Yang–Baxter operators on an object

Let $\mathcal{C}$ be a monoidal category and let $X$ be an object of $\mathcal{C}$. Write $X^{\otimes 3}$ for the triple tensor product and write $1_X^{\otimes k}$ for the $k$-fold tensor power of ${\mathrm{id}}_X$. The equation below is interpreted in a strict model of $\mathcal{C}$: fix a monoidal equivalence from $\mathcal{C}$ to a strict monoidal category, as in Mac Lane strictification, with tensor constraint $J_{A,B}:E(A)\otimes E(B)\to E(A\otimes B)$. Put $X^{\mathrm{\prime }}=E(X)$ and $R^{\mathrm{\prime }}=J_{X,X}^{-1}E(R)J_{X,X}$, and read the displayed equation with $X^{\mathrm{\prime }},R^{\mathrm{\prime }}$ in that strict category. When $\mathcal{C}$ is strict the display below is literal.

A Yang–Baxter operator on $X$ is an invertible morphism $R\text{\hspace{0.17em}⁣}:X\otimes X\to X\otimes X$ such that

$(R\otimes 1_X)(1_X\otimes R)(R\otimes 1_X)=(1_X\otimes R)(R\otimes 1_X)(1_X\otimes R)\mathrm{.}$

Both sides are endomorphisms of $X^{\otimes 3}$ in the strict model, so the display is a well-formed equality of morphisms of the strict model; transported back along the equivalence it is a well-formed statement about $X$. It is the Yang–Baxter equation, and an invertible solution $R$ is also called an $R$-matrix on $X$ in the categorical sense. Invertibility is part of the data: a solution of the cubic equation that is not invertible is not a Yang–Baxter operator in this sense.

In a strict braided monoidal category the braiding gives the basic example $R=c_{X,X}$ on any object $X$ (Braided monoidal category): the Yang–Baxter equation for the braiding is In a strict braided monoidal category the braiding satisfies the Yang-Baxter equation, and $c_{X,X}$ is invertible with inverse $c_{X,X}^{-1}$ because each component of a braiding is an isomorphism. In a general braided monoidal category the same example is read in a strict model via the braided strictification of the category.