For an object of a braided monoidal category, take its self-braiding
It is invertible by definition. Suppress canonical associators using the monoidal coherence theorem, and put , . The hexagon identity gives
Apply naturality of this braiding to the morphism in its second argument. It says
Substitution gives
This 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.

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