Let $(\mathcal{C}, \otimes, I)$ a monoidal category, and let $\mathbb{B}(\mathcal{C})$ the bicategory (with only one object and $(\mathcal{C}, \otimes, I)$ as (monoidal) category of morphisms and cells). Let $\mathbb{B}(\mathcal{C})^{op}$ the dual respect morphisms (i.e. twisting $\otimes$, and inverting canonical isomorphisms, in the monoidal category $\mathcal{C}$). A pseudo-functor (see ncatlab) $(Q, \theta): \mathbb{B}(\mathcal{C})\to \mathbb{B}(\mathcal{C})^{op}$ that fixes objects, morphisms, and cells, and is unitary (i.e. the canonical isomorphisms $Q(I)\cong I$ is the identity) is a family of isomorphism $q_{A, B}^{-1}: B\otimes A= Q(B)\otimes Q(A)\cong Q(A\otimes B)= A\otimes B $ such that:

U1) $q_{A, I}\circ r_A= l_A$

U2) $q_{I, A}\circ l_A= r_A$

where $r_A: A \cong A\otimes I$, $l_A: A \cong I\otimes A$ canonically.

and with:

$a\circ q_{C\otimes B, A}^{-1}\circ A\otimes q_{C, B}^{-1}= q_{C, B\otimes A}^{-1}\circ q_{B, A}^{-1}\otimes C\circ a^{-1}: A\otimes (B\otimes C) \to C\otimes (B\otimes A) $

or equivalently:

$A\otimes q_{C, B}\circ q_{C\otimes B, A}= a\circ q_{B, A}\otimes C\circ q_{C, B\otimes A}\circ a : (C\otimes B)\otimes A) \to A\otimes (B\otimes C) $

If we further require that $Q^{op}\circ Q=1_{\mathbb{B}(\mathcal{C})}$ then: $q_{A,B}\circ q_{B, A}=1$.

I ask:

1) is a such pseudofunctor $Q$ a symmetry?

2) What about a generalization to braids (i.e. how describe a braid as a pseudo-functor)?

Note that exist a coherence theorem: any diagram of canonical isomorphisms (from monoidal structure or from pseudofunctor data) commute

  • $\begingroup$ Opps, I forget a comma "," between "symmetry" and "braids" in the title. $\endgroup$ – Buschi Sergio Dec 8 '12 at 18:03
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    $\begingroup$ You can edit the title! $\endgroup$ – MTS Dec 9 '12 at 0:37
  • $\begingroup$ Ok, I have the answere now. $\endgroup$ – Buschi Sergio Jan 6 '13 at 13:04
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    $\begingroup$ Would you mind posting the answer then? $\endgroup$ – Manuel Bärenz Oct 29 '13 at 18:08
  • $\begingroup$ I agree with Turion. This is a reasonable question and it would be better to have it answered here than for this to linger without an answer $\endgroup$ – David White Oct 29 '13 at 18:44

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