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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 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 natural tensor interchange isomorphism in a braided monoidal category, constrained by the two hexagon axioms.

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.

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.

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.