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Judging from the axioms of a symmetric monoidal category, can we say anything about the left unitor being related to the right unitor?

We have the morphisms (using notation as nlab) $$ \lambda_1 :1 \otimes 1 \rightarrow 1 $$ $$ \rho_1 : 1 \otimes 1 \rightarrow 1$$ It seems desirable to me that $$ \lambda_1 =\rho_1 b_{1,1}$$ holds. But this is doesn't seemed to be implied.

The reason for this is that: wouldn't one want a canonical choice of isomorphism $$ 1 \otimes 1 \simeq 1?$$

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  • $\begingroup$ @Arnaud You should probably post this as an answer, because it is an answer (and a good one imo) $\endgroup$ Jul 30 '19 at 9:49
  • $\begingroup$ @MarkKamsma The nLab doesn't give a proof for the braided case unfortunately. Since the diagrams are a bit too big to reproduce, I ended up posting the relevant references as an answer. $\endgroup$
    – Arnaud D.
    Jul 30 '19 at 10:58
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Actually, $\lambda_I$ and $\rho_I$ are equal in any monoidal category, and $\lambda_X=\rho_X B_{1,X}$ in any symmetric monoidal category, although this is not entirely obvious. In fact, Mac Lane originally required these as axioms, and also that $\lambda_{A\otimes B}\circ \alpha_{I,A,B}=\lambda_A\otimes B$ and $ A\otimes\rho_B \circ \alpha_{A,B,I} =\rho_{A\otimes B}$, but Kelly showed that all these identities could be deduced from the triangle, pentagon and hexagon diagrams :

On MacLane's conditions for coherence of natural associativities, commutativities, etc., G.M. Kelly, 1964, Journal of Algebra 1, pp 397-402

The argument can also be found on the nLab.

Later, Joyal and Street proved that $\lambda_X=\rho_X B_{1,X}$ even holds in braided monoidal categories :

Braided tensor categories, A. Joyal and R. Street, 1993, Advances in Mathematics 102, pp 20-78

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