , \phi )$ as above is an object $\mathbf{1}$ of $\mathcal{C}$ together with functorial isomorphisms
\[ l : \mathbf{1} \otimes X \to X \quad \text{and}\quad r : X \otimes \mathbf{1} \to X \]
such that for all objects $X, Y$ the diagram
4.43.0.2
\begin{equation} \label{categories-equation-triangle} \vcenter { \xymatrix{ X \otimes (\mathbf{1} \otimes Y) \ar[rr]_\phi \ar[rd]_{\text{id} \otimes l} & & (X \otimes \mathbf{1}) \otimes Y \ar[ld]^{r \otimes \text{id}} \\ & X \otimes Y } } \end{equation}
is commutative. We will often think of units as pairs $(\mathbf{1}, 1)$ as in the following lemma.
Proof. Given a unit $(\mathbf{1}, l, r)$ we get an isomorpism $r : \mathbf{1} \otimes \mathbf{1} \to \mathbf{1}$ and $L$ and $R$ are equivalences as they are isomorphic to the identity functor. Conversely, suppose given $(\mathbf{1}, 1)$ such that $L$ and $R$ are equivalences. We obtain functorial isomorphisms $l_ X : \mathbf{1} \otimes X \to X$ and $r_ X : X \otimes \mathbf{1} \to X$ characterized by $L(l_ X) = 1 \otimes \text{id}_ X$ and $R(r_ X) = \text{id}_ X \otimes 1$. Then we have to show that the two arrows $\text{id}_ X \otimes l_ Y$ and $r_ X \otimes \text{id}_ Y$ from $X \otimes \mathbf{1} \otimes Y$ to $X \otimes Y$ are the same for all $X$ and $Y$. This property only depends on the isomorphism classes of $X$ and $Y$. Since $R$ and $L$ are equivalences, it suffices to do this for $X = Z \otimes \mathbf{1}$ and $Y = \mathbf{1} \otimes W$ for some objects $Z$ and $W$. In other words, we have to show that
\[ \text{id}_ Z \otimes \text{id}_\mathbf {1} \otimes l_{\mathbf{1} \otimes W} = r_{Z \otimes \mathbf{1}} \otimes \text{id}_\mathbf {1} \otimes \text{id}_ Y \]
By construction these maps are equal to $\text{id}_ Z \otimes 1 \otimes \text{id}_\mathbf {1} \otimes \text{id}_ W$ and $\text{id}_ Z \otimes \text{id}_\mathbf {1} \otimes 1 \otimes \text{id}_ W$. Thus it suffices to show that $1 \otimes \text{id}_\mathbf {1} = \text{id}_\mathbf {1} \otimes 1$.
We have $\text{id}_\mathbf {1} \otimes 1 = r_\mathbf {1} \otimes \text{id}_\mathbf {1}$ and $l_\mathbf {1} \otimes \text{id}_\mathbf {1} = \text{id}_\mathbf {1} \otimes 1$. We may write $r_\mathbf {1} = a \circ 1$ and $l_\mathbf {1} = b \circ 1$ for some $a, b$ automorphisms of $\mathbf{1}$. Thus we have $\text{id}_\mathbf {1} \otimes 1 = (a \otimes \text{id}_\mathbf {1}) \circ (1 \otimes \text{id}_\mathbf {1})$ and $1 \otimes \text{id}_\mathbf {1} = (\text{id}_\mathbf {1} \otimes b) \circ (\text{id}_\mathbf {1} \otimes 1)$. Then we can write
\begin{align*} (1 \otimes \text{id}_\mathbf {1}) \circ (\text{id}_\mathbf {1} \otimes 1 \otimes \text{id}_\mathbf {1}) & = (1 \otimes \text{id}_\mathbf {1}) \circ (\text{id}_\mathbf {1} \otimes \text{id}_\mathbf {1} \otimes b) \circ (\text{id}_\mathbf {1} \otimes \text{id}_\mathbf {1} \otimes 1) \\ & = (\text{id}_\mathbf {1} \otimes b) \circ (1 \otimes \text{id}_\mathbf {1}) \circ (\text{id}_\mathbf {1} \otimes \text{id}_\mathbf {1} \otimes 1) \\ & = (\text{id}_\mathbf {1} \otimes b) \circ (1 \otimes 1) \end{align*}
and we also have
\begin{align*} (1 \otimes \text{id}_\mathbf {1}) \circ (\text{id}_\mathbf {1} \otimes 1 \otimes \text{id}_\mathbf {1}) & = (\text{id}_\mathbf {1} \otimes b) \circ (\text{id}_\mathbf {1} \otimes 1) \circ (a \otimes \text{id}_\mathbf {1} \otimes \text{id}_\mathbf {1}) \circ (1 \otimes \text{id}_\mathbf {1} \otimes \text{id}_\mathbf {1}) \\ & = (\text{id}_\mathbf {1} \otimes b) \circ (a \otimes \text{id}_\mathbf {1}) \circ (1 \otimes 1) \end{align*}
This proves that $a \otimes \text{id}_\mathbf {1}$ is the identity and hence $a$ is the identity as desired.
To finish the proof, we note that the rules above determine inverse equivalences of categories between the category of units (suitably defined) and the category of pairs $(\mathbf{1}, 1)$. $\square$
Lemma 4.43.2. Let $(\mathcal{C}, \otimes , \phi )$ be as above. Let $(\mathbf{1}, 1)$ be a unit (see Lemma 4.43.1). Then
$1 \otimes \text{id}_\mathbf {1} = \text{id}_\mathbf {1} \otimes 1$
$\Gamma = \text{Mor}(\mathbf{1}, \mathbf{1})$ is a commutative monoid,
$a = 1 \circ (a \otimes \text{id}_\mathbf {1}) \circ 1^{-1} = 1 \circ (\text{id}_\mathbf {1} \otimes a) \circ 1^{-1}$ for all $a \in \Gamma $,
any other unit is isomorphic to $(\mathbf{1}, 1)$ by a unique isomorphism.
Proof. Part (1) was shown in the proof of Lemma 4.43.1. For $a \in \Gamma $ we have
\begin{align*} (1 \circ (a \otimes \text{id}_\mathbf {1})) \otimes \text{id}_\mathbf {1} & = (1 \otimes \text{id}_\mathbf {1}) \circ (a \otimes \text{id}_\mathbf {1} \otimes \text{id}_\mathbf {1}) \\ & = (\text{id}_\mathbf {1} \otimes 1) \circ (a \otimes \text{id}_\mathbf {1} \otimes \text{id}_\mathbf {1}) \\ & = (a \otimes \text{id}_\mathbf {1}) \circ (\text{id}_\mathbf {1} \otimes 1) \\ & = (a \otimes \text{id}_\mathbf {1}) \circ (1 \otimes \text{id}_\mathbf {1}) \\ & = (a \circ 1) \otimes \text{id}_\mathbf {1} \end{align*}
Thus $1 \circ (a \otimes \text{id}_\mathbf {1}) = a \circ 1$ and this implies one half of (3). The other half follows in the same way. To see (2) we observe that for $a, b \in \Gamma $ we have
\[ a \circ b = 1 \circ (a \otimes \text{id}_\mathbf {1}) \circ 1^{-1} \circ 1 \circ (\text{id}_\mathbf {1} \otimes b) \circ 1^{-1} = 1 \circ (a \otimes b) \circ \circ 1^{-1} \]
and $b \circ a$ evaluates to the same expression. Thus (2) holds. To see (4) suppose that $(\mathbf{1}', 1')$ is a second unit. Using $r$ and $l'$ there are isomorphisms $\mathbf{1}' \otimes \mathbf{1} \to \mathbf{1}'$ and $\mathbf{1}' \otimes \mathbf{1} \to \mathbf{1}$. Thus there exists an isomorphism $t : \mathbf{1}' \to \mathbf{1}$. Then the diagram
\[ \xymatrix{ \mathbf{1}' \otimes \mathbf{1}' \ar[r]_{t \otimes t} \ar[d]_{1'} & \mathbf{1} \otimes \mathbf{1} \ar[d]^1 \\ \mathbf{1}' \ar[r]^ t & \mathbf{1} } \]
commutes up to an automorphism $a$ of $\mathbf{1}$. After replacing $t$ by $a \circ t$ the diagram will commute (hint: use (3) to see that $1 \circ (a \otimes a) = a^2 \circ 1$). $\square$
Definition 4.43.3. 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 $\mathbf{1}$.
We always write $\mathbf{1}$ to denote a unit of a monoidal category and we denote $1 : \mathbf{1} \otimes \mathbf{1} \to \mathbf{1}$ a chosen isomorphism; as the pair $(\mathbf{1}, 1)$ is determined up to unique isomorphism (Lemma 4.43.2) there is no harm in choosing one.
Lemma 4.43.4. In a monoidal category $\mathcal{C}, \otimes , \phi , \mathbf{1}, 1$ and with notation as in the proof of Lemma 4.43.1 we have
the arrows $1, r_\mathbf {1}, l_\mathbf {1} : \mathbf{1} \otimes \mathbf{1} \to \mathbf{1}$ agree,
the arrows $l_ X \otimes \text{id}_ Y, l_{X \otimes Y} : \mathbf{1} \otimes X \otimes Y \to X \otimes Y$ agree, and
the arrows $\text{id}_ X \otimes r_ Y , r_{X \otimes Y} : X \otimes Y \otimes \mathbf{1} \to X \otimes Y$ agree.
A monoidal category satisfies the assumptions of [Theorem 5.2, associativity].
Proof. We have seen (1) in the proof of Lemma 4.43.1. We have seen in the proof of Lemma 4.43.1 that $l_ X$ and $l_{X \otimes Y}$ are the unique morphisms such that $\text{id}_\mathbf {1} \otimes l_{X \otimes Y} = 1 \otimes \text{id}_{X \otimes Y}$ and $\text{id}_\mathbf {1} \otimes l_ X = 1 \otimes \text{id}_ X$. Part (2) follows immediately. Part (3) is proved in a similar manner. Jointly with the commutativity of (4.43.0.1) and (4.43.0.2) this means the final statement of the lemma holds. $\square$
In [Theorem 5.2, associativity] it is explained how a quintuple $(\mathcal{C}, \otimes , \phi , \mathbf{1}, 1)$ as in the lemma above is coherent. A reformulation is that in a monoidal category (in our sense) for every $n \geq i \geq 0$ we have functorial isomorphisms
\[ X_1 \otimes \ldots \otimes X_ i \otimes \mathbf{1} \otimes X_{i + 1} \otimes \ldots \otimes X_ n \to X_1 \otimes \ldots \otimes X_ n \]
such that all possible diagrams formed from these isomorphisms commute. So for example starting with two insertions of $\mathbf{1}$ and using the isomorphisms in different order would result in the same morphism. Besides the convention on removing parentheses when writing iterates of $\otimes $, from now on we identify $X \otimes \mathbf{1}$ and $\mathbf{1} \otimes X$ with $X$ without further mention. Moreover, we will say “let $\mathcal{C}$ be a monoidal category” with $\otimes , \phi , \mathbf{1}$ understood.
Definition 4.43.5. 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 an 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 $\mathcal{C}'$.
\[ F(X) \otimes F(Y) \to F(X \otimes Y) \]
\[ \xymatrix{ F(X) \otimes (F(Y) \otimes F(Z)) \ar[r] \ar[d] & F(X) \otimes F(Y \otimes Z) \ar[r] & F(X \otimes (Y \otimes Z)) \ar[d] \\ (F(X) \otimes F(Y)) \otimes F(Z) \ar[r] & F(X \otimes Y) \otimes F(Z) \ar[r] & F((X \otimes Y) \otimes Z) } \]
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 4.43.6. 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 4.43.7. 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 4.43.6 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 4.43.8. 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 $Y$.
\[ \vcenter { \xymatrix{ X \ar[rd]_1 \ar[r]_-{\eta \otimes 1} & X \otimes Y \otimes X \ar[d]^{1 \otimes \epsilon } \\ & X } } \quad \text{and}\quad \vcenter { \xymatrix{ Y \ar[rd]_1 \ar[r]_-{1 \otimes \eta } & Y \otimes X \otimes Y \ar[d]^{\epsilon \otimes 1} \\ & Y } } \]
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 4.43.9. Let $\mathcal{C}$ be a monoidal category. If $Y$ is a left dual to $X$, then functorially in $Z$ and $Z'$.
\[ \mathop{\mathrm{Mor}}\nolimits (Z' \otimes X, Z) = \mathop{\mathrm{Mor}}\nolimits (Z', Z \otimes Y) \quad \text{and}\quad \mathop{\mathrm{Mor}}\nolimits (Y \otimes Z', Z) = \mathop{\mathrm{Mor}}\nolimits (Z', X \otimes Z) \]
Proof. Consider the maps
\[ \mathop{\mathrm{Mor}}\nolimits (Z' \otimes X, Z) \to \mathop{\mathrm{Mor}}\nolimits (Z' \otimes X \otimes Y, Z \otimes Y) \to \mathop{\mathrm{Mor}}\nolimits (Z', Z \otimes Y) \]
where we use $\eta $ in the second arrow and the sequence of maps
\[ \mathop{\mathrm{Mor}}\nolimits (Z', Z \otimes Y) \to \mathop{\mathrm{Mor}}\nolimits (Z' \otimes X, Z \otimes Y \otimes X) \to \mathop{\mathrm{Mor}}\nolimits (Z' \otimes X, Z) \]
where we use $\epsilon $ in the second arrow. To show these arrows are mutually inverse, consider a map $a : Z' \to Z \otimes Y$. We have to show that
\[ Z' \xrightarrow {\text{id}_{Z'} \otimes \eta } Z' \otimes X \otimes Y \xrightarrow {a \otimes \text{id}_{X \otimes Y}} Z \otimes Y \otimes X \otimes Y \xrightarrow {\text{id}_ Z \otimes \epsilon \otimes \text{id}_ Y} Z \otimes Y \]
is equal to $a$. The composition of the first two arrows equals $(\text{id}_{Z \otimes Y} \otimes \eta ) \circ a : Z' \to Z \otimes Y \to Z \otimes Y \otimes X \otimes Y$. Then the composition of $\text{id}_{Z \otimes Y} \otimes \eta $ and $\text{id}_ Z \otimes \epsilon \otimes \text{id}_ Y$ equals the identity by definition of the dual. Similarly for the other composition. We omit the proof of the second equality. $\square$
Remark 4.43.10. Lemma 4.43.9 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 commutes. 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.
\[ \mathop{\mathrm{Mor}}\nolimits (Z' \otimes X, Z) \longrightarrow \mathop{\mathrm{Mor}}\nolimits (Z', Z \otimes Y) \]
\[ \eta _ Z : Z \to Z \otimes X \otimes Y \]
\[ \epsilon _{Z'} : Z' \otimes Y \otimes X \to Z' \]
\[ (\epsilon \otimes \text{id}_ Y) \circ \eta _ Y = \text{id}_ Y \quad \text{and}\quad \epsilon _ X \circ (\eta \otimes \text{id}_ X) = \text{id}_ X \]
\[ \xymatrix{ \mathop{\mathrm{Mor}}\nolimits (Z' \otimes X, Z) \ar[r] \ar[d]_{\text{id}_ W \otimes -} & \mathop{\mathrm{Mor}}\nolimits (Z', Z \otimes Y) \ar[d]^{\text{id}_ W \otimes -} \\ \mathop{\mathrm{Mor}}\nolimits (W \otimes Z' \otimes X, W \otimes Z) \ar[r] & \mathop{\mathrm{Mor}}\nolimits (W \otimes Z', W \otimes Z \otimes Y) } \]
Lemma 4.43.11. 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 4.43.10. $\square$
A commutativity constraint for $(\mathcal{C}, \otimes )$ is a functorial isomorphism
\[ \psi : X \otimes Y \longrightarrow Y \otimes X \]
such that the composition
\[ X \otimes Y \xrightarrow {\psi } Y \otimes X \xrightarrow {\psi } X \otimes Y \]
is the identity. We say $\psi $ is compatible with a given associativity constraint $\phi $ if for all objects $X, Y, Z$ the diagram
4.43.11.1
\begin{equation} \label{categories-equation-hexagon} \vcenter { \xymatrix{ X \otimes (Y \otimes Z) \ar[r]_\phi \ar[d]^\psi & (X \otimes Y) \otimes Z \ar[r]_\psi & Z \otimes (X \otimes Y) \ar[d]^\phi \\ X \otimes (Z \otimes Y) \ar[r]^\phi & (X \otimes Z) \otimes Y \ar[r]^\psi & (Z \otimes X) \otimes Y } } \end{equation}
commutes.
Definition 4.43.12. 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 and we may use the language and notation discussed above.
Lemma 4.43.13. In a symmetric monoidal category $\mathcal{C}, \otimes , \phi , \psi , \mathbf{1}, 1$ we have
the arrows $1 \circ \psi , 1 : \mathbf{1} \otimes \mathbf{1} \to \mathbf{1}$ agree,
the arrows $\text{id}_ X \otimes l_ Y, (l_ X \otimes \text{id}) \circ (\psi \otimes \text{id}_ Y): X \otimes \mathbf{1} \otimes Y \to X \otimes Y$ agree,
A symmetric monoidal category satisfies the assumptions of [Theorem 5.1, associativity].
Proof. We may write $\psi = a \otimes \text{id}_\mathbf {1}$ for a unique isomorphism $a : \mathbf{1} \to \mathbf{1}$. Lemma 4.43.2 implies that $a \otimes \text{id}_\mathbf {1} = \text{id}_\mathbf {1} \otimes a$. Functoriality of $\psi $ says that the diagram
\[ \xymatrix{ (\mathbf{1} \otimes \mathbf{1}) \otimes \mathbf{1} \ar[r]_\psi \ar[d]_{1 \otimes \text{id}_\mathbf {1}} & \mathbf{1} \otimes (\mathbf{1} \otimes \mathbf{1}) \ar[d]^{\text{id}_\mathbf {1} \otimes 1} \\ \mathbf{1} \otimes \mathbf{1} \ar[r]^\psi & \mathbf{1} \otimes \mathbf{1} } \]
commutes. Thus the top arrow is equal to $a \otimes \text{id}_\mathbf {1} \otimes \text{id}_\mathbf {1}$. Thus (4.43.11.1) for $X = Y = Z = \mathbf{1}$ says that $a \otimes \text{id}_\mathbf {1} \otimes \text{id}_\mathbf {1}$ is equal to its own square. Hence $a = \text{id}_\mathbf {1}$. This proves (1).
Part (2) states that $\psi : X \otimes \mathbf{1} \to \mathbf{1} \otimes X$ is the identity, if we identify the source and the target with $X$ in our monoidal category. This follows from the commutativity of (4.43.11.1) for $\mathbf{1}, X, \mathbf{1}$, namely
\[ \xymatrix{ \mathbf{1} \otimes (X \otimes \mathbf{1}) \ar[r]_\phi \ar[d]^\psi & (\mathbf{1} \otimes X) \otimes \mathbf{1} \ar[r]_\psi & \mathbf{1} \otimes (\mathbf{1} \otimes X) \ar[d]^\phi \\ \mathbf{1} \otimes (\mathbf{1} \otimes X) \ar[r]^\phi & (\mathbf{1} \otimes \mathbf{1}) \otimes X \ar[r]^\psi & (\mathbf{1} \otimes \mathbf{1}) \otimes X } \]
commutes and we know all but one of the morphisms $\psi $ in this diagram are equal to the identity.
In addition to (1) and (2) the commutativity of the diagrams (4.43.0.1), (4.43.0.2), (4.43.11.1) and the results of Lemma 4.43.4 imply the final statement of the lemma. $\square$
In [Theorem 5.1, associativity] it is explained how a sextuple $(\mathcal{C}, \otimes , \phi , \psi , \mathbf{1}, 1)$ as in the lemma above is coherent. A reformulation is that in a symmetric monoidal category (in our sense) for every $n \geq 1$ and permutation $\sigma $ of $\{ 1, \ldots , n\} $ we have functorial isomorphisms
\[ X_1 \otimes \ldots \otimes X_ n \to X_{\sigma (1)} \otimes \ldots \otimes X_{\sigma (n)} \]
such that all possible diagrams formed from these isomorphisms commute and these isomorphisms are compatible with the structure of a monoidal category (e.g., with the isomorphisms when we insert a $\mathbf{1}$ in a slot).
Lemma 4.43.14. 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 4.43.8. 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 4.43.15. 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.
\[ F(X) \otimes F(Y) \to F(X \otimes Y) \]
\[ \xymatrix{ F(X) \otimes F(Y) \ar[r] \ar[d] & F(X \otimes Y) \ar[d] \\ F(Y) \otimes F(X) \ar[r] & F(Y \otimes X) } \]
Remark 4.43.16. 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.
\[ \mathop{\mathrm{Mor}}\nolimits (X, hom(Y, Z)) = \mathop{\mathrm{Mor}}\nolimits (X \otimes Y, Z) \]
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
\[ hom(Y, Z) \otimes hom(X, Y) \otimes X \longrightarrow Z \]
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
\[ Z \otimes hom(X, Y) \otimes X \to Z \otimes Y \]
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
\[ hom(Y, Z) \otimes X \otimes hom(X, Y) \longrightarrow Z \]
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 $hom(Y, Z) \otimes Y \to Z$.
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (2)
Comment #6068 by Reid Barton on
Comment #6197 by Johan on