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I'm learning about symmetric monoidal closed categories from Borceux's handbook and found myself asking a question similar to this already asked question :

Morphisms in a symmetric monoidal closed category.

However, I'm struggling to understand the accepted answer, nor do I understand why $$\mathrm{ev}_{IC} \circ \rho_{[I, C]}^{-1} \circ i_c =1$$

Is this some basic fact about adjunctions that I'm missing?Any explanation would be helpful.

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    I think you should try to understand the accepted answer, rather than do an awful explicit computation with units and such. – 2017-01-31
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    Thank you for the advice. After some thought, the fact that the morphism is an isomorphism via Yoneda is clear. It's the explicit computation that I'm interested to know. Why is that morphism the inverse? – 2017-01-31
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    Call the natural isomorphism witnessing the adjunction $\varphi$ such that $i = \varphi(\rho)$. $\varphi$ can be expressed in terms of the unit of the adjunction and the functorial action of $[I,-]$. Use that definition and calculate. Don't forget naturality. (Alternatively, use naturality to show that this is essentially $\varphi^{-1}(\varphi(\rho)) \circ \rho^{-1}$. Alternatively, directly use the naturality of $\varphi$ and that $\text{ev} = \varphi^{-1}(id)$.) – 2017-01-31

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