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Stream: theory: category theory

Topic: At what point do we start needing ends in hom alg

Patrick Nicodemus (Nov 08 2025 at 20:35):

Suppose I am interested in specifically Ab-enriched category theory as this is an important example that I care about. There is no such thing as the Ab-enriched limit of a diagram $F : J \to A$, where both $J,A$ are Ab-enriched and $F$ respects the additive structure. However, for any $a \in A$ I can still talk about un-enriched cones from $a$ to $F$, and this $Cone(a,F)$ forms an Abelian group under addition. We can form an (ordinary, not enriched) category of cones and talk about the terminal object in this category; and call this a limit.

How far can we go with this before I run into a problem with some construction not being additive?
I thought initially that the limit functor $(J\to A)\to A$ might not be additive on natural transformations, but it seems that it is?

Patrick Nicodemus (Nov 08 2025 at 20:36):

By the way, I already know of one answer to my question: if you want to compute an Kan extension between two additive functors, and you use the naive/ordinary definition of limit instead of an enriched end, you may get a functor that is not additive. But it would be nice if there were a simpler example.

Kevin Carlson (Nov 11 2025 at 00:33):

I'm a bit confused about why the limit weighted by the constant functor at $\mathbb{Z}$ is not such a thing as the Ab-enriched limit of $F.$

Patrick Nicodemus (Nov 11 2025 at 00:35):

Suppose the source category has one object, i.e., it is a not-necessarily commutative ring $R$. Then "the constant functor at $\mathbb{Z}$" should be equivalent to "the canonical ring homomorphism $R\to \mathbb{Z}$". But if $R$ is, say, $\mathbb{Z}/n\mathbb{Z}$ then there may be no such ring homomorphism at all. (We require 1 to be sent to 1 here, 1 is the identity morphism and not the 0 of the Abelian group)

Patrick Nicodemus (Nov 11 2025 at 00:44):

Constant functors exist in general when the monoidal product in the category you're enriching over is the Cartesian product, in which case the unit for the monoidal product agrees with the terminal object. I chose the Ab example as the simplest example of enriching over a non-Cartesian monoidal category.

Kevin Carlson (Nov 11 2025 at 02:50):

Oh, right, I had forgotten that conical enriched limits only make sense when the indexing category is free on an ordinary category (or the monoidal product is cartesian, as you say.)

Mike Shulman (Nov 11 2025 at 07:37):

Patrick Nicodemus said:

There is no such thing as the Ab-enriched limit of a diagram $F : J \to A$, where both $J,A$ are Ab-enriched and $F$ respects the additive structure. However, for any $a \in A$ I can still talk about un-enriched cones from $a$ to $F$.

Do you mean un-enriched cones from $a$ to the underlying ordinary functor of $F$? So that such a thing would consist of, for each object $x\in J$, a morphism $c_x : a\to F(x)$, such that for every morphism $f$ in the abelian group $\hom_J(x,y)$, we have $c_y = F(f) \circ c_x$? It seems to me that the only such cone consists of zero morphisms, since for any $x$ we could take $f = 0 \in \hom_J(x,x)$ and get $c_x = F(0) \circ c_x = 0 \circ c_x = 0$.

Patrick Nicodemus (Nov 11 2025 at 14:11):

Oh, yes. That's a good point, Mike. That's evidently a problem