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Stream: learning: questions

Topic: Products in enriched categories

Mike Stay (Oct 01 2020 at 18:42):

Under what conditions on V can a V-enriched category have products?

Mike Stay (Oct 01 2020 at 19:18):

Really? For example, let V = ℘(A*), the poset of sets of words in the alphabet A. This has a monoidal product • by taking all pairwise concatenations. A V-enriched category C makes sense: taking a graph whose edges are labeled by elements of A, the set of paths between two vertices gives a set of words in A and C(x,y)•C(y,z) ⊆ C(x,z).

But I don't see how we could possibly have duplication followed by projection onto the first element be the identity unless both duplication and deletion are the empty word. How do you define V-enriched products in an arbitrary monoidal category V?

Nathanael Arkor (Oct 01 2020 at 19:32):

The standard definition seems to require V to have finite products itself, so that a V-enriched product of $X$ and $Y$ in $\mathbf C$ is an object $X \times Y$ such that there is a isomorphism $\mathbf C(A, X \times Y) \cong \mathbf C(A, X) \times \mathbf C(A, Y)$ in V.

Mike Stay (Oct 01 2020 at 20:06):

Thanks! I presume that's natural in A?

Dan Doel (Oct 01 2020 at 20:37):

Yes. If you rub out the $A$ then it is the 'representable presheaf' characterization of products, and an isomorphism of presheaves is natural in their parameter (or whatever it's called).

sarahzrf (Oct 01 2020 at 20:48):

i guess you could swap out × for ⊗ there to define tensor products in any V-enriched category :thinking:

sarahzrf (Oct 01 2020 at 20:49):

does that give you anything interesting in, say, the Ab-enriched case?

sarahzrf (Oct 01 2020 at 20:51):

oh wait lol that doesnt make any sense :sweat_smile:

sarahzrf (Oct 01 2020 at 20:52):

wrong kind of distributivity of ⊸ over ⊗...

sarahzrf (Oct 01 2020 at 20:53):

one would expect A ⊸ B ⊗ C ≅ (A ⊸ B) ⊗ C or something—oh wait that's a strength :eyes:

sarahzrf (Oct 01 2020 at 20:53):

huhh

Dan Doel (Oct 01 2020 at 20:57):

Well, certainly you could talk about representing the tensor product.

sarahzrf (Oct 01 2020 at 21:06):

yeah but i meant it doesnt seem like a very natural kind of thing to occur—for example, the actual tensor product doesnt satisfy it back in the enriching category

John Baez (Oct 01 2020 at 21:18):

It looks like someone deleted their comment before Mike's comment "Really?"

It's not good to be too shy about making mistakes here. It's just a conversation; fixing mistakes can be helpful but deleting comments can make it harder to tell what's going on.

John Baez (Oct 01 2020 at 21:20):

If V is a monoidal category, there's a category VCat.
If V is braided monoidal, VCat gets to be monoidal.
If V is symmetric monoidal, VCat gets to be braided monoidal.
If V is symmetric monoidal, VCat gets to be symmetric monoidal.
If V is symmetric monoidal, VCat gets to be symmetric monoidal.

This looks a bit repetitive but the point is that we're going down one column of the periodic table; VCat is always "one up" from V in this column.

John Baez (Oct 01 2020 at 21:22):

periodic table of n-categories

John Baez (Oct 01 2020 at 21:22):

This lets you guess how things might work for enriched 2-categories, etc.

John Baez (Oct 01 2020 at 21:23):

It's fun to see why V needs to be not merely monoidal but braided to define the tensor product of V-categories.

To tensor two V-categories you take the cartesian product of their sets of objects, and then tensor their hom-objects. But think about how you'll define composition in the tensor product of two V-categories!

John Baez (Oct 01 2020 at 21:25):

Also:

If V is cartesian monoidal, then VCat gets to be cartesian.

Dan Doel (Oct 01 2020 at 21:26):

Are the last two lines in the sequence supposed to be actually duplicated?

Dan Doel (Oct 01 2020 at 21:27):

Also with the premise being the same as the line above them?

John Baez (Oct 01 2020 at 21:31):

Yes, that's why I said "This looks a bit repetitive..." and then explained what's going on here.

John Baez (Oct 01 2020 at 21:32):

The chart helps - that's why I included it.

Dan Doel (Oct 01 2020 at 21:32):

Oh I see. It's symmetric all the way up after you get there.

Dan Doel (Oct 01 2020 at 21:33):

Like h-levels.

John Baez (Oct 01 2020 at 21:34):

Yes. It's "stabilization".

John Baez (Oct 01 2020 at 21:34):

So I predict that if V is a sylleptic monoidal 2-category then the 2-category VCat is braided monoidal, etc.

John Baez (Oct 01 2020 at 21:35):

I don't know if anyone has worked such stuff out.

Mike Stay (Oct 01 2020 at 21:37):

There's something rather Yoneda-like about the definition of an enriched product.

Nathanael Arkor (Oct 01 2020 at 21:39):

This is what Dan was referring to when pointing out this was a definition via representability.

Dan Doel (Oct 01 2020 at 21:40):

Yeah, you can also write it as: $y(A×B) \cong yA × yB$, and that works for a lot of other things.

Dan Doel (Oct 01 2020 at 21:41):

Even $y(A^B) \cong yA^{yB}$

Mike Stay (Oct 01 2020 at 21:45):

Is the notion of enriched Lawvere Theory expressible this way? The definition on the nLab mentions neither enriched products nor presheaves.

Nathanael Arkor (Oct 01 2020 at 21:47):

Enriched Lawvere theories are defined in terms of finite cotensor, rather than finite (enriched) products. John Power mentions why this is at the start of his paper Enriched Lawvere theories. Essentially, finite cotensor are the right generalisation to get a monad–theory correspondence, as in the unenriched setting.

Dan Doel (Oct 01 2020 at 21:50):

The representability thing is orthogonal to enrichment, too.

Dan Doel (Oct 01 2020 at 21:55):

It's just often a pleasant way of presenting categorical constructions. Something to add to the list of ways of doing so ((co)limits, initial/terminal objects, adjunctions, ...).

Dan Doel (Oct 01 2020 at 22:03):

Oh, and I guess there's a caveat with enrichment, because if $V$ is not symmetric monoidal, you can't necessarily use the $y$ version. At least, so says nlab.

Reid Barton (Oct 02 2020 at 01:26):

Normally something that's going to be called the product $X \times Y$ should be equipped with maps (in the enriched case, maps in the "underlying category") $X \times Y \to X$ and $X \times Y \to Y$, and then being the product is the statement that the particular map $\mathbf{C}(A, X \times Y) \to \mathbf{C}(A, X) \times \mathbf{C}(A, Y)$ is an isomorphism for each $A$. In this setting, you don't need to worry about naturality in $A$ because the map always exists and is natural in $A$. (Furthermore, you don't really need to assume that $V$ has products, because you can rephrase the condition as "$\mathbf{C}(A, -)$ takes the diagram to a product diagram in $V$ for every $A$"--the representability form Dan mentioned. If $V$ doesn't have products then it's just less likely that you will be able to find them in a $V$-enriched category.)

Reid Barton (Oct 02 2020 at 01:29):

On the other hand, you can also take "an isomorphism $\mathbf{C}(A, X \times Y) \to \mathbf{C}(A, X) \times \mathbf{C}(A, Y)$" as equipping $X \times Y$ with the structure of the product of $X$ and $Y$, and reconstruct the maps $X \times Y \to X$ and $X \times Y \to Y$ by feeding in the identity of $X \times Y$ to this isomorphism, but in that case I think the isomorphism had better be specifically $V$-natural in $A$.