For a monoidal category $\mathcal{C}$ with $\alpha_{a,b,c}: a \otimes (b \otimes c) \rightarrow (a \otimes b) \otimes c$, $\rho_a : a \otimes 1 \rightarrow a$, and $\lambda_a: 1 \otimes a \rightarrow a$, I want to show that $\lambda_1 = \rho_1$.
From the triangular identities, we have,
$\bullet$ $id_1 \otimes \rho_1 = \rho_{1 \otimes 1} \circ \alpha_{1,1,1}$
$\bullet$ $id_1 \otimes \lambda_1 = \rho_1 \otimes id_1 \circ \alpha_{1,1,1}$
$\bullet$ $\lambda_{1 \otimes 1} = (\lambda_1 \otimes id_1) \circ \alpha_{1,1,1}$
I was hoping to somehow show that $\rho_{1 \otimes 1} = \rho_1 \otimes id_1$...because if I can do that then the result is immediate from the first and second triangular identities. However, I can't really show that so I'm kind of stuck...I would appreciate any hints.
Consider the following diagram
The right hand triangle commutes if every other rectangle/triangle in the diagram does. But making extensive use of the pentagonal and the triangle identities and the naturality of $\alpha$, we can show that they do. Now since $-\otimes e$ is naturally isomorphic to the identity functor, commutativity of the rightmost triangle implies that $\varrho\alpha=1\otimes\varrho$. In a similar way, we get $\lambda\alpha^{-1}=\lambda\otimes1$.
Now $$(\lambda_e⊗1_e)\alpha=\lambda_a=1_e⊗\lambda_e=(\rho_e⊗1_e)\alpha\ :\ e⊗(e⊗e)\to e⊗e$$ The first identity has just been proven, the second one follows from the naturality of the monomorphism $\lambda$, and the third one is one of the axioms. We conclude that $\lambda_e⊗1_e=\rho_e⊗1_e$, and therefore $\lambda_e=\rho_e$