= Solution
For an object $X$ of a <braided monoidal category>, take its self-<braiding>
$$
\boxed{y=c_{X,X}:X\otimes X\longrightarrow X\otimes X.}
$$
It is invertible by definition. Suppress canonical <associators> using the <monoidal coherence theorem>, and put $R_{12}=y\otimes1$, $R_{23}=1\otimes y$. The hexagon identity gives
$$
c_{X,X\otimes X}=R_{23}R_{12}.
$$
Apply <naturality> of this <braiding> to the morphism $y:X\otimes X\to X\otimes X$ in its second argument. It says
$$
c_{X,X\otimes X}(1\otimes y)=(y\otimes1)c_{X,X\otimes X}.
$$
Substitution gives
$$
\boxed{R_{23}R_{12}R_{23}=R_{12}R_{23}R_{12}.}
$$
This is the equation for a <Yang–Baxter operator>. Restoring the uniquely determined <associators> gives the non-strict diagram in the paper. \b[Every object therefore has the canonical Yang–Baxter operator supplied by its self-braiding.]
Back to article page
