$\newcommand{\M}{\mathcal{M}}\newcommand{\ML}{\underline{\mathcal{M}}}\newcommand{\N}{\mathcal{N}}\newcommand{\NL}{\underline{\mathcal{N}}}\newcommand{\V}{\mathscr{V}}\newcommand{\VL}{\underline{\mathscr{V}}}\newcommand{\hom}{\mathsf{hom}}\newcommand{\op}{^{\mathsf{op}}}\require{AMScd}$For a closed monoidal category $\V$, there is a correspondence between closed $\V$-modules and strongly tensored (and cotensored) $\V$-enriched categories. My question concerns whether or not we can weakened the hypotheses of this theorem. I put the definitions of all used terms at the end.

Now the correspondence, as far as I'm aware of it/have been able to verify myself, contains the following points:

  • Any closed $\V$-module $\M$ can be canonically enriched to a $\V$-category $\ML$
  • The module tensor $\odot$ is, in this correspondence, always promotable to a strong tensor making $\ML$ a strongly tensored $\V$-category
  • Any strongly tensored $\V$-category has a canonical $\V$-module structure given by the same tensor (where the hypothesis of being strongly, as opposed to weakly, tensored is used in a crucial way)
  • $\V$-module functors $\M\to\N$ of closed $\V$-modules always lift to enriched $\V$-functors $\ML\to\NL$ under the above correspondences
  • Enriched $\V$-functors $\ML\to\NL$ of always descend to $\V$-module functors $\M\to\N$ if we assume the cotensor exists i.e. assume the modules are "doubly" closed
  • Enriched natural transformations are in correspondence between 'module' transformations of module/enriched functors between closed $\V$-modules/strongly tensored $\V$-categories

Only one of these points seemed to require $\pitchfork$ when I checked the details. Moreover, $\pitchfork$ was only required for one little detail and was used exactly once in exactly one diagram. This makes me wonder if we can state the correspondence without referencing "doubly" closed modules at all. My question is whether or not this is true.

Specifically, here was the thing I required $\pitchfork$ to show:

We take a doubly closed $\V$-module $\M$ and a closed $\V$-module $\N$. We suppose $F:\ML\to\NL$ is a $\V$-enriched functor, and want to make it a $\V$-module functor. We define $\alpha^F:v\odot Fm\to F(v\odot m)$ as the $\hom_\N(Fm,-)$ adjunct of: $$v\overset{\eta_v}{\longrightarrow}\hom_\M(m,v\odot m)\overset{F_{m,v\odot m}}{\longrightarrow}\hom_\N(Fm,F(v\odot m))$$Where $\eta$ denotes the relevant adjunction unit.

If we want $(F;\alpha^F)$ to define a module functor we want $\alpha^F$ to be natural in both variables. It suffices to show naturality in each variable separately; I was able to show that $\alpha^F$ is natural in the first variable $v$, but in order to show it is natural in $m$ I required $\pitchfork$: I required the following diagram to always commute in $\M$: $$\begin{CD}v@>\eta^{m'}_v>>\hom_\M(m',v\odot m')\\@V\eta^m_vVV@VV\hom_\M(f,1)V\\\hom_\M(m,v\odot m)@>>\hom_\M(1,1\odot f)>\hom_\M(m,v\odot m')\end{CD}$$If the two variable adjunction involving $\pitchfork$ exists, this is possible to show. However, I could not show this commutes without that assumption. Question: is it possible to show the diagram commutes without using $\pitchfork$?

I hope this is true, because it is mildly annoying that we need $\pitchfork$ once and only once. Alternatively, I'd be interested if someone had a good proof that $\alpha^F$ is natural that doesn't require the above diagram to commute.


I use definition $6$ of this for '$\V$-module'.

Here, a "doubly" closed $\V$-module is a $\V$-module $\M$ equipped with a two variable adjunction: $$\M(v\odot m,n)\cong\V(v,\hom(m,n))\cong\M(m,v\pitchfork n)$$Where $\odot:\V\times\M\to\M,\pitchfork:\V\op\times\M\to\M,\hom:\M\op\times\M\to\V$ are functors. $\odot$ is referred to as the tensor and $\pitchfork$ as the cotensor. A closed $\V$-module is the same, just without the data of $\pitchfork$.

A strongly tensored $\V$-enriched category $\underline{\M}$ is a $\V$-enriched category together with a $\V$-functor (in the sense that it is a $\V$-functor in each variable separately) $\odot:\VL\times\ML\to\ML$ and isomorphisms $\kappa:\ML(v\odot m,n)\cong\VL(v,\ML(m,n))$ which are enriched natural transformations in $m$ and $n$. That this isomorphism is enrichable distinguishes it from what I'd call weakly tensored $\V$-categories where $\kappa$ is merely an isomorphism of ordinary functors.

$\V$-module functors are defined in the paper. I made up a definition of "module" transformation myself, but I won't repeat it here as it is not relevant.


1 Answer 1


$\newcommand{\M}{\mathcal{M}}\newcommand{\ML}{\underline{\mathcal{M}}}\newcommand{\N}{\mathcal{N}}\newcommand{\NL}{\underline{\mathcal{N}}}\newcommand{\V}{\mathscr{V}}\newcommand{\VL}{\underline{\mathscr{V}}}\newcommand{\hom}{\mathsf{hom}}\newcommand{\op}{^{\mathsf{op}}}\require{AMScd}$I asked under what conditions the below diagram commutes.

$$\begin{CD}v@>\eta^{m'}_v>>\hom_\M(m',v\odot m')\\@V\eta^m_vVV@VV\hom_\M(f,1)V\\\hom_\M(m,v\odot m)@>>\hom_\M(1,1\odot f)>\hom_\M(m,v\odot m')\end{CD}$$

It is actually reasonably easy to see (once you know the answer, which I discovered for myself by lucky accident) that this diagram commutes (always) if and only if the adjunction: $$\kappa:\M(v\odot m,n)\cong\V(v,\hom_{\M}(m,n))$$Is natural in all variables (a priori the adjunction is only natural in $v$ and $n$). For such an adjunction with parameter we can always arrange for the adjunction to be natural in all variables by re-defining one of the two functors involved. However, in applications where both $\odot$ and $\hom_\M$ are already defined as bifunctors, we should do due diligence and check $\kappa$ is natural in $m$ as well as in $v$ and $n$.

For many reasons which I won't go into here (unless someone is interested) demanding that $\kappa$ be natural in all variables is extremely convenient and is indeed necessary for a lot of 'obvious' and ignored details/theorems to actually be true in the context of working with $\V$-modules and their associated tensored categories.

The business with $\pitchfork$ - if we have a two variable adjunction then that forces $\kappa$ to be natural in all variables, which is why I initially mistook the commutativity of the diagram as meaningfully linked to the existence of $\pitchfork$. But in fact the equivalences I mentioned, and a few others which I didn't, all carry through if $\pitchfork$ doesn't exist (but obviously they dualise perfectly to $\V$-comodules etc. if one considers $\pitchfork$ over $\odot$).


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