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An object that is both a monoid object and a comonoid in a braided monoidal category, with $\Delta ,\epsilon$ preserving multiplication and unit. The product on its tensor square uses the ambient braiding.

The braided monoidal category with objects $n\ge 0$, endomorphism groups $B_n$, no arrows between unequal objects, tensor given by juxtaposition, and block-crossing braiding. Its tensor is strict.

A monoidal category with natural invertible braidings $c_{X,Y}:X\otimes Y\to Y\otimes X$ satisfying the two hexagon axioms. The equation $c_{Y,X}{c}_{X,Y}=1$ is an additional symmetric condition, not a braided axiom.

A monoidal functor between braided monoidal categories compatible with their braidings: $F(c_{X,Y})F_2(X,Y)=F_2(Y,X)c_{FX,FY}$.

The braided monoidal category generated by one object: objects are parenthesized tensor expressions in that object and the unit, and morphisms are structural isomorphisms and crossings subject to precisely the monoidal category and braiding 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.

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.

For an object $X$ of a braided monoidal category, take its self-braidingIt is invertible by definition. Suppress canonical associators using the monoidal coherence theorem, and put $R_{12}=y\otimes 1$, $R_{23}=1\otimes y$. The hexagon identity givesApply naturality of this braiding to the morphism $y:X\otimes X\to X\otimes X$ in its second argument. It saysSubstitution givesThis is the equation for a Yang–Baxter operator. Restoring the uniquely determined associators gives the non-strict diagram in the paper. Every object therefore has the canonical Yang–Baxter operator supplied by its self-braiding.

We use right comodules and write their coactions as $u\mapsto \sum u_{(0)}\otimes u_{(1)}$. A coquasitriangular structure gives the braiding on these comodulesIn an arbitrary symmetric monoidal category this notation abbreviates a composite of the two coactions, the ambient symmetry, and $\gamma$; it does not assume that the objects have elements. The coquasitriangular axioms ensure that this composite is a comodule morphism, is invertible using the convolution inverse of $\gamma$, and satisfies the two hexagon laws. More explicitly, in scalar notation those laws come fromwhile the comodule-morphism condition isUnit normalizations give the unit constraints. These descriptions are identities of morphisms in the ambient symmetric category, with the displayed reordering carried by its symmetry.

Apply the preceding self-braiding result to the regular right comodule $(H,\Delta )$. Its Yang–Baxter operator isCompose its braid equation with the three counits. Expanding $R_{23}{R}_{12}{R}_{23}$ and cancelling the leading counit factors givesExpanding $R_{12}{R}_{23}{R}_{12}$ instead givesThey are equal by the Yang–Baxter operator equation. This is the required scalar identity:Indeed, after ${\Delta }^{\otimes 3}$ the ordered factors are $a_{(1)},a_{(2)},b_{(1)},b_{(2)},c_{(1)},c_{(2)}$. The two ambient symmetries on the left produce the three pairings $(a_{(1)},b_{(1)})$, $(a_{(2)},c_{(1)})$, $(b_{(2)},c_{(2)})$. On the right, the central symmetry produces $a_{(1)},a_{(2)},b_{(2)},b_{(1)},c_{(1)},c_{(2)}$, yielding exactly the other three pairings. Hence the calculation identifies the actual morphisms requested, also when the category is not a category of vector spaces.

For the given bimonoids, use right comodules. The corestriction functor for comodules associated to a comonoid morphism $f:H\to K$ keeps underlying objects and morphisms, and replaces a coaction ${\rho }_M$ by $(1\otimes f){\rho }_M$. The comonoid-morphism axioms ensure that this is a $K$-coaction.

The tensor coaction for two $H$-comodules in the ambient braided monoidal category isand the unit coaction is $j_H:I\to H$. If $f$ is also a monoid morphism, then $f{m}_H=m_K(f\otimes f)$ and $f{j}_H=j_K$. Substituting these identities, and using naturality of the ambient braiding, shows that corestriction preserves both tensor and unit coactions exactly. Its structural maps are identities, so it is a strict monoidal functor.

Conversely, suppose this induced functor is strict monoidal. Apply equality of the tensor coactions to the two regular right comodules $(H,{\Delta }_H)$. Then apply ${\epsilon }_H\otimes {\epsilon }_H$ to their two underlying $H$ factors. The counit laws and naturality of the braiding remove those factors and leaveEquality on the unit comodule similarly gives $f{j}_H=j_K$. Thus $f$ is a monoid morphism. The criterion is exactlyThe regular-comodule argument uses only the counit laws; it requires no elementwise or finite-dimensional assumption on the ambient category.