# In the coherence theorem for braided monoidal categories, why does it suffice to show the result for strict monoidal categories only?

$$\newcommand{\M}{\mathcal{M}}\newcommand{\B}{\mathfrak{B}}\newcommand{\hom}{\operatorname{Hom}}\newcommand{\BM}{\mathsf{BM}}\newcommand{\SBM}{\mathsf{SBM}}\newcommand{\S}{\mathcal{S}}$$I refer to the chapter: "Symmetry and braiding in monoidal categories" from CWM.

I've just finished this chapter, but I am unsatisfied with the statements concerning braided coherence.

The so-called "braided coherence theorem":

For any braided monoidal $$\M$$, $$\hom_{\BM}(\B,\M)\simeq\M$$Via an equivalence that assigns an $$F$$ in the LHS to the object $$F(1)$$ in the RHS.

Mac Lane proves this theorem by instead saying:

We know any braided monoidal $$\M$$ is equivalent to a strict braided monoidal $$\S$$ via functors which are strong monoidal in both directions. We will show then that: $$\hom_{\SBM}(\B,\S)\cong\S$$

I don't see why this is sufficient. It seems as if Mac Lane is implying: $$\hom_{\BM}(\B,\M)\simeq\hom_{\BM}(\B,\S)\simeq\hom_{\SBM}(\B,\S)$$If this is true, then his proof would indeed be sufficient. While I can believe the first equivalence, as we need only post-compose functors on either side with the equivalences $$\M\to\S,\,\S\to\M$$, I can't quite believe the second. I can't believe that we can easily promote a strong braided monoidal functor $$F:\B\to\S$$ to a strict $$F'$$, as the axiom on a monoidal natural isomorphism would here be (see the definitions below) - since $$\mu_2=1$$ by strictness - $$\theta\otimes\theta=\theta\circ\mu_1$$. But the obvious choice to make an equivalence of categories would just put $$F'$$ as the same functor $$F$$, but with a different $$\mu$$. So $$\theta$$ would be chosen to be the identity transformation (what other choice is there, that I'm missing?) and the axiom for it to be a monoidal transformation would not be satisfied in general.

Question: Why is it sufficient to prove $$\hom_{\SBM}(\B,\S)\cong\S$$?

Definitions:

I define "braiding" via the nLab definition, since Mac Lane's definitions are unfortunately outdated.

If $$\M,\M'$$ are braided categories, with associators, left and right unitors, unit and braidings $$(\alpha,\lambda,\rho,e,\gamma)$$ and $$(\alpha',\lambda',\rho',e',\gamma')$$ respectively, then we say: $$F:\M\to\M'$$ is a strong braided functor when there is a natural isomorphism: $$\mu:F(-)\otimes F(-)\implies F(-\otimes-)$$ and an isomorphism $$\epsilon:e'\to F(e)$$ which satisfy:

For all $$x,y,z\in\M$$, the composite: $$F(\alpha)\circ\mu\circ(1\otimes\mu):F(x)\otimes(F(y)\otimes F(z))\to F(x)\otimes F(y\otimes z)\to F(x\otimes (y\otimes z))\to F((x\otimes y)\otimes z)$$

Is equal to: $$\mu\circ(\mu\otimes1)\circ\alpha:F(x)\otimes (F(y)\otimes F(z))\to (F(x)\otimes F(y))\otimes F(z)\to F(x\otimes y)\otimes F(z)\to F((x\otimes y)\otimes z)$$

So that $$F$$ associates. Moreover we require: $$\lambda'=F(\lambda)\circ\mu\circ(\epsilon\otimes1):e'\otimes F(x)\to F(e)\otimes F(x)\to F(e\otimes x)\to F(x)$$And: $$\rho'=F(\rho)\circ\mu\circ(1\otimes\epsilon):F(x)\otimes e'\to F(x)\otimes F(e)\to F(x\otimes e)\to F(x)$$

We also require: $$\mu\circ\gamma':F(x)\otimes F(y)\to F(y)\otimes F(x)\to F(y\otimes x)$$To equal: $$F(\gamma)\circ\mu:F(x)\otimes F(y)\to F(x\otimes y)\to F(y\otimes x)$$

A natural transformation $$\theta:(F,\mu_1,\epsilon_1)\implies(G,\mu_2,\epsilon_2)$$ of (braided) monoidal functors is said to be monoidal when: $$\theta\circ\mu_1:F(x)\otimes F(y)\to F(x\otimes y)\to G(x\otimes y)$$Equals: $$\mu_2\circ(\theta\otimes\theta):F(x)\otimes F(y)\to G(x)\otimes G(y)\to G(x\otimes y)$$

Let $$\B$$ be the braid category. Its object class is $$\Bbb N_0$$, and the arrow class $$\B(n,m)$$ is empty if $$n\neq m$$. When $$n=m$$, $$\B(n,n)$$ is the $$n$$th Artin braid group (if $$n=0$$, we leave $$\B(0,0)=\{1\}$$) with the same composition and identities. $$\B$$ is a strict braided monoidal category, through the product $$n\otimes m=n+m$$, and: $$f\otimes g:n\otimes m\to n'\otimes m'$$ is defined to be the braiding which is $$f$$ on the first $$n$$ strings and $$g$$ on the subsequent $$m$$ strings (lay them side by side). The unit is $$0$$, and the braiding on $$\B$$, $$\gamma:n\otimes m\to m\otimes n$$ assigns to every braid the 'swizzled' (wording my own) braid where the first $$n$$ strings are braided to the final $$n$$ strings in $$m+n$$, and the final $$m$$ strings in $$n+m$$ are braided to the first $$m$$ strings.

Now, we denote for any braided monoidal $$\M$$ the category of strong braided monoidal functors $$\B\to\M$$ as: $$\hom_{\BM}(\B,\M)$$, and if $$\S$$ is a strict braided monoidal category then $$\hom_{\SBM}(\B,\S)$$ denotes the category of strict braided monoidal functors; in both, the arrows are the monoidal natural transformations.

• Cross-posted on MO because I feel my question might be too obscure - as some of my previous questions seemingly have been - to receive an answer (in timely fashion) on MSE. Oct 20, 2022 at 17:03