# Are the left unitor and right unitor in a symmetric monoidal category related?

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 $$1 \otimes 1 \simeq 1?$$

• @Arnaud You should probably post this as an answer, because it is an answer (and a good one imo) Jul 30 '19 at 9:49
• @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. Jul 30 '19 at 10:58

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