 Solution

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.