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
\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
\xymatrix{ X \otimes (Y \otimes ( Z \otimes W)) \ar[r] \ar[d] & (X \otimes Y) \otimes (Z \otimes W) \ar[r] & ((X \otimes Y) \otimes Z) \otimes W \\ X \otimes ((Y \otimes Z) \otimes W) \ar[rr] & & (X \otimes (Y \otimes Z)) \otimes W \ar[u] }
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
\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.
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
\mathbf{1} \otimes X \to X \quad \text{and}\quad X \otimes \mathbf{1} \to X
such that for all objects X, Y the diagram
\xymatrix{ X \otimes (\mathbf{1} \otimes Y) \ar[rr]_\phi \ar[rd] & & (X \otimes \mathbf{1}) \otimes Y \ar[ld] \\ & X \otimes Y }
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
F(X) \otimes F(Y) \to F(X \otimes Y)
functorial in X and Y such that for all objects X, Y, and Z the diagram
\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) }
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
\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 } }
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
\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)
functorially in Z and 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. 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
\mathop{\mathrm{Mor}}\nolimits (Z' \otimes X, Z) \longrightarrow \mathop{\mathrm{Mor}}\nolimits (Z', Z \otimes Y)
functorial in both Z and Z'. The unit of the adjunction produces maps
\eta _ Z : Z \to Z \otimes X \otimes Y
functorial in Z and the counit of the adjoint produces maps
\epsilon _{Z'} : Z' \otimes Y \otimes X \to Z'
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
(\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
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
\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) }
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
\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
\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 }
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
F(X) \otimes F(Y) \to F(X \otimes Y)
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
\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) }
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
\mathop{\mathrm{Mor}}\nolimits (X, hom(Y, Z)) = \mathop{\mathrm{Mor}}\nolimits (X \otimes Y, Z)
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
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.
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