Let $S$ be a set. Is there a monoidal category $TS$ that we can construct from $S$ such that monoidal functors $F: TS \to M$ (up to monoidal natural isomorphism) correspond to $M$-enriched categories with underlying object set $S$ (up to $M$-enriched equivalence)?

The construction I had in mind was to take $S^2$, add elements $(a,b)\otimes (b,c)$, add maps $\circ_{a,b,c} : (a,b)\otimes (b,c)\to (a,c)$, add a unit element $\mathbf{1}$ and maps $\mathbf{1}_{x}:\mathbf{1}\to (x,x)$, freely add the rest of the structure needed for it to be monoidal, and then impose commutativity of diagrams corresponding to associativity of the composition maps and unit axioms. My hope is that a monoidal functor out of this category into some monoidal category $M$ designates all of the structure necessary to specify an $M$-enriched category with objects $S$. Is such a construction known? I am not specifying any particular level of strictness on the monoidal functors involved here, but I suspect that has an impact on the answer to my question.


1 Answer 1


I don't think you can construct $T(S)$ as a monoidal category. It's more common to construct $T(S)$ as a bicategory with object set $S$. (see the end for comments on regarding it as a monoidal category)

Let $I(S)$ denote the indiscrete category on $S$. Then a lax 2-functor $I(S) \to M$ (where $M$ is considered as a 1-object bicategory -- more generally $M$ could be an arbitrary bicategory) is the same as an $M$-enriched category with object set $S$ (this goes back to Benabou's original paper on bicategories).

For any bicategory $B$ there is a lax morphism classifier $B'$ such that a weak 2-functor $B' \to C$ is the same as a lax 2-functor $B \to C$ (where $C$ is a bicategory) [actually, I'm not quite certain of this -- if we were talking about strict 2-categories and strict 2-functors rather than weak ones I'd be more certain. You can fit into this framework by replacing $M$ with a monoidal category that is strictly associative]. So $I(S)'$ has the property you're looking for. But it's not exactly a monoidal category -- it's bicategory with many objects.

I think this construction is "well-known". For example, a construction with a similar flavor comes in Gepner and Haugseng's definition of an enriched $\infty$-category: it $S$ is a Kan complex, then let $I(S)$ be the "indiscrete simplicial Kan complex on $S$", with $I(S)_n = S^{n+1}$. A monoidal $\infty$-category $M$ is also sort of simplicial quasicategory, and an $M$-enriched category with object space $S$ is a map of bisimplicial sets $I(S) \to M$ (well, I think they speak in terms of cartesian fibrations over $\Delta$ rather simplicial objects, but these are equivalent by straightening/unstraightening).

If you want $TS$ to be a monoidal category rather than a bicategory, then I think you're in luck because the inclusion from monoidal categories into bicategories should have a left adjoint, just like the inclusion of monoids into category. So just apply this left adjoint to $I(S)'$ above. This adjoint is kind of weird -- you freely add composites for morphisms with non-composable domain and codomain. But in the case of such a simple bicategory as this, I suppose it's not so bad -- and should look basically like you describe.


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