This is my first, and probably my last, (for a while) posting on MO. I am very much a student and I don't claim to be a research mathematician, at all, but I have seen that sometimes "regular" MSE users ask questions here if they feel their question is too obscure to receive a good answer on MSE.


I think my question is such a question; the original MSE question has been cross-posted [here](https://math.stackexchange.com/q/4557529/815585). If this is not appropriate, I will readily delete this post on MO. I apologise if the answer is very trivial, but I just don't see it. I've also spotted a fair few mistakes in Mac Lane's book before, so I'm partly motivated to ask here in case Mac Lane's approach is just plain wrong...




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$\newcommand{\M}{\mathcal{M}}\newcommand{\B}{\mathfrak{B}}\newcommand{\hom}{\operatorname{Hom}}\newcommand{\BM}{\mathsf{BM}}\newcommand{\SBM}{\mathsf{SBM}}\newcommand{\S}{\mathcal{S}}$I refer to the chapter: "Symmetry and braiding in monoidal categories" from [CWM](http://www.mtm.ufsc.br/%7Eebatista/2016-2/maclanecat.pdf).


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.
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Mac Lane proves this theorem by instead saying:



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> 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: $$\hom\_{\SBM}(\B,\S)\cong\S$$
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> 
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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:




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I define "braiding" via the [nLab definition](https://ncatlab.org/nlab/show/braiding), since [Mac Lane's definitions are unfortunately outdated](https://math.stackexchange.com/a/537648/815585).


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 $\epsilon:e'\to F(e)$ that satisfy certin axioms to follow. Mac Lane calls $\mu$ by "$F\_2$", and $\epsilon$ by "$F\_0$". The axioms:



> 
> For all $x,y,z\in\M$, the composite: $$F(\alpha)\circ\mu\circ(1\otimes\mu):F(x)\otimes(F(y)\otimes F(z))\to F(x)\otimes F(y\otimes z)\to F(x\otimes (y\otimes z))\to F((x\otimes y)\otimes z)$$
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> Is equal to: $$\mu\circ(\mu\otimes1)\circ\alpha:F(x)\otimes (F(y)\otimes F(z))\to (F(x)\otimes F(y))\otimes F(z)\to F(x\otimes y)\otimes F(z)\to F((x\otimes y)\otimes z)$$
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> 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: $$\rho'=F(\rho)\circ\mu\circ(1\otimes\epsilon):F(x)\otimes e'\to F(x)\otimes F(e)\to F(x\otimes e)\to F(x)$$
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> We also require: $$\mu\circ\gamma':F(x)\otimes F(y)\to F(y)\otimes F(x)\to F(y\otimes x)$$To equal: $$F(\gamma)\circ\mu:F(x)\otimes F(y)\to F(x\otimes y)\to F(y\otimes x)$$
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> 


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.




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Why is the category of strong braided functors from the braid category to a braided monoidal $M$ equivalent to the subcategory of *strict* functors?
It is indeed true that any strong braided monoidal functor from the braid category to a strict braided monoidal category is equivalent to a strict braided monoidal functor, however that comes out of the proof.


MacLane's argument is the following.


(1) There is an equivalence


$$Hom\_{BM}(\mathfrak{B}, M) \simeq Hom\_{BM}(\mathfrak{B}, S).$$


[this follows for exactly the reason explained by the OP]


(2) there is a functor $S\_0 \to Hom\_{BM}(\mathfrak{B}, S)$, where $S\_0$ is the underlying category of $S$.


This takes an object $a$ of $S\_0$ and constructs a functor $F\_a$ whose value on $n \in \mathfrak{B}$ is $F\_a(n) = a^{\otimes n}$. This makes sense because $S$ is strict, and we have to verify that it is a (strong) monoidal functor. This is what MacLane does on page 264 and the very top of 265.


It turns out that $F\_a$ is actually a *strict* braided monoidal functor.


(3) There is also a functor back $ Hom\_{BM}(\mathfrak{B}, S) \to S\_0$ which is "evaluate at 1".


(4) The composite $S\_0 \to Hom\_{BM}(\mathfrak{B}, S) \to S\_0$ is the identity. Thus to show that $ Hom\_{BM}(\mathfrak{B}, S) \to S\_0$ is an equivalence, it suffices to show that it is fully-faithful.


This is what is proven on the rest of page 265, however I think there are some typos. The two functors $F$ and $G$ should be *strong* braided monoidal functors, despite the fact that MacLane writes *strict*. I think the argument that is written there only uses that they are strong monoidal. This is clear if you allow yourself the usual coherence theorem for non-braided monoidal categories (which gives the morphisms labeled "$F\_w$" and "$G\_w$").


So in the end we prove that every strong monoidal fucntor is equivalent to one in the image of $S\_0$, and in particular to a strict monoidal functor. Specifically the strong monoidal functor $G$ would be equivalent to $F\_{G(1)}$. Note, however, that the underlying functor of $F\_{G(1)}$ is *not* the same as $G$. The "strictification" potentially changes the value of the functor on objects.


