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Stream: learning: questions

Topic: Question about copowers

Bruno Gavranović (Nov 26 2023 at 13:56):

Let $\mathcal{V}$ be a monoidal closed category. Let $\mathcal{C}$ be a $\mathcal{V}$-category.

Then, a copower of of $\mathcal{C}$ over $\mathcal{V}$ is defined to be a functor

$$\bullet : \mathcal{V} \times \mathcal{C} \to \mathcal{C}$$

such that $\underline{\mathcal{V}}(A, \mathcal{C}(B, C)) \cong \mathcal{C}(A \bullet B, C)$ holds naturally in all variables, where $\underline{\mathcal{V}}$ represents the internal hom of $\mathcal{V}$. This is according to the nLab page on copowers and Coend calculus book(Def. 2.2.3).
But isn't this wrong?

$\mathcal{C}$ isn't a ($\mathbf{Set}$-)category, so we can't have a functor $\bullet$ - only a $\mathcal{V}$-functor.

This seems like an innocuous mistake, so I set out to look online for a reference which actually states that $\bullet$ is a $\mathcal{V}$-functor explicitly. To my big suprise, I couldn't.

I'm pretty sure that it has to be, because otherwise the definition doesn't make sense, but out of an abundance of caution I want to ask about it here. I'd be surprised either way - either I got something very wrong or there's no easily found reference which states that $\bullet$ is a $\mathcal{V}$-functor :)

Reid Barton (Nov 26 2023 at 14:37):

If $\mathcal{C}$ is a $\mathcal{V}$-category then it is an ordinary category, via $\mathrm{Hom}(X,Y) = \mathcal{V}(1, \mathcal{C}(X, Y))$.

Reid Barton (Nov 26 2023 at 14:46):

This notion of tensored category comes up in model category theory and in that setting it's common to regard an "enriched model category" as an ordinary category together with extra enrichment/tensored/cotensored structure, each of which determines the other two. So in that case, the tensor is viewed as an ordinary (bi)functor between ordinary categories, and in fact, it can be taken as the primary structure, with the enrichment derived from it.

Reid Barton (Nov 26 2023 at 14:56):

If you're really not assuming that $\mathcal{V}$ is symmetric monoidal, then what I wrote above still works, but (as far as I know) the tensor product can't be a $\mathcal{V}$-functor, because there is no tensor product of $\mathcal{V}$-categories to serve as its domain.

Reid Barton (Nov 26 2023 at 15:02):

Assuming you want $\mathcal{V}$ to be symmetric monoidal, then it's also possible to stay entirely within the world of $\mathcal{V}$ categories and to define the tensor $A \otimes B$ ($A \in \mathcal{V}$, $B \in \mathcal{C}$) by its universal property, as a weighted colimit. I assume it is also automatically $\mathcal{V}$-functorial in $A$ and $B$.

Reid Barton (Nov 26 2023 at 15:18):

So in summary, I would say that this notion of "tensoring over a category" mostly comes up when we indeed regard $\mathcal{C}$ primarily as an ordinary category, rather than as a $\mathcal{V}$-category.

Mike Shulman (Nov 26 2023 at 18:33):

Reid Barton said:

Assuming you want $\mathcal{V}$ to be symmetric monoidal, then it's also possible to stay entirely within the world of $\mathcal{V}$ categories and to define the tensor $A \otimes B$ ($A \in \mathcal{V}$, $B \in \mathcal{C}$) by its universal property, as a weighted colimit. I assume it is also automatically $\mathcal{V}$-functorial in $A$ and $B$.

You can do that even if $\mathcal{V}$ isn't symmetric. You just have to be careful to use the correct internal-hom in $\mathcal{V}$ when defining the universal property, since a non-symmetric (bi)closed monoidal category has two different internal-homs.

Todd Trimble (Nov 26 2023 at 22:43):

Bruno, did you look in Kelly's book? See section 3.7.

Bruno Gavranović (Nov 27 2023 at 10:32):

Reid Barton said:

So in summary, I would say that this notion of "tensoring over a category" mostly comes up when we indeed regard $\mathcal{C}$ primarily as an ordinary category, rather than as a $\mathcal{V}$-category.

Your argument seems to be that many ordinary categories $\mathcal{C}$ are also additionally enriched, justifying the use of the copower functor. That doesn't sound compelling to me.
It unnecessarily ties your notion of copower to $\mathbf{Set}$-enrichment and ordinary categories: why should that be, in lieu of a valid definition of copower as a $\mathcal{V}$-functor?
The functor $\mathcal{V}(I, -) : \mathcal{V} \to \mathbf{Set}$ by which we treat enriched categories as ordinary ones is not faithful in general, thus meaning your notion of copower will not see all the structure of the original enrichment in $\mathcal{C}$.

Bruno Gavranović (Nov 27 2023 at 10:34):

Todd Trimble said:

Bruno, did you look in Kelly's book? See section 3.7.

I did. This book (and many other references) define copowers on objects (as assigning to every $v : \mathcal{V}$ and $c : \mathcal{C}$ an object $v \bullet c$ such that ...), but it doesn't actually specify whether copowers are treated as a $\mathcal{V}$- or a $\mathbf{Set}$-functor.

Todd Trimble (Nov 27 2023 at 13:06):

To resolve this question, what you need to do is keep in mind what Kelly says earlier in the book. This is a special case of indexed or weighted limits. Earlier, where he discusses generalities of indexed limits/colimits, he says explicitly that taking indexed limits/colimits is $V$-functorial: see the top of page 39.

Also read carefully the paragraph at the top of page 11 on terminological and notational conventions. If Kelly had meant "ordinary functor", he either would have said so or indicated so through his choice of notation.

Dylan McDermott (Nov 27 2023 at 13:36):

It's worth noting that Kelly's comment about $V$-functoriality relies on symmetry, and not only because it involves the tensor of $V$-categories. Even if $x$ is fixed, to make the copowers $v \bullet x$ into a $V$-functor $({-}) \bullet x : V \to C$ you need symmetry.

Reid Barton (Nov 27 2023 at 14:54):

Bruno Gavranović said:

It unnecessarily ties your notion of copower to $\mathbf{Set}$-enrichment and ordinary categories: why should that be, in lieu of a valid definition of copower as a $\mathcal{V}$-functor?

Alternatively: it avoids unnecessarily bringing in the whole subject of $\mathcal{V}$-enriched categories to a situation that can be completely described in a simpler way without them: The tensor is an action of $\mathcal{V}$ on $\mathcal{C}$, and for each $c \in \mathcal{C}$, the action $- \bullet c$ has a right adjoint, the "$\mathcal{V}$-valued Hom" $\mathcal{C}(c, -)$.

But it's partly a matter of taste, and also depends on what you're doing.

The functor $\mathcal{V}(I, -) : \mathcal{V} \to \mathbf{Set}$ by which we treat enriched categories as ordinary ones is not faithful in general, thus meaning your notion of copower will not see all the structure of the original enrichment in $\mathcal{C}$.

That would be a problem. But it's not the case, since the object $\mathcal{C}(c, d)$ of $\mathcal{V}$ is determined (by Yoneda) by all $\mathrm{Hom}_{\mathcal{V}}(a, \mathcal{C}(c, d))$, $a \in \mathcal{V}$, and by assumption, these are given in terms of the tensor by $\mathrm{Hom}_{\mathcal{C}}(a \bullet c, d)$.

Bruno Gavranović (Nov 28 2023 at 11:02):

Todd Trimble said:

To resolve this question, what you need to do is keep in mind what Kelly says earlier in the book. This is a special case of indexed or weighted limits. Earlier, where he discusses generalities of indexed limits/colimits, he says explicitly that taking indexed limits/colimits is $V$-functorial: see the top of page 39.

Also read carefully the paragraph at the top of page 11 on terminological and notational conventions. If Kelly had meant "ordinary functor", he either would have said so or indicated so through his choice of notation.

Got it - it's a $\mathcal{V}$-functor.

That's what I suspected, but I was wary of playing the game of "the author would've written it if they meant it". Especially when there's other references which explicitly say "functor", and not a "$\mathcal{V}$-functor" (see my original post).

Bruno Gavranović (Nov 28 2023 at 11:09):

Dylan McDermott said:

It's worth noting that Kelly's comment about $V$-functoriality relies on symmetry, and not only because it involves the tensor of $V$-categories. Even if $x$ is fixed, to make the copowers $v \bullet x$ into a $V$-functor $({-}) \bullet x : V \to C$ you need symmetry.

Ah, where do I need symmetry here? Or more generally, as per @Mike Shulman's comment, braiding? (assuming I interpreted his comment correctly)

Bruno Gavranović (Nov 28 2023 at 11:10):

Reid Barton said:

That would be a problem. But it's not the case, since the object $\mathcal{C}(c, d)$ of $\mathcal{V}$ is determined (by Yoneda) by all $\mathrm{Hom}_{\mathcal{V}}(a, \mathcal{C}(c, d))$, $a \in \mathcal{V}$, and by assumption, these are given in terms of the tensor by $\mathrm{Hom}_{\mathcal{C}}(a \bullet c, d)$.

What you're saying sounds strange to me so I'll try to repeat it back: are you saying that it's enough to define an ordinary copower functor, and that this determines the $\mathcal{V}$-one?

Dylan McDermott (Nov 28 2023 at 12:33):

Bruno Gavranović said:

Dylan McDermott said:

It's worth noting that Kelly's comment about $V$-functoriality relies on symmetry, and not only because it involves the tensor of $V$-categories. Even if $x$ is fixed, to make the copowers $v \bullet x$ into a $V$-functor $({-}) \bullet x : V \to C$ you need symmetry.

Ah, where do I need symmetry here? Or more generally, as per Mike Shulman's comment, braiding? (assuming I interpreted his comment correctly)

If $V$ is biclosed monoidal, then you end up mixing left and right closure. When $V$ enriches over itself using right closure, the definition of copower gives you a universal property involving left closure. So to define a morphism $V(v, v') \to C(v \bullet x, v' \bullet x)$ you need a map between the two. Maybe braiding is enough, I've never thought about it.

Mike Shulman (Nov 28 2023 at 15:38):

You don't need a braiding either, not just to define copowers (although making them $V$-functorial is a different question). Copowers in a non-symmetric monoidal category are sketched in section 6 of my paper Contravariance through enrichment; I don't know offhand of another reference specifically about monoidal categories, but Street's paper "Absolute colimits in enriched categories" does the even more general case of enriching over a bicategory.