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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.

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Use the Yoneda lemma!

By adjunction there is a natural isomorphism $$\mathcal{C}(X \otimes I, C) \to \mathcal{C}(X, [I, C])$$ for all objects $X$ in $\mathcal{C}$. Being an isomorphism is preserved by all functors, so the natural transformation $$\mathcal{C}(X, C) \to \mathcal{C}(X \otimes I, C)$$ induced by the right unitor $\rho_X : X \otimes I \to X$ is a natural isomorphism. Now compose this with the first natural isomorphism to get $$\mathcal{C}(X, C) \cong \mathcal{C}(X, [I, C])$$ and this means $C \cong [I, C]$ via the morphism $C \to [I, C]$ that is the adjoint transpose of $\rho_C : C \otimes I \to C$.

I'll also remark that we didn't need to know that $\mathcal{C}$ is symmetric monoidal – just monoidal closed is good enough.

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  • $\begingroup$ I just have one more question. I see that the natural iso $\mathcal C (-, C) \cong \mathcal C (-,[I, C])$ corresponds to the adjoint of $\rho_C$ under Yoneda. but how can I be sure that element itself is an iso? $\endgroup$ – Paul Slevin Sep 7 '12 at 21:15
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    $\begingroup$ You can either find an inverse explicitly by taking $X = [I, C]$, or you can use the fact that the Yoneda embedding is fully faithful (and hence reflects isomorphisms). $\endgroup$ – Zhen Lin Sep 8 '12 at 1:51

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