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?


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.

  • $\begingroup$ 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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