Timeline for Category with few endofunctors?
Current License: CC BY-SA 4.0
29 events
| when toggle format | what | by | license | comment | |
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| Aug 28, 2021 at 12:29 | vote | accept | Tim Campion | ||
| Aug 28, 2021 at 10:15 | answer | added | Harry West | timeline score: 5 | |
| Aug 24, 2021 at 18:33 | comment | added | Tim Campion | @HarryWest That's beautiful, and really deserves to be an answer! | |
| Aug 24, 2021 at 16:44 | comment | added | markvs | I do not know what an arrow category is and why it came up here. If my suggestion does not work, its ok with me. | |
| Aug 24, 2021 at 16:33 | history | edited | Tim Campion | CC BY-SA 4.0 |
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| Aug 24, 2021 at 16:31 | comment | added | Tim Campion | @MarkSapir I see -- you're suggesting using a bipartite graph admitting a surjective homomorphism to the arrow category. After a bit of thought, I think I've convinced myself that if such a graph is countably infinite, then it admits at least continuum-many endomorphisms, so unfortunately doesn't fit the bill here. I haven't carefully thought through the case of higher cardinalities, but I'm a bit pessimistic. | |
| Aug 24, 2021 at 16:17 | comment | added | markvs | Take a bipartite directed graph where all edges go from part $A$ to part $B$. The free category is not much different from the graph. You add constant endomorphisms, that is it. | |
| Aug 24, 2021 at 14:37 | comment | added | markvs | Yes take the free category. | |
| Aug 24, 2021 at 14:08 | comment | added | Tim Campion | @MarkSapir I believe you when you say that graphs with weird endomorphism monoids can be constructed. But a graph is not a category— most graphs don’t admit even one category structure! You can always take the free category on a graph, but it seems this will generally make the endomorphism monoid grow. So unless you can clarify further, I really don’t understand your suggestion. | |
| Aug 24, 2021 at 12:52 | comment | added | markvs | Yes. Constructing such graphs is an exercise. | |
| Aug 23, 2021 at 14:28 | comment | added | markvs | Endofunctors for graphs are endomorphisms. It is easy to create graphs with weird endomorphism semigroups. | |
| Aug 23, 2021 at 14:16 | comment | added | Tim Campion | @MarkSapir Thanks, I'm still not sure what you're proposing. Every category has an underlying directed graph, so I'm continuing to guess that you mean to take the free category on a directed graph. Could you say something about why you think such a category might be constructed to have few endofunctors? I'm not seeing it right now. | |
| Aug 23, 2021 at 14:13 | comment | added | Tim Campion | @NeilStrickland Thanks, I think that answers most of my questions! | |
| Aug 23, 2021 at 14:13 | comment | added | Tim Campion | @GabrielC.Drummond-Cole Thanks again. I now believe the correct thing to say is that a morphism in $\mathcal C^{\mathcal C}$ is a functor $[1] \times \mathcal C \to \mathcal C$, so the number thereof is bounded by $\kappa^{\kappa'}$ where $\kappa' = 3 \times \kappa$ is the number of morphisms in $[1] \times \mathcal C$. | |
| Aug 23, 2021 at 14:11 | history | edited | Tim Campion | CC BY-SA 4.0 |
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| Aug 22, 2021 at 23:22 | comment | added | Gabriel C. Drummond-Cole | @TimCampion maybe I'm miscalculating, but I think for $\mathcal{C}$ the one object category with a single nontrivial idempotent, $\mathcal{C}^{\mathcal{C}}$ has two objects (the identity and the constant functor) and six morphisms (two identities and 4 non-isomorphisms, each with component the idempotent). | |
| Aug 22, 2021 at 22:20 | comment | added | markvs | @TimCampion: You want an example. I suggested that some graph may be an example. Of course, not any graph. | |
| Aug 22, 2021 at 21:50 | comment | added | Neil Strickland | What if $\mathcal{C}$ is $\mathbb{N}$, considered as a category with one object? Then $\mathcal{C}^{\mathcal{C}}$ is equivalent to the discrete category with objects $\mathbb{N}$, so both $\mathcal{C}$ and $\mathcal{C}^{\mathcal{C}}$ have countably many morphisms. | |
| Aug 22, 2021 at 21:38 | comment | added | Andrej Bauer | Ah, sorry, I missed the first occurrence of "skeleton". | |
| Aug 22, 2021 at 21:35 | comment | added | Tim Campion | @MarkSapir I’m not sure what you’re suggesting but the free category on a directed graph can have more endofunctors than objects. Consider the finite cardinals for instance. | |
| Aug 22, 2021 at 21:33 | comment | added | Tim Campion | @AndrejBauer I don’t understand— we’re talking about 1-categories so there’s only one notion of isomorphism. And I thought I did say to take the skeleton before any relevant cardinality counts. | |
| Aug 22, 2021 at 21:30 | comment | added | Tim Campion | @GabrielC.Drummond-Cole fixed, thanks | |
| Aug 22, 2021 at 21:29 | history | edited | Tim Campion | CC BY-SA 4.0 |
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| Aug 22, 2021 at 19:22 | comment | added | Gabriel C. Drummond-Cole | Why is $\kappa^\lambda$ an upper bound? If $\lambda$ is 1 it seems like there can be more than $\kappa$ many morphisms (indeed $\kappa+1>\kappa$ if $\kappa$ is finite) | |
| Aug 22, 2021 at 19:10 | comment | added | Andrej Bauer | "Skeleton" up to natural isomorphism or natural equivalence? It probably does not matter. And another thought: should we not be looking at objects up to isomorphism, or at least count objects in the skeleton of $\mathcal{C}$? | |
| Aug 22, 2021 at 18:52 | comment | added | markvs | A directed graph? | |
| Aug 22, 2021 at 18:36 | history | edited | Tim Campion | CC BY-SA 4.0 |
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| Aug 22, 2021 at 17:55 | history | edited | Tim Campion | CC BY-SA 4.0 |
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| Aug 22, 2021 at 17:36 | history | asked | Tim Campion | CC BY-SA 4.0 |