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The braid group $B_n$ has generators ${\sigma }_1,\dots ,{\sigma }_{n-1}$ and relationsIn the strict monoidal category $\mathcal{C}$, put $X^{\otimes 0}=I$ and sendThe images are invertible because the Yang–Baxter operator $y$ is invertible. Distant images commute by the tensor interchange law, and adjacent images satisfy the braid group relations by the Yang–Baxter operator equation tensored with the remaining identities. Thus these assignments define group homomorphisms $B_n\to \mathrm{Aut}(X^{\otimes n})$, and hence a functor $\mathbf{B}\to \mathcal{C}$.

Juxtaposition of braids puts generators into the corresponding two blocks of tensor factors, so their images are the tensor products of their images. Strictness of $\mathcal{C}$ gives $X^{\otimes (m+n)}=X^{\otimes m}\otimes X^{\otimes n}$ and $X^{\otimes 0}=I$ exactly. The resulting functor is strict monoidal and sends $1$ to $X$.

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 a two-body operator $R(u)$ acting on a tensor product of particle species spaces, the spectral equation is $R_{12}(u)R_{13}(u+v)R_{23}(v)=R_{23}(v)R_{13}(u+v)R_{12}(u)$. It guarantees agreement between the two ways to reorder three particles in factorized scattering. A Yang–Baxter operator is a related braid-form operator; multiplying by the permutation operator converts between the braided and unbraided conventions.