7
$\begingroup$

We will suppose, for the sake of simplicity, that everything is happening within a fixed 'metacategory' $\textbf{SET}$ of sets and functions. So, from now on, a 'category' just means a category object in $\textbf{SET}$ - i.e. a small category.

Let $\mathscr{V}$ be a monoidal category. A $\mathscr{V}$-enriched category $\mathscr{C}$ consists of:

  • Objects: A set Ob($\mathscr{C}$).
  • Morphisms: For each pair of $\mathscr{C}$-objects $(X, Y)$, a $\mathscr{V}$-object Hom$(X, Y)$.
  • Composition: For each triple of $\mathscr{C}$-objects $(X, Y, Z)$, a $\mathscr{V}$-morphism $\circ$ : Hom$(X, Y)$ $\otimes$ Hom$(Y, Z)$ $\rightarrow$ Hom$(X, Z)$.
  • Identities: For each $\mathscr{C}$-object $X$, a $\mathscr{V}$-morphism id$_X$: $I$ $\rightarrow$ Hom$(X, X)$ (where $I \in \mathscr{V}$ is the unit of $\otimes$).

This data is then subject to the usual associativity and unitality axioms which are expressed via the commutativity of certain diagrams in $\mathscr{V}$. From this enriched category, we can extract an underlying category $\mathscr{C}_0$ by defining $\mathscr{C}(X, Y) = \mathscr{V}(I, \text{Hom}(X, Y))$.

My question is about if this is reversible - namely, can we define a $\mathscr{V}$-enriched category to be a category $\mathscr{C}$ equipped with a 'hom-functor' to $\mathscr{V}$? I'm having some trouble finding a reference for this but it seems like there should be a fairly obvious definiton. A $\mathscr{V}$-atlas on a category $\mathscr{C}$ consists of:

  • Morphisms: A functor Hom: $\mathscr{C}^{op} \times \mathscr{C} \rightarrow \mathscr{V}$.
  • Composition: For each triple of $\mathscr{C}$-objects $(X, Y, Z)$, a $\mathscr{V}$-morphism $\circ$ : Hom$(X, Y)$ $\otimes$ Hom$(Y, Z)$ $\rightarrow$ Hom$(X, Z)$.
  • Parametrisation: For each pair of $\mathscr{C}$-objects $(X, Y)$, an isomorphism $\eta: \mathscr{C}(X, Y) \xrightarrow{\sim} \mathscr{V}(I, \text{Hom}(X, Y))$ such that for all $X \xrightarrow{f} Y \xrightarrow{g} Z$ in $\mathscr{C}$, $\eta(g \circ f) = \eta(g)\circ\eta(f)$ (where on the left we have compositon in $\mathscr{C}$ and on the right we have composition in $\mathscr{V}$).

I'm unsure though if this gives associativity and unitality as in the usual defintion of a $\mathscr{V}$-enriched category, or if we only get associativity and unitality for $I$-shaped elements of the hom-objects. Could this be remedied by just requiring the associativity and unitality laws to hold as in the usual definiton? Any help or references would be much appreciated.

$\endgroup$
2
  • 1
    $\begingroup$ You can require associativity and unitality to hold as in the usual definition but then the ordinary homsets of $C$ aren't really doing anything and you might as well just use the usual definition... why not just say that you're looking at an enriched category together with an equivalence between the underlying category of that one and $C$? $\endgroup$ Commented Sep 25, 2020 at 19:48
  • $\begingroup$ @QiaochuYuan Yeah, just requiring associativity and unitality as in the usual defintion is just a cop out and kind of trivialises the whole endeavor anyway. The motivation for looking at it this way is as follows: Suppose you have a category Cat which axiomatises the (a) category of categories (something like Lawvere's ETCC or, more contemporarily, a nice 2-category with a duality involution and a Yoneda structure). Then how can you talk about a category C in Cat being enriched over another category V in Cat without needing to pass through discrete categories and so on? $\endgroup$
    – Fawzi
    Commented Sep 25, 2020 at 19:55

2 Answers 2

10
$\begingroup$

When your enriched categories are bicomplete enough (specifically, tensored and cotensored over $\mathscr{V}$), you can view the extra structure of the enrichment as a kind of action of $\mathscr{V}$ on them: this is called a closed $\mathscr{V}$-module in Definition 10.1.3 of Riehl's Categorical homotopy theory (with comparison in Proposition 10.1.4). The point is that the adjunction between the tensors and the internal hom (which is a better-behaved form of your "parametrisation") will allow you to formulate the associativity and unitality for the hom composition very nicely in terms of associativity of the action.

If you want to consider more general (not necessarily co/tensored) $\mathscr{V}$-enriched categories, you can pass to a weaker form of enrichment by relaxing the closed $\mathscr{V}$-module structure to a simple (weak) $\mathscr{V}$-module structure (viewing $\mathscr{V}$ as a weak monoid in the monoidal $2$-categories of categories); this corresponds to enriching over the category of presheaves on $\mathscr{V}$. Then the $\mathscr{V}$-enrichment is a condition of representability of the action.

I do not know a reference for this story for $1$-categories, but it is essentially what I understand of the construction in Definitions 4.2.1.25 and 4.2.1.28 of Lurie's Higher algebra and the explanations in the introduction of Heine's "An equivalence between enriched $\infty$-categories and $\infty$-categories with weak action" which compares these two points of view on enriched $(\infty,1)$-categories.

$\endgroup$
1
  • $\begingroup$ Ah this is perfect. Thank you, will read more into it. $\endgroup$
    – Fawzi
    Commented Sep 26, 2020 at 10:41
7
$\begingroup$

An answer has already been accepted, but I believe it does not really answer the question as given, so here is another.

You can define $\mathcal V$-enrichment as structure on a category $\mathcal C$, in terms of a functor $\mathcal C^{\text{op}} \times \mathcal C \to \mathcal V$ satisfying several compatibility laws. This definition explicitly appears, for example, as Definition 5.1 of McDermott–Uustalu's What Makes a Strong Monad?. This idea also appears in the setting of skew-enrichment in Campbell's Skew-enriched categories (Definition 2.1), where ordinary enrichment is recovered when the enrichment is normal (also Definition 2.1).

$\endgroup$
1
  • $\begingroup$ Thank you so much varkor! $\endgroup$ Commented Sep 14, 2022 at 14:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .