(Whitney trick) Suppose given a braided monoidal category , and an object which has a right dual in the sense that there is a counit and a unit satisfying the familiar triangular equations. Then also is a right dual of , where the counit may be defined by the composite
and the unit may be defined by the composite
Proof
Using Mac Lane’s coherence theorem, we may make the simplifying assumption that the underlying monoidal category of is strict monoidal. (The hexagonal coherence conditions for the braiding then collapse to “braiding triangles”.) We must verify the two triangular equations for ; here we will verify just one, that
is the identity , where are unit isomorphisms. The proof of the other triangular equation follows a wholly similar pattern.
We will write compositions of morphisms in left-to-right order, so the composite of and is written . Thus, spelled out, we want to show
The moves below mostly just copy the moves of the Whitney trick familiar from knot theory, which avoids Reidemeister I moves (the Reidemeister II and III moves correspond to regular isotopies). The Reidemeister II move translates to invertibility of the braiding. The Reidemeister III move translates to the “Yang-Baxter equation” (see Braided Monoidal Categories by Joyal and Street); we won’t bother proving this equation here.
Here we go:
Claim: . This is seen by contemplating the diagram
where the parallelogram commutes by naturality of , and the right triangle commutes as a disguised version of a braiding triangle condition.
Remark
The naming “Whitney trick” comes from the illustration of the Whitney trick for 1-knots given in Louis Kauffman’s On Knots, said to be “superior to the bending of spoons”. Of course there is a much more general (and powerful) geometric maneuver called the “Whitney trick” which plays an important role in, for example, the proof of the s-cobordism theorem, as expounded here. I don’t know yet what if any algebraic maneuver in higher category theory would correspond to this geometric lemma; it would be worth finding out I’m sure.
Remark
By this proposition, a left rigid braided monoidal category is rigid. Even better, if we choose a left dual for each , then one obtains a functor that is an adjoint equivalence: there is a natural isomorphism which realizes an adjoint equivalence .
What is not true is that is (necessarily) a monoidal equivalence. A rigid braided monoidal category is by definition pivotal if indeed it is a monoidal equivalence.
Revised on September 8, 2015 at 09:38:40
by
Todd Trimble