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The structure map of a comodule; a right coaction satisfies $(\rho \otimes 1)\rho =(1\otimes \Delta )\rho$ with coherent parentheses, and $(1\otimes \epsilon )\rho$ is the inverse right unitor.

The structure map $\epsilon :C\to I$ of a comonoid, with $(\epsilon \otimes 1)\Delta$ and $(1\otimes \epsilon )\Delta$ equal to the inverse unitors. This is distinct from the counit of an adjunction.

A category equipped with a monoidal tensor product, a monoidal unit object, and natural associators and unitors satisfying the pentagon and triangle axioms.

Every diagram formed solely from the canonical associators, unitors and their inverses commutes. Consequently structural reparenthesizations can be suppressed in calculations; this statement does not make distinct braidings equal.

Here the unqualified term allows lax comparison maps $F_2:FX\otimes FY\to F(X\otimes Y)$ and $F_0:I\to FI$, natural and compatible with associators and unitors. Invertible comparison maps give a strong monoidal functor. The opposite direction gives an opmonoidal functor.

The distinguished object $I$ in a monoidal category, with the natural unitors $I\otimes X\cong X\cong X\otimes I$.

An object with multiplication $m:A\otimes A\to A$ and unit $j:I\to A$ satisfying associativity and unit diagrams using the ambient associators and unitors.

For the Category of sets, take the monoidal tensor product to be the Cartesian product, $A\otimes B=A\times B$, with unit a singleton $1$. On morphisms use $(f\times g)(a,b)=(f(a),g(b))$. Define the associator by $((a,b),c)\mapsto (a,(b,c))$ and the unitors by $(\ast ,a)\mapsto a$ and $(a,\ast )\mapsto a$. These are natural isomorphisms, with inverses obtained by rebracketing or adjoining $\ast$. Both paths in the pentagon send $(((a,b),c),d)$ to $(a,(b,(c,d)))$, and both paths in the triangle send $((a,\ast ),b)$ to $(a,b)$. This proves the axioms elementwise.

There is a second structure: use disjoint union $A⨿B$ and unit $\varnothing$. Define the associator by preserving each element and its summand while changing the nested tags, and define the unitors by removing the empty summand. These are natural and invertible. In the pentagon, either path takes an element in any of the four summands to that same element in the same summand of $A⨿(B⨿(C⨿D))$; the triangle removes an empty intermediate summand on either path. Thus both axioms hold.

These product and coproduct monoidal structures on sets have different units, $1$ and $\varnothing$, which are not isomorphic. They are even inequivalent as monoidal categories: an equivalence of the underlying Category of sets preserves terminal objects, so it cannot take the singleton unit of the product structure to the empty unit of the coproduct structure. Hence the monoidal structure on sets is not unique.

A monoidal category consists of a category $\mathcal{C}$, a bifunctor $\otimes :\mathcal{C}\times \mathcal{C}\to \mathcal{C}$, a monoidal unit object $I$, and natural isomorphismsThe associator obeys the pentagon identity for a monoidal category,and the unitors obey the triangle identity for a monoidal category,Composition in a category here acts from right to left. The first equation compares two paths from $((A\otimes B)\otimes C)\otimes D$ to $A\otimes (B\otimes (C\otimes D))$; the second compares two ways of removing the intervening unit.

It is precisely a bicategory with one object $\ast$. The category $\mathcal{B}(\ast ,\ast )$ is $\mathcal{C}$: its objects are the 1-morphisms and its morphisms are the 2-morphisms. Horizontal composition in a category is $\otimes$, the identity 1-morphism is $I$, and the associator and unitors give the weak associativity and unit constraints. Vertical composition in a category is ordinary composition in a category in $\mathcal{C}$; the interchange law follows because $\otimes$ is a bifunctor. The bicategory pentagon and triangle axioms are exactly the two equations above. Conversely the sole hom-category of any one-object bicategory gives these data, with the convention that horizontal composition in a category uses the indicated ordering of tensor factors.

A monoidal category consists of a category $\mathcal{C}$, a bifunctor $\otimes :\mathcal{C}\times \mathcal{C}\to \mathcal{C}$, a monoidal unit object $I$, and natural isomorphismsThe associator and unitors satisfy the pentagon identity for a monoidal categoryand the triangle identity for a monoidal categoryComposition is written from right to left. Naturality of these isomorphisms is part of the definition.

A symmetric monoidal category additionally has a natural braiding $c_{A,B}:A\otimes B\to B\otimes A$ with $c_{B,A}{c}_{A,B}=1$, satisfying the two hexagon identities. Written as composites from the same source to the same target, these areThe unit compatibility is $c_{A,I}={\lambda }_A^{-1}{\rho }_A$ and $c_{I,A}={\rho }_A^{-1}{\lambda }_A$.

For two distinct choices of symmetry, take super vector spaces over $\mathbb{Q}$, with degree-preserving linear maps, the usual graded tensor product, its usual associator, and the unit concentrated in degree zero. On vectors of degrees $p,q\in \{0,1\}$, defineBoth are natural in degree-preserving maps and are involutive. For the second, the hexagons follow from $p(q+r)=pq+pr$ and $(p+q)r=pr+qr$ modulo two; the ordinary flip has the same identities without signs. Degree zero of the unit gives unit compatibility. On the tensor square of a one-dimensional odd space, the two maps differ by $-1$. Thus the same monoidal structure can carry two different symmetries; the second uses the Koszul sign rule.

We prove the monoidal coherence theorem by a terminating normalization argument. Its precise formal assertion is that any two composites of associators, unitors, their inverses, and their tensor products with identities, between the same formal tensor expressions, are equal. The expressions must have the same ordered list of object letters after unit symbols are removed. This does not assert that different symmetries are equal, nor does it treat an accidental equality of actual objects as a new structural identification.

Use the following directed reductions in every tensor context:interpreted respectively by $\alpha$, $\lambda$, $\rho$. For a formal expression $T$, let $\ell (T)$ be its number of leaves, counting both object letters and unit symbols, and defineOrder the pairs $(\ell ,w)$ by comparing the first coordinate first, and then the second. A unit reduction decreases the first coordinate. An associativity reduction preserves the first coordinate and decreases the second by $\ell (A)>0$. The same calculation holds in any context, since associativity preserves the size of the rewritten subtree. Hence there is no infinite directed reduction path.

An irreducible expression has no unit next to another tensor factor and no composite left tensor factor. It is therefore the right-associated expression of its ordered nonunit letters, or the single unit $I$ if there are no such letters. In particular the normal form is uniquely determined by the ordered word.

We next show that every pair of first reductions has a commuting completion. Reductions in disjoint subexpressions commute by bifunctoriality of the monoidal tensor product. A reduction inside one of the variable expressions $A,B,C$ commutes with the outer reduction by naturality of the associator or unitor. The genuine overlaps are exactly the following cases.

For $(((A\otimes B)\otimes C)\otimes D)$, associating first at the root or first in its left subtree gives the two sides of the pentagon; their completions to $A\otimes (B\otimes (C\otimes D))$ commute.

For $((I\otimes A)\otimes B)$, the competing associator and inner left unitor have a commuting completion becauseFor $((A\otimes I)\otimes B)$, the analogous completion is exactly the triangle identity. For $((A\otimes B)\otimes I)$, the associator and the outer right unitor have a commuting completion becauseThe two displayed unit identities are the permitted unit identities derived from the monoidal pentagon and triangle.

The final overlap is $I\otimes I$, where both unitors apply. They agree as a consequence of the same identities. Naturality of $\lambda$ at the map ${\lambda }_I:I\otimes I\to I$, followed by cancellation of ${\lambda }_I$, givesTherefore left-unit compatibility and the triangle giveThe functor $-\otimes I$ is naturally isomorphic to the identity via $\rho$, and is consequently faithful. Thus ${\lambda }_I={\rho }_I$, so this overlap commutes too. There are no other nonvariable overlaps among the three reduction patterns.

Termination and these commuting completions imply equality of all normalization arrows, not merely uniqueness of the normal-form object. Here is the induction. Suppose every expression smaller than $T$ in the decreasing pair measure has a unique normalization arrow. Two normalization paths from $T$ start with arrows $r:T\to T_1$ and $s:T\to T_2$. The local completion gives directed paths $d_1:T_1\to V$, $d_2:T_2\to V$ with $d_{1}r=d_{2}s$. Append any normalization ${\nu }_V$ of $V$. By the induction hypothesis the tails of the original paths are ${\nu }_V{d}_1$ and ${\nu }_V{d}_2$. Their composites from $T$ are equal. If $T$ is already normal, its only directed normalization is the identity. This proves the induction.

Write ${\nu }_T:T\to N(T)$ for the resulting unique normalization. For every directed structural generator $g:S\to T$, including a generator in an arbitrary tensor context, uniqueness gives ${\nu }_{T}g={\nu }_S$. Since all structural generators are invertible, the corresponding equation holds for their inverses as well. Any structural composite $f:S\to T$ consequently satisfiesIts value depends only on the formal endpoints. Thus every structural diagram with the specified ordered word commutes, proving the monoidal coherence theorem.

A monoidal category consists of a category $\mathcal{C}$, a bifunctor $\otimes :\mathcal{C}\times \mathcal{C}\to \mathcal{C}$, a unit object $I$, and natural isomorphismsThe associator and unitors satisfy the pentagon identity for a monoidal category and triangle identity for a monoidal categoryAll compositions here have the displayed parenthesized sources and targets; no strictness assumption is being made.

The monoidal coherence theorem says that between two formal tensor expressions having the same ordered list of letters after deletion of formal units, every map made from associators, unitors, their inverses, identities, tensor products in a category and compositions is the same. Thus every diagram of such structural maps commutes. It does not permit permutation of letters or identify different additional maps such as braidings.

We prove this by canonical normalization. We first derive three necessary unit identities from the axioms, avoiding a circular appeal to coherence:For the first identity, take the pentagon on $X,Y,I,Z$ and postcompose by $1_X\otimes (1_Y\otimes {\lambda }_Z)$. Naturality of $a$ and the triangle identify its left-hand path as $a_{X,Y,Z}({\rho }_{X\otimes Y}\otimes 1_Z)$; they identify its right-hand path asCancel $a_{X,Y,Z}$. Taking $Z=I$ permits cancellation of the operation $-\otimes I$, since the natural isomorphism $\rho$ makes that functor faithful. This proves the first identity. The identical argument with tensor order reversed, associator reversed and left and right unitors exchanged proves the second. Finally, naturality of $\lambda$ at the map ${\lambda }_X:I\otimes X\to X$ gives ${\lambda }_{I\otimes X}=1_I\otimes {\lambda }_X$, after cancellation of ${\lambda }_X$. Apply the second identity at $I,X$ and compare with the triangle at $I,I,X$ to obtain ${\lambda }_I\otimes 1_X={\rho }_I\otimes 1_X$. Set $X=I$ and use the same faithfulness to conclude ${\lambda }_I={\rho }_I$. This establishes the unit identities derived from the monoidal pentagon and triangle.

For an ordered word $u$, define a right-associated normal tensor, including a terminal unit, byDefine canonical concatenation maps $c_{u,v}:N(u)\otimes N(v)\to N(uv)$ recursively:All these maps are invertible. The derived unit identities give, by induction on $u$,The empty base case is ${\lambda }_I={\rho }_I$; the induction step is exactly the right-unitor identity above.

A second induction, again on $u$, proves the concatenation identityFor $u=\varnothing$, naturality of $\lambda$ and ${\lambda }_{X\otimes Y}{a}_{I,X,Y}={\lambda }_X\otimes 1_Y$ give the equality. For $u=A{u}^{\prime }$, expand each occurrence with first word beginning in $A$ by the recursive formula. Naturality moves the maps $c_{u^{\prime },v}$ and $c_{v,w}$ through the associators. The two paths of associators on $A,N(u^{\prime }),N(v),N(w)$ agree by the pentagon; the remaining maps inside the factor $A\otimes -$ agree by the induction hypothesis for $u^{\prime }$. This proves the concatenation identity for every triple of words, including empty words.

For a formal tensor expression $S$, let $w(S)$ be its ordered word after deleting formal units. Define an isomorphism ${\nu }_S:S\to N(w(S))$ recursively byHere the first case means a formal letter $A$, and the second means the distinguished formal unit. Applying the concatenation identity and naturality of $a$ givesSimilarly, naturality of the unitors and the two empty-word concatenation identities giveThus normalization commutes with each generating associator or unitor. It also commutes with their inverses. The recursive tensor formula shows the same for a generating map applied inside any larger tensor expression, and composition preserves the property. Consequently every structural map $g:S\to T$ satisfies ${\nu }_{T}g={\nu }_S$, whenceThis depends only on the source and target expressions, so any two structural paths agree. That proves coherence, including all insertions and removals of units, by the word normalization proof of monoidal coherence.

Let $x$ denote the generating object, the image of the unique object of the terminal category. The free braided monoidal category on one object can be described syntactically. Its objects are all fully parenthesized expressions built from $x$, the monoidal unit object $I$, and a binary monoidal tensor product. Thus $x\otimes (I\otimes x)$ and $(x\otimes I)\otimes x$ are distinct objects, though canonically isomorphic. Its morphisms are generated by the associators, unitors, braidings and their inverses, closed under composition and tensor product, subject to the pentagon, triangle, naturality and hexagon axioms. No symmetry relation is imposed on the braiding. The embedding of the terminal category selects $x$.

The braid category $\mathbf{B}$ has objects the nonnegative integers, withwhere $B_n$ is the braid group on $n$ strands and $B_0,B_1$ are trivial. Composition is stacking braids, with the first morphism followed by the second. The monoidal tensor product is addition on objects and side-by-side juxtaposition of braids; its monoidal unit object is $0$. The associators and unitors are identities, so it is a strict monoidal category.

Choose the positive crossing convention once and for all. Its braiding $c_{m,n}:m+n\to n+m$ is the block braid moving the first $m$ strands over the next $n$ strands while retaining the order within each block. In particular $c_{1,1}={\sigma }_1$. The usual braid group relations express the naturality and hexagon laws for these block braids. The generating functor $1\to \mathbf{B}$ selects $1$.

The distinction is parenthesized tensor expressions versus strand counts: the first construction keeps the structural isomorphisms visible, while $\mathbf{B}$ has strict tensor arithmetic. These descriptions give the two requested free constructions without needing to establish their universal properties.

Define the length of a formal tensor expression recursively byThe canonical strict monoidal functor $Q:\mathcal{F}br\to \mathbf{B}$ sends $u$ to $|u|$, sends every associator and unitor to an identity braid, and sends $c_{u,v}$ to the block braiding $c_{|u|,|v|}$. The pentagon and triangle become identity equations; the braiding axioms become the corresponding block-braid equations. Thus the assignment respects the defining relations, andon the nose. It is a braided monoidal functor with identity comparison maps.

To prove that $Q$ is an equivalence of categories, choose a standard parenthesization $x^{[n]}$ of $n$ copies of $x$, with $x^{[0]}=I$. On $x^{[n]}$, define the image of ${\sigma }_i$ by canonically exposing the $i$th and $(i+1)$st factors, applying $c_{x,x}$ there, and restoring the chosen parentheses. The monoidal coherence theorem makes this independent of the structural rebracketing. Crossings on disjoint pairs commute by the tensor interchange law. The adjacent braid group relations follows from the hexagon laws and naturality of the braiding, as in the Yang–Baxter calculation below. Hence these assignments give group homomorphisms $B_n\to \mathrm{Aut}(x^{[n]})$ and a functor $S:\mathbf{B}\to \mathcal{F}br$.

Equip $S$ with the canonical rebracketing maps $x^{[m]}\otimes x^{[n]}\to x^{[m+n]}$. Their monoidal functor axioms follow from the monoidal coherence theorem; the block-braiding compatibility follows by iterating the two hexagon laws. Therefore $S$ is a strong monoidal functor compatible with the braiding.

We have $QS=1_{\mathbf{B}}$, including its comparison maps. For every formal expression $u$, there is a canonical structural isomorphismobtained by rebracketing and inserting the units appearing in $u$. These maps form a natural isomorphism $SQ\Rightarrow 1_{\mathcal{F}br}$. To check naturality, it suffices to check the generating morphisms: for associators and unitors it is exactly monoidal coherence; for $c_{u,v}$ it follows from the hexagon expansion into the elementary crossings defining $S$. Composition and tensor product then preserve the equation. The same structural coherence shows that $\theta$ is monoidal.

Thus $Q$ has a specified quasi-inverse and is the required equivalence:Only ordinary monoidal coherence, together with the defining braiding axioms, has been used; a separate braided coherence theorem is unnecessary.

An opmonoidal functor $F:(\mathcal{C},\otimes ,I)\to (\mathcal{D},\otimes ,J)$ consists of a functor, a natural transformationand a morphism $F_0:FI\to J$. These maps need not be invertible. If $\alpha ,\lambda ,\rho$ are the respective associators and unitors, its axioms areThe first equation has domain $F((X\otimes Y)\otimes Z)$ and codomain $FX\otimes (FY\otimes FZ)$, which fixes the direction of every arrow. This is also called a colax monoidal functor.

An opmonoidal natural transformation $\tau :F\Rightarrow G$ between opmonoidal functors is a natural transformation satisfyingIt respects both the tensor and unit comparison maps. These are commutative diagrams and a unit commutative diagram; the arrow directions are opposite to those for a lax monoidal functor.

An opmonoidal monad on a monoidal category is a monad $(T,\eta ,\mu )$ whose endofunctor is an opmonoidal functor and whose unit and multiplication of a monad are opmonoidal natural transformations. Suppress only the canonical parentheses. Explicitly,The composite opmonoidal functor $T^2$ has tensor comparison $T_2(TX,TY)T(T_2(X,Y))$ and unit comparison $T_{0}T(T_0)$, explaining the last two equations.

For two algebras for a monad $(X,a)$ and $(Y,b)$, defineLet $c=(a\otimes b)T_2(X,Y)$. The unit law for a monad algebra follows at once from the opmonoidality of $\eta$:For the multiplication law, naturality of $T_2$, the algebra laws, and the opmonoidality of $\mu$ giveThe two unit-comparison equations above similarly make $(I,T_0)$ a algebra for a monad. If $f,g$ are morphisms of algebras for a monad, naturality of $T_2$ shows that $f\otimes g$ is an algebra morphism.

For a third algebra $(Z,d)$, the base associator is also an algebra morphism: its intertwining equation is precisely the opmonoidal associativity axiom, followed by $a\otimes b\otimes d$. The two base unitors are algebra morphisms by the opmonoidal unit axioms. Their pentagon and triangle commute because they commute after the faithful forgetful functor, and the lifted maps have exactly the same underlying morphisms.

The Eilenberg-Moore category is therefore monoidal, with these lifted constraints. Its forgetful functor $U:{\mathcal{C}}^T\to \mathcal{C}$ preserves the tensor product, unit object and constraints exactly, so it is a strict monoidal functor.

On the monoidal category of modules over the commutative ring $k$, consider the monad $T=H{\otimes }_k-$ coming from the unit and multiplication of the bialgebra $H$. Its opmonoidal functor structure has comparison mapsHere and below Sweedler notation abbreviates the comultiplication $\Delta h=\sum h_{(1)}\otimes h_{(2)}$. The opmonoidal associativity and unit axioms are the coassociativity and counit laws of the coalgebra. The unit and multiplication of a monad are opmonoidal natural transformations because the bialgebra axioms sayThe Eilenberg-Moore category of this opmonoidal monad is the category of left $H$-modules: a monad-algebra map $H{\otimes }_{k}M\to M$ is exactly a unital associative action.

Applying the preceding construction gives the diagonal bialgebra action and the unit actionThe usual associators and unitors for the tensor product of modules are $H$-linear, and the underlying tensor product is exactly ${\otimes }_k$. The forgetful functor into $k$-modules is strict monoidal. The base need not be a field; the modules need be neither flat modules nor finitely generated modules.

By the monoidal coherence theorem, calculations may suppress the canonical associators and unitors, restoring them uniquely afterwards. In this notation the two Frobenius monoidal functor identities readWriteWe check both snake identities for this prospective dual pair in a monoidal category. For $FX$, expand and use the first Frobenius identity:The third line uses naturality of $\psi$ and $\phi$. The last line uses $(e\otimes 1)(1\otimes n)=1_X$, followed by the opmonoidal and monoidal unit axioms.

For $FY$, the second Frobenius identity gives the other calculation:Thus both triangular identities hold, andare the evaluation morphism and coevaluation morphism of a dual pair in a monoidal category. Notice that none of the comparison maps was assumed invertible.

The two triangular identities $(e\otimes 1_X)(1_X\otimes n)=1_X$ and $(1_Y\otimes e)(n\otimes 1_Y)=1_Y$, with canonical associators and unitors inserted, for a dual pair in a monoidal category.

A monoidal category whose associators and unitors are identities; the tensor and unit laws are then equalities of objects and morphisms.

The associator and unitors of a monoidal category satisfyThus the two ways to remove a unit between two tensor factors agree. Together with the pentagon identity for a monoidal category this implies all unit identities derived from the monoidal pentagon and triangle and the monoidal coherence theorem.