# Todd Trimble Whitney trick

###### Lemma

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

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

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

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

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

$B \stackrel{\rho_B^{-1}}{\cong} B \otimes I \stackrel{1_B \otimes \eta'}{\to} B \otimes A \otimes B \stackrel{\epsilon' \otimes 1_B}{\to} I \otimes B \stackrel{\lambda_B}{\cong} B$

is the identity $1_B$, where $\rho, \lambda$ 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 $f: X \to Y$ and $g: Y \to Z$ is written $f g: X \to Z$. Thus, spelled out, we want to show

$(\rho_B^{-1})(1_B \otimes \eta)(1_B \otimes c_{A, B}^{-1})(c_{B, A} \otimes 1_B)(\epsilon \otimes 1_B)(\lambda_B) \;\; = \;\; 1_B.$

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:

$\array{ (\rho_B^{-1})(1_B \otimes \eta)(1_B \otimes c_{A, B}^{-1})(c_{B, A} \otimes 1_B)(\epsilon \otimes 1_B)(\lambda_B) & = & (\rho_B^{-1})(1_B \otimes \eta)(c_{B, B} \otimes 1_A)(c_{B, B}^{-1} \otimes 1_A)(1_B \otimes c_{A, B}^{-1})(c_{B, A} \otimes 1_B)(\epsilon \otimes 1_B)(\lambda_B) \;\;\;\; (Reidemeister\; II) \\ & = & (\lambda_B^{-1})(\eta \otimes 1_B)(1_B \otimes c_{B, A}^{-1})(c_{B, B}^{-1} \otimes 1_A)(1_B \otimes c_{A, B}^{-1})(c_{B, A} \otimes 1_B)(\epsilon \otimes 1_B)(\lambda_B) \;\;\;\; (Claim \; below)\\ & = & (\lambda_B^{-1})(\eta \otimes 1_B)(c_{A, B}^{-1} \otimes 1_B)(1_A \otimes c_{B, B}^{-1})(c_{B, A}^{-1} \otimes 1_B)(c_{B, A} \otimes 1_B)(\epsilon \otimes 1_B)(\lambda_B) \;\;\;\; (Reidemeister\; III) \\ & = & (\lambda_B^{-1})(\eta \otimes 1_B)(c_{A, B}^{-1} \otimes 1_B)(1_A \otimes c_{B, B}^{-1})(\epsilon \otimes 1_B)(\lambda_B) \;\;\;\; (Reidemeister\; II) \\ & = & (\lambda_B^{-1})(\eta \otimes 1_B)(c_{A \otimes B, B}^{-1})(\epsilon \otimes 1_B)(\lambda_B) \;\;\;\; (Braiding\; triangle) \\ & = & (\lambda_B^{-1})(\eta \otimes 1_B)(1_B \otimes \epsilon)(c_{I, B}^{-1})(\lambda_B) \;\;\;\; (Naturality\; of\; c) \\ & = & (\lambda_B^{-1})(\eta \otimes 1_B)(1_B \otimes \epsilon)(\rho_B) \;\;\;\; (Unit\; coherence) \\ & = & 1_B \;\;\;\; (Triangular\; equation\; for\; (\eta, \epsilon)) }$

Claim: $(\rho_B^{-1})(1_B \otimes \eta)(c_{B, B} \otimes 1_A) = (\lambda_B^{-1})(\eta \otimes 1_B)(1_B \otimes c_{B, A}^{-1})$. This is seen by contemplating the diagram

$\array{ B & \stackrel{\lambda_B^{-1}}{\to} & I \otimes B & \stackrel{\eta \otimes 1_B}{\to} & B \otimes A \otimes B \\ _\mathllap{\rho_B^{-1}} \downarrow & \swarrow _\mathrlap{c_{B, I}^{-1}} & & \swarrow _\mathrlap{c_{B, B \otimes A}^{-1}} & \downarrow _\mathrlap{1 \otimes c_{B, a}^{-1}} \\ B \otimes I & \underset{1_B \otimes \eta}{\to} & B \otimes B \otimes A & \underset{c_{B, B} \otimes 1_A}{\to} & B \otimes B \otimes A }$

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

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