Todd Trimble
Whitney trick

Lemma

(Whitney trick) Suppose given a braided monoidal category B\mathbf{B}, and an object AA which has a right dual BB in the sense that there is a counit ϵ:ABI\epsilon: A \otimes B \to I and a unit η:IBA\eta: I \to B \otimes A satisfying the familiar triangular equations. Then also AA is a right dual of BB, where the counit ϵ\epsilon' may be defined by the composite

BAc B,AABϵIB \otimes A \stackrel{c_{B, A}}{\to} A \otimes B \stackrel{\epsilon}{\to} I

and the unit η\eta' may be defined by the composite

IηBAc A,B 1AB.I \stackrel{\eta}{\to} B \otimes A \stackrel{c_{A, B}^{-1}}{\to} A \otimes B.
Proof

Using Mac Lane’s coherence theorem, we may make the simplifying assumption that the underlying monoidal category of B\mathbf{B} is strict monoidal. (The hexagonal coherence conditions for the braiding then collapse to “braiding triangles”.) We must verify the two triangular equations for (η,ϵ)(\eta', \epsilon'); here we will verify just one, that

Bρ B 1BI1 BηBABϵ1 BIBλ BBB \stackrel{\rho_B^{-1}}{\cong} B \otimes I \stackrel{1_B \otimes \eta'}{\to} B \otimes A \otimes B \stackrel{\epsilon' \otimes 1_B}{\to} I \otimes B \stackrel{\lambda_B}{\cong} B

is the identity 1 B1_B, where ρ,λ\rho, \lambda 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 f:XYf: X \to Y and g:YZg: Y \to Z is written fg:XZf g: X \to Z. Thus, spelled out, we want to show

(ρ B 1)(1 Bη)(1 Bc A,B 1)(c B,A1 B)(ϵ1 B)(λ B)=1 B.(\rho_B^{-1})(1_B \otimes \eta)(1_B \otimes c_{A, B}^{-1})(c_{B, A} \otimes 1_B)(\epsilon \otimes 1_B)(\lambda_B) \;\; = \;\; 1_B.

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:

(ρ B 1)(1 Bη)(1 Bc A,B 1)(c B,A1 B)(ϵ1 B)(λ B) = (ρ B 1)(1 Bη)(c B,B1 A)(c B,B 11 A)(1 Bc A,B 1)(c B,A1 B)(ϵ1 B)(λ B)(ReidemeisterII) = (λ B 1)(η1 B)(1 Bc B,A 1)(c B,B 11 A)(1 Bc A,B 1)(c B,A1 B)(ϵ1 B)(λ B)(Claimbelow) = (λ B 1)(η1 B)(c A,B 11 B)(1 Ac B,B 1)(c B,A 11 B)(c B,A1 B)(ϵ1 B)(λ B)(ReidemeisterIII) = (λ B 1)(η1 B)(c A,B 11 B)(1 Ac B,B 1)(ϵ1 B)(λ B)(ReidemeisterII) = (λ B 1)(η1 B)(c AB,B 1)(ϵ1 B)(λ B)(Braidingtriangle) = (λ B 1)(η1 B)(1 Bϵ)(c I,B 1)(λ B)(Naturalityofc) = (λ B 1)(η1 B)(1 Bϵ)(ρ B)(Unitcoherence) = 1 B(Triangularequationfor(η,ϵ))\array{ (\rho_B^{-1})(1_B \otimes \eta)(1_B \otimes c_{A, B}^{-1})(c_{B, A} \otimes 1_B)(\epsilon \otimes 1_B)(\lambda_B) & = & (\rho_B^{-1})(1_B \otimes \eta)(c_{B, B} \otimes 1_A)(c_{B, B}^{-1} \otimes 1_A)(1_B \otimes c_{A, B}^{-1})(c_{B, A} \otimes 1_B)(\epsilon \otimes 1_B)(\lambda_B) \;\;\;\; (Reidemeister\; II) \\ & = & (\lambda_B^{-1})(\eta \otimes 1_B)(1_B \otimes c_{B, A}^{-1})(c_{B, B}^{-1} \otimes 1_A)(1_B \otimes c_{A, B}^{-1})(c_{B, A} \otimes 1_B)(\epsilon \otimes 1_B)(\lambda_B) \;\;\;\; (Claim \; below)\\ & = & (\lambda_B^{-1})(\eta \otimes 1_B)(c_{A, B}^{-1} \otimes 1_B)(1_A \otimes c_{B, B}^{-1})(c_{B, A}^{-1} \otimes 1_B)(c_{B, A} \otimes 1_B)(\epsilon \otimes 1_B)(\lambda_B) \;\;\;\; (Reidemeister\; III) \\ & = & (\lambda_B^{-1})(\eta \otimes 1_B)(c_{A, B}^{-1} \otimes 1_B)(1_A \otimes c_{B, B}^{-1})(\epsilon \otimes 1_B)(\lambda_B) \;\;\;\; (Reidemeister\; II) \\ & = & (\lambda_B^{-1})(\eta \otimes 1_B)(c_{A \otimes B, B}^{-1})(\epsilon \otimes 1_B)(\lambda_B) \;\;\;\; (Braiding\; triangle) \\ & = & (\lambda_B^{-1})(\eta \otimes 1_B)(1_B \otimes \epsilon)(c_{I, B}^{-1})(\lambda_B) \;\;\;\; (Naturality\; of\; c) \\ & = & (\lambda_B^{-1})(\eta \otimes 1_B)(1_B \otimes \epsilon)(\rho_B) \;\;\;\; (Unit\; coherence) \\ & = & 1_B \;\;\;\; (Triangular\; equation\; for\; (\eta, \epsilon)) }

Claim: (ρ B 1)(1 Bη)(c B,B1 A)=(λ B 1)(η1 B)(1 Bc B,A 1)(\rho_B^{-1})(1_B \otimes \eta)(c_{B, B} \otimes 1_A) = (\lambda_B^{-1})(\eta \otimes 1_B)(1_B \otimes c_{B, A}^{-1}). This is seen by contemplating the diagram

B λ B 1 IB η1 B BAB ρ B 1 c B,I 1 c B,BA 1 1c B,a 1 BI 1 Bη BBA c B,B1 A BBA\array{ B & \stackrel{\lambda_B^{-1}}{\to} & I \otimes B & \stackrel{\eta \otimes 1_B}{\to} & B \otimes A \otimes B \\ _\mathllap{\rho_B^{-1}} \downarrow & \swarrow _\mathrlap{c_{B, I}^{-1}} & & \swarrow _\mathrlap{c_{B, B \otimes A}^{-1}} & \downarrow _\mathrlap{1 \otimes c_{B, a}^{-1}} \\ B \otimes I & \underset{1_B \otimes \eta}{\to} & B \otimes B \otimes A & \underset{c_{B, B} \otimes 1_A}{\to} & B \otimes B \otimes A }

where the parallelogram commutes by naturality of cc, 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 M\mathbf{M} is rigid. Even better, if we choose a left dual A A^\dagger for each AA, then one obtains a functor () :M opM(-)^\dagger: \mathbf{M}^{op} \to \mathbf{M} that is an adjoint equivalence: there is a natural isomorphism uA:AA u A: A \stackrel{\sim}{\to} A^{\dagger\dagger} which realizes an adjoint equivalence (() ) op() ((-)^\dagger)^{op} \dashv (-)^\dagger.

What is not true is that uA:AA u A: A \stackrel{\sim}{\to} A^{\dagger\dagger} 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