= Solution
Define the length of a formal tensor expression recursively by
$$
|I|=0,\qquad |x|=1,\qquad |u\otimes v|=|u|+|v|.
$$
The canonical <strict monoidal functor> $Q:\mathcal Fbr\to\mathbf B$ 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, and
$$
Q(u\otimes v)=Qu+Qv,\qquad QI=0
$$
on 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> $B_n\to\operatorname{Aut}(x^{[n]})$ and a <functor> $S:\mathbf B\to\mathcal Fbr$.
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 isomorphism
$$
\theta_u:x^{[|u|]}\longrightarrow u
$$
obtained by rebracketing and inserting the units appearing in $u$. These maps form a <natural isomorphism> $SQ\Rightarrow1_{\mathcal Fbr}$. 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.
\b[Thus $Q$ has a specified quasi-inverse and is the required equivalence:]
$$
\boxed{Q:\mathcal Fbr\simeq\mathbf B\quad\text{is braided and strict monoidal}.}
$$
Only ordinary <monoidal coherence>, together with the defining <braiding> axioms, has been used; a separate braided coherence theorem is unnecessary.
Back to article page
