This question is about $n$-categories, or perhaps $(\infty,n)$-categories, or ... My guess is that the answer will not depend sensitively on the model of higher categories, so rather than have me force you to work in my favorite model, I ask only that you say with some precision what model you are using, and that your model is not too strict.

You may assume that I have, and am comfortable with, some chosen symmetric monoidal $n$-category $(S,\otimes)$. I am trying to understand the notion of "($n$-)category enriched in $S$". By this I would like to mean the following. Roughly, $A$ is enriched in $S$ if $A$ has a set of $0$-morphisms, and between any pair of $0$-morphisms $X,Y \in A$, there is an $S$-object $A(X,Y)$ of 1-morphisms between them. The composition should be a morphism $A(X,Y) \otimes A(Y,Z) \to A(X,Z)$ in $S$.

I am fairly satisfied that, at least in my example, I can write all of this down explicitly. Recall that $S$ comes with a chosen object $1 \in S$. Then the corepresentable functor $S(1,-) : S \to (n-1)\text{-Cat}$ is symmetric monoidal in an essentially unique way. The usual thing is to use this functor, and define the de-enrichment $A_\delta$ of $A$ to be the $n$-category with hom-$(n-1)$-categories given by $A_\delta(-,-) = S(1,A(-,-))$. Certainly in my example I can work out this $(n-1)$-category. (De-enrichment is often a highly lossy operation, and I am OK with that.)

Suppose that $A$ and $B$ are both $S$-enriched categories. This is where I start to run into trouble. I understand what an $S$-enriched functor $A \to B$ is. But I'm having trouble figuring out what is the correct definition of "natural transformation", and in general of the higher morphisms.

Question: Given a symmetric monoidal $n$-category $S$, what is the $(n+1)$-category of $S$-enriched ($n$-)categories? In particular, what are the higher morphisms?

I recognize that this "$(n+1)$-category of $S$-enriched categories" is likely itself enriched in $S$-enriched categories. But I am interested simply in writing down its de-enrichment — I'm looking for it just as an $(n+1)$-category.

A final remark: The notion of enrichment in this question is not the same as in n-categories enriched in an (n+1)-category. That question concerned $n$-categories in which the collection of $n$-morphisms was an object of the enriching category. I would like the collection of $1$-morphisms to be an object of the enriching category.

  • 1
    $\begingroup$ Are you using your assumption that the enriching $n$-category $S$ is symmetric monoidal instead of merely plain monoidal? $\endgroup$ Sep 1, 2012 at 14:20
  • $\begingroup$ Consider an (n+k)-category $C$ whose 0- though k-morphisms are trivial. Let $T_m(C)$ denote the (n+k-m)-category obtained by dropping the 0- through (m-1)-morphisms of $C$. (So the objects of $T_m(C)$ are the m-morphisms of $C$.) For $k$ large, $T_{k+1}(C)$ is a symmetric monoidal $n$-category; call it $S$. Now consider $T_k(C)$. I would be tempted to say that $T_k(C)$ is an (n+1)-category enriched in $S$ in a sense similar to yours. Does this count as an example of your version of enrichment in $S$, or am I missing something? ... $\endgroup$ Sep 1, 2012 at 15:20
  • $\begingroup$ ... (continued) If the above makes sense in your context, then can one model the definition of functors, natural transformations, etc. of $S$-enriched categories on the definition of ordinary functors, NTs etc. from $C$ to itself? Also, does the assumption that $k$ is large play any important role? Or could we, for example assume that $k=1$? $\endgroup$ Sep 1, 2012 at 15:26
  • $\begingroup$ @Kevin: The reason I want my enriching $n$-category $S$ to be symmetric monoidal is because then the category of $S$-enriched categories is itself symmetric monoidal, and the motivation for my question comes from a related rinse-and-repeat exercise. I agree that it is perfectly reasonable to enrich in less-symmetric categories. $\endgroup$ Sep 1, 2012 at 16:51

2 Answers 2


For $F,G:C\to D$, an $S$-enriched transformation $F\to G$ will consist of, for each $x\in C$, a morphism $1\to D(F x, G x)$ in $S$, together with for each $x,y\in C$, an equivalence between the composites $$ C(x,y) \to D(F x, F y) \to D(F y, G y) \otimes D(F x, F y) \to D(F x, G y)$$ and $$ C(x,y) \to D(G x, G y) \to D(G x, G y) \otimes D(F x, G x) \to D(F x, G y)$$ plus some higher coherence cells until you run out of room. Does that help?

  • 1
    $\begingroup$ Awesome. Are you aware of anywhere that spells out the higher coherences (e.g. in terms of associahedra)? I would be interested in e.g. the n=3 case. Probably I just have to work it out myself. $\endgroup$ Sep 9, 2012 at 13:51
  • 3
    $\begingroup$ Just look at the definition of "tritransformation" in Gordon-Power-Street Coherence for tricategories and interpret the diagrams as taking place in $S$ rather than in $Bicat$. $\endgroup$ Sep 10, 2012 at 3:11

Another way to get the definition of an $S$-transformation, if you know what an $S$-functor is and also the tensor product of $S$-categories (so $S$ must be symmetric) and $S$ has an initial object preserved by $\otimes$ in each variable, is as an $S$-functor

$$ C \otimes \mathbf{2} \to D$$

where $\mathbf{2}$ is the $S$-category with two objects $a$ and $b$, hom-objects $1$ from $a$ to $a$, $b$ to $b$, and $a$ to $b$, and the initial object of $S$ from $b$ to $a$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.