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Let $\mathcal{V}$ be a symmetric monoidal category and $\underline{\mathcal{M}}$ and $\underline{\mathcal{N}}$ be cotensored and tensored $\mathcal{V}$-categories. Now, say that we have an adjunction $$F: \mathcal{M} \leftrightarrows \mathcal{N} : G$$ between the underlying categories . I want to show that if $F$ is actually a $\mathcal{V}$-functor and F preserves tensors in the sense that we have natural isomorphisms $F(v \otimes m) \cong v \otimes Fm$, then we can actually turn this into an enriched adjunction.

Here is what I tried. Let us define $s_{m,n}:\underline{\mathcal{M}}(m,Gn) \rightarrow \underline{\mathcal{N}}(Fm,n)$ as the composite. $$\underline{\mathcal{M}}(m,Gn) \xrightarrow{F_{m,Gn}} \underline{\mathcal{N}}(Fm,FGn) \xrightarrow{\epsilon^n_\ast} \underline{\mathcal{N}}(Fm,n)$$ which is enriched natural. Now, let us define an inverse $t_{m,n} : \underline{\mathcal{N}}(Fm,n) \rightarrow \underline{\mathcal{M}}(m,Gn)$ by using the adjunctions and the fact that $F$ preserves tensors by requiring it to be be the morphism mapping to $id_{\underline{\mathcal{N}}(Fm,n)}$ under the chain of isomorphisms
$$\mathcal{V}(\underline{\mathcal{N}}(Fm,n),\underline{\mathcal{M}}(m,Gn)) \cong \mathcal{M}(\underline{\mathcal{N}}(Fm,n) \otimes m, Gn) \cong \mathcal{N}(\underline{\mathcal{N}}(Fm,n) \otimes Fm,n) \cong \mathcal{V}(\underline{\mathcal{N}}(Fm,n), \underline{\mathcal{N}}(Fm,n).$$

One now wants to prove that these are mutually inverse, so i did what to me looked like the obvious thing - tried to compose them and show that they cancelled, but I got no luck (I got expressions that were very messy and I couldn't reduce them further). How could I go about? Would anyone be so kind to spell out in some more detail how one actually shows that these two are mutually inverse?

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This doesn't really answer your precise question, but the way I know how to prove this is via the Yoneda embedding in the enriching category $\mathcal{V}$. For any $m \in \mathcal{M}$ and $n \in \mathcal{N}$, consider the presheaf of sets on $\mathcal{V}$ given by

$$ v \mapsto {\cal{V}}(v, \underline{\cal{M}}(m, Gn)) $$

Then we can use the unenriched adjunction, tensoring, and the assumed compatibility between them to obtain natural isomorphisms

$$ {\cal{V}}(v, \underline{\cal{M}}(m, Gn)) \cong {\cal{M}}(v \otimes m, Gn) \cong {\cal{N}}(F(v \otimes m), n) \cong {\cal{N}}(v \otimes Fm, n) \cong {\cal{V}}(v, \underline{\cal{N}}(Fm, n)) $$

So, the objects $\underline{\cal{M}}(m, Gn)$ and $\underline{\cal{N}}(Fm, n)$ represent the same presheaf on $\cal{V}$, and hence are isomorphic. Moreover, this is natural in $m, n$. There's more work to do, checking coherences and such, but this is basically how it goes.

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  • $\begingroup$ Thank you for your answer. Right, I have tried to do it that way as well, but to actually show in detail that this isomorphism is enriched natural requires some (to me) non-trivial diagram-chasing. $\endgroup$ – user101036 Apr 27 '14 at 18:59
  • $\begingroup$ Right, these are the details I was referring to. I think if you are willing to assume that $\cal{V}$ is closed, then another similar presheaf/Yoneda argument suffices. It's been a while since I wrote everything down, though, so I could be misremembering it. $\endgroup$ – Aaron Royer Apr 28 '14 at 0:45

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