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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.

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.