Let $\mathcal C$ be a symmetric monoidal closed category. This means that every functor $- \otimes B$ has a right adjoint $[B, -]$. Let $I$ be the unit and let $\rho \colon - \otimes I \to 1_{\mathcal C}$ be the right unitor. There are isomorphisms $ \mathcal C (C \otimes I, C ) \cong \mathcal (C , [I,C])$ because there is an adjunction. If we let $i_C \colon C \to [I,C]$ be the morphism corresponding to $\rho_C$ under the adjunction, then $i_C$ should be an isomorphism. I am having trouble proving this however.
If we let $\mathrm{ev}_{IC} \colon [I,C] \otimes I \to C$ be the morphism corresponding to the identity under the adjunction (i.e. the counit of the adjunction $- \otimes I \dashv [I, - ]$ ) then the inverse for $i_C$ is supposed to be given by the morphism
$ [I,C] \xrightarrow{\rho_{[I,C]}^{-1}} [I,C] \otimes I \xrightarrow{\mathrm{ev}_{IC}} C. $
I can see quite easily that $\mathrm{ev}_{IC} \circ \rho_{[I,C]}^{-1} \circ i_c = 1 $ but I can't for the life of me figure out why $ i_c \circ \mathrm{ev}_{IC} \circ \rho_{[I,C]}^{-1} =1 $. Could someone give me a hand? Thanks.