Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 122 / 4 / a

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.

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 $X$.