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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.

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Are you using your assumption that the enriching $n$-category $S$ is symmetric monoidal instead of merely plain monoidal? –  Kevin Walker Sep 1 '12 at 14:20
    
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? ... –  Kevin Walker Sep 1 '12 at 15:20
    
... (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 $k=1$? –  Kevin Walker Sep 1 '12 at 15:26
    
@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. –  Theo Johnson-Freyd Sep 1 '12 at 16:51
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2 Answers

up vote 5 down vote accepted

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?

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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. –  Theo Johnson-Freyd Sep 9 '12 at 13:51
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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 $Bicat$. –  Mike Shulman Sep 10 '12 at 3:11
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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 $a$.

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