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. 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...

$\newcommand{\M}{\mathcal{M}}\newcommand{\B}{\mathfrak{B}}\newcommand{\hom}{\operatorname{Hom}}\newcommand{\BM}{\mathsf{BM}}\newcommand{\SBM}{\mathsf{SBM}}\newcommand{\S}{\mathcal{S}}$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 $\hom_{\SBM}(\B,\S)\cong\S$?


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 $\epsilon:e'\to F(e)$ 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: $$F(\alpha)\circ\mu\circ(1\otimes\mu):F(x)\otimes(F(y)\otimes F(z))\to F(x)\otimes F(y\otimes z)\to F(x\otimes (y\otimes z))\to F((x\otimes y)\otimes z)$$

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.

  • 1
    $\begingroup$ Generally speaking you should pick one site to post to first and then wait at least a few days before cross-posting. That said, I think aside from cross-posting issues this is an appropriate MO question. Certainly what's written in CWM is pretty confusing. (For example, he doesn't seem to assume that the functor is strict in the proof because he explicitly discusses F_2 and F_0.) $\endgroup$ Oct 20, 2022 at 19:12
  • $\begingroup$ @NoahSnyder Noted. Though he does later assume the $F$ have to be strict, e.g. when he writes statements similar to: "$F(\sigma)$ must be the braiding $\gamma$..." $\endgroup$
    – FShrike
    Oct 20, 2022 at 19:52

1 Answer 1


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.

  • $\begingroup$ Thank you for this, I need a moment to parse this fully. Are you saying that Mac Lane should have not bothered with the category of strict monoidal functors at all? $\endgroup$
    – FShrike
    Oct 20, 2022 at 21:52
  • $\begingroup$ To elaborate on my first comment: you seem to be showing that $Hom_{BM}(B,M)\simeq Hom_{BM}(B,S)\simeq S_0\simeq M$ directly, without even needing to pass to $Hom_{SBM}(B,S)$ $\endgroup$
    – FShrike
    Oct 20, 2022 at 22:22
  • $\begingroup$ Accepted. Thank you for pointing out explicitly that we identify $G\cong F_{G(1)}$ $\endgroup$
    – FShrike
    Oct 21, 2022 at 10:08

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