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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.

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    $\begingroup$ See ncatlab.org/nlab/show/monoidal+category Lemma 2 $\endgroup$
    – user148177
    Nov 30 '14 at 1:17
  • $\begingroup$ @user148177 Thanks, but I'm not really sure how this answers my question. The proof says that the equation $(1 \lambda_1) \circ \alpha_{1,1,1} = \lambda_1 1$ follows from the lemma (which says $(\lambda_1 1) = \lambda_{11} \circ \alpha_{1,1,1}$. So again we have a (similar) problem...we need to show that $\lambda_{11} = 1 \lambda_1$, right? $\endgroup$
    – user196415
    Nov 30 '14 at 12:08
  • $\begingroup$ in the proof of lemma 2, invert $\alpha_{1,1,1}$, find the RHS are equal, and use transitivity of equality. don't really see the problem with their proof. the statement of lemma 2 is exactly your question, right? $\endgroup$
    – user148177
    Dec 1 '14 at 2:09
  • $\begingroup$ @user148177 Yes it is the same question...but in the proof of Lemma 2, they say that they use lemma 1 to get the first equation (which is the problem that I'm having). In other words...I'm not sure how the first equation follows from Lemma 1. $\endgroup$
    – user196415
    Dec 1 '14 at 2:24
  • $\begingroup$ hmm yeah im not sure then, let me think about it a little. btw here is the original paper sciencedirect.com/science/article/pii/0021869364900183 (combination of thm 6, 7). they don't do anything different though $\endgroup$
    – user148177
    Dec 1 '14 at 3:35
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Consider the following diagram
enter image description here
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 $\lambda_e=\rho_e$

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