# The Unapologetic Mathematician

## The Braided Coherence Theorem

I find myself up earlier than expected this morning, so I thought I’d bang out the proof I’d promised of the “coherence theorem” for braided monoidal categories.

Recall that we’re considering a braided monoidal category $\mathcal{M}$ with underlying “ordinary category” $\mathcal{M}_0$ — the same category, just forgetting that it’s braided and monoidal. Then the statement is that the category of braided monoidal functors from $\mathcal{B}raid$ to $\mathcal{M}$ is equivalent to $\mathcal{M}_0$.

To make our lives a bit easier, we’ll use the strictification theorem to pass from the (weak) monoidal category $\mathcal{M}$ to a monoidally equivalent strict monoidal category $\mathcal{S}$, with underlying category $\mathcal{S}_0$. Now the braiding on $\mathcal{M}$ induces (via the equivalence) one on $\mathcal{S}$. The new statement is that the category of strictly monoidal functors from $\mathcal{B}raid$ to $\mathcal{S}$ is isomorphic to $\mathcal{S}_0$. Then the result we really want (about $\mathcal{M}$) will follow.

One direction is easy: we take a functor $F$ and evaluate it at the object $1$ to get an object $F(1)\in\mathcal{S}_0$. A monoidal natural transformation $\eta:F\rightarrow G$ has, in particular, the component $\eta_1:F(1)\rightarrow G(1)$, which makes “evaluate at $1$” a functor.

For the other direction, we’ll pick an object $S$ of $\mathcal{S}_0$ and use it to construct a strictly braided monoidal functor from $\mathcal{B}raid$ to $\mathcal{S}$ that sends $1$ to $S$, and then show that an arrow $f:S\rightarrow S'$ in $\mathcal{S}_0$ induces a natural transformation between the functors built on $S$ and $S'$. We’ll see that everything in sight is unique, so this construction actually provides an “on the nose” inverse functor to “evaluate at $1$“.

So, we’ve got an object $S$ and we set $F_S(1)=S$. Then for $F$ to be monoidal we must set $F(n)=S^{\otimes n}$, since $n$ is $1^{\otimes n}$ (remember that the “tensor power” is defined just like regular exponentiation: take the tensor product of $n$ copies of the object in question). This completely defines the functor $F_S$ on objects.

Now any morphism in $\mathcal{B}raid$ is an element of some braid group $B_n$. Let’s look at simply crossing the left strand over the right in $B_2$: this is $\beta_{1,1}$, so to have a strictly braided functor we must set $F_S(\beta_{1,1})=\beta_{S,S}$. In $B_n$, the crossings of one strand over the one to its right generate the whole group. If we cross strand $i$ over strand $i+1$ we’ll call the generator $\sigma_i$. In terms of the category $\mathcal{B}raid$, we can write $\sigma_i$ as $1_{i-1}\otimes\beta_{1,1}\otimes1_{n-i-1}$, so monoidality forces us to set $F_S(\sigma_i)=1_{S^{\otimes i-1}}\otimes\beta_{S,S}\otimes1_{S^{\otimes n-i-1}}$.

We do have to check that the relations of the braid group are satisfied. But if $|i-j|>1$ then we have $F_S(\sigma_i\sigma_j)=F_S(\sigma_j\sigma_i)$ because the monoidal product in $\mathcal{S}$ is a functor. And we see that $F_S(\sigma_i\sigma_{i+1}\sigma_i)=F_S(\sigma_{i+1}\sigma_i\sigma_{i+1})$ because this is exactly an instance of the triangle relation for the braiding on $\mathcal{S}$!

So once we pick the single object $S=F_S(1)$ everything else about $F_S$ is uniquely determined. If we have an arrow $f:S\rightarrow S'$ then we must set the component $\eta_1$ of the induced natural transformation to be $f:F_S(1)\rightarrow F_{S'}(1)$. Then monoidality forces $\eta_n:F_S(n)\rightarrow F_{S'}(n)$ to be $f^{\otimes n}$. The construction is indeed a functor sending $\mathcal{S}_0$ to the category of strictly monoidal functors from $\mathcal{B}raid$ to $\mathcal{S}$, and it is clearly inverse to evaluation at $1$.

July 6, 2007