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 with underlying “ordinary category”
— the same category, just forgetting that it’s braided and monoidal. Then the statement is that the category of braided monoidal functors from
to
is equivalent to
.
To make our lives a bit easier, we’ll use the strictification theorem to pass from the (weak) monoidal category to a monoidally equivalent strict monoidal category
, with underlying category
. Now the braiding on
induces (via the equivalence) one on
. The new statement is that the category of strictly monoidal functors from
to
is isomorphic to
. Then the result we really want (about
) will follow.
One direction is easy: we take a functor and evaluate it at the object
to get an object
. A monoidal natural transformation
has, in particular, the component
, which makes “evaluate at
” a functor.
For the other direction, we’ll pick an object of
and use it to construct a strictly braided monoidal functor from
to
that sends
to
, and then show that an arrow
in
induces a natural transformation between the functors built on
and
. We’ll see that everything in sight is unique, so this construction actually provides an “on the nose” inverse functor to “evaluate at
“.
So, we’ve got an object and we set
. Then for
to be monoidal we must set
, since
is
(remember that the “tensor power” is defined just like regular exponentiation: take the tensor product of
copies of the object in question). This completely defines the functor
on objects.
Now any morphism in is an element of some braid group
. Let’s look at simply crossing the left strand over the right in
: this is
, so to have a strictly braided functor we must set
. In
, the crossings of one strand over the one to its right generate the whole group. If we cross strand
over strand
we’ll call the generator
. In terms of the category
, we can write
as
, so monoidality forces us to set
.
We do have to check that the relations of the braid group are satisfied. But if then we have
because the monoidal product in
is a functor. And we see that
because this is exactly an instance of the triangle relation for the braiding on
!
So once we pick the single object everything else about
is uniquely determined. If we have an arrow
then we must set the component
of the induced natural transformation to be
. Then monoidality forces
to be
. The construction is indeed a functor sending
to the category of strictly monoidal functors from
to
, and it is clearly inverse to evaluation at
.