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Define the length of a formal tensor expression recursively byThe canonical strict monoidal functor $Q:\mathcal{F}br\to \mathbf{B}$ sends $u$ to $|u|$, sends every associator and unitor to an identity braid, and sends $c_{u,v}$ to the block braiding $c_{|u|,|v|}$. The pentagon and triangle become identity equations; the braiding axioms become the corresponding block-braid equations. Thus the assignment respects the defining relations, andon the nose. It is a braided monoidal functor with identity comparison maps.

To prove that $Q$ is an equivalence of categories, choose a standard parenthesization $x^{[n]}$ of $n$ copies of $x$, with $x^{[0]}=I$. On $x^{[n]}$, define the image of ${\sigma }_i$ by canonically exposing the $i$th and $(i+1)$st factors, applying $c_{x,x}$ there, and restoring the chosen parentheses. The monoidal coherence theorem makes this independent of the structural rebracketing. Crossings on disjoint pairs commute by the tensor interchange law. The adjacent braid group relations follows from the hexagon laws and naturality of the braiding, as in the Yang–Baxter calculation below. Hence these assignments give group homomorphisms $B_n\to \mathrm{Aut}(x^{[n]})$ and a functor $S:\mathbf{B}\to \mathcal{F}br$.

Equip $S$ with the canonical rebracketing maps $x^{[m]}\otimes x^{[n]}\to x^{[m+n]}$. Their monoidal functor axioms follow from the monoidal coherence theorem; the block-braiding compatibility follows by iterating the two hexagon laws. Therefore $S$ is a strong monoidal functor compatible with the braiding.

We have $QS=1_{\mathbf{B}}$, including its comparison maps. For every formal expression $u$, there is a canonical structural isomorphismobtained by rebracketing and inserting the units appearing in $u$. These maps form a natural isomorphism $SQ\Rightarrow 1_{\mathcal{F}br}$. To check naturality, it suffices to check the generating morphisms: for associators and unitors it is exactly monoidal coherence; for $c_{u,v}$ it follows from the hexagon expansion into the elementary crossings defining $S$. Composition and tensor product then preserve the equation. The same structural coherence shows that $\theta$ is monoidal.

Thus $Q$ has a specified quasi-inverse and is the required equivalence:Only ordinary monoidal coherence, together with the defining braiding axioms, has been used; a separate braided coherence theorem is unnecessary.

By the monoidal coherence theorem, calculations may suppress the canonical associators and unitors, restoring them uniquely afterwards. In this notation the two Frobenius monoidal functor identities readWriteWe check both snake identities for this prospective dual pair in a monoidal category. For $FX$, expand and use the first Frobenius identity:The third line uses naturality of $\psi$ and $\phi$. The last line uses $(e\otimes 1)(1\otimes n)=1_X$, followed by the opmonoidal and monoidal unit axioms.

For $FY$, the second Frobenius identity gives the other calculation:Thus both triangular identities hold, andare the evaluation morphism and coevaluation morphism of a dual pair in a monoidal category. Notice that none of the comparison maps was assumed invertible.

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.