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Let $\mathcal C$ be a category whose skeleton has $\lambda$-many objects and $\kappa$-many morphisms. Then the skeleton of the endofunctor category $\mathcal C^{\mathcal C}$ has at most $\kappa^{3 \times \kappa}$ many morphisms.

My guess is that in most cases, this upper bound is achieved.

Question: What is an example of a category $\mathcal C$ (other than the terminal category or equivalents) whose skeleton has $\lambda$ many objects and $\kappa$ many morphisms, but the skeleton of whose endofunctor category $\mathcal C^{\mathcal C}$ has

  1. Strictly fewer than $\kappa^\kappa$ many morphisms?

  2. Strictly fewer than $\kappa^\kappa$ many objects?

  3. As few as $\kappa$-many morphisms?

  4. As few as $\lambda$-many objects?

EDIT: Neil Strickland's example of $B\mathbb N$ in the comments below affirmatively answers (1), (2), and (3). So it remains to see about (4).

Note that $\mathcal C^{\mathcal C}$ always has (skeletally) at least $\lambda+1$ many objects and $\kappa+1$-many morphisms, given by constant morphisms between constant endofunctors, along with the identity functor.

Note also that if $\mathcal C$ is accessible, and has as many objects and morphisms as the size of the universe, then the number of accessible endofunctors and morphisms between them is also the size of the universe. So if my guess is correct, then most endofunctors are usually non accessible.

See also here for an argument which shows that if $\mathcal C$ has small products and coproducts and is not a preorder, then $\mathcal C^{\mathcal C}$ has (skeletally) at least $2^\kappa$-many objects, where $\kappa$ is the size of the universe.

Tim Campion
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  • A directed graph? – markvs Aug 22 '21 at 18:52
  • "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 $\mathcal{C}$? – Andrej Bauer Aug 22 '21 at 19:10
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    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) – Gabriel C. Drummond-Cole Aug 22 '21 at 19:22
  • @GabrielC.Drummond-Cole fixed, thanks – Tim Campion Aug 22 '21 at 21:30
  • @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. – Tim Campion Aug 22 '21 at 21:33
  • @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. – Tim Campion Aug 22 '21 at 21:35
  • Ah, sorry, I missed the first occurrence of "skeleton". – Andrej Bauer Aug 22 '21 at 21:38
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    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. – Neil Strickland Aug 22 '21 at 21:50
  • @TimCampion: You want an example. I suggested that some graph may be an example. Of course, not any graph. – markvs Aug 22 '21 at 22:20
  • @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). – Gabriel C. Drummond-Cole Aug 22 '21 at 23:22
  • @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 $[1] \times \mathcal C$. – Tim Campion Aug 23 '21 at 14:13
  • @NeilStrickland Thanks, I think that answers most of my questions! – Tim Campion Aug 23 '21 at 14:13
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    @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. – Tim Campion Aug 23 '21 at 14:16
  • Endofunctors for graphs are endomorphisms. It is easy to create graphs with weird endomorphism semigroups. – markvs Aug 23 '21 at 14:28
  • Yes. Constructing such graphs is an exercise. – markvs Aug 24 '21 at 12:52
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    @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. – Tim Campion Aug 24 '21 at 14:08
  • Yes take the free category. – markvs Aug 24 '21 at 14:37
  • 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. – markvs Aug 24 '21 at 16:17
  • @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. – Tim Campion Aug 24 '21 at 16:31
  • 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. – markvs Aug 24 '21 at 16:44
  • @HarryWest That's beautiful, and really deserves to be an answer! – Tim Campion Aug 24 '21 at 18:33

1 Answers1

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An example of (1),(2),(3),(4) with $\kappa=\lambda=|\mathbb R|$ is to take $\mathcal C$ to be the posetal category $\mathbb R.$ This is already skeletal. Its category of endofunctors is the poset of non-decreasing functions $f:\mathbb R\to\mathbb R,$ with the pointwise order. There are at most $|\mathbb R^{\mathbb Q}|=|\mathbb R|$ such functions: $f$ is determined by the preimages $f^{-1}((-\infty,q))$ for $q\in\mathbb Q,$ each of these intervals is convex, and there are at most $|\mathbb R|$ convex subsets of $\mathbb R.$

(I don't know if there are, unconditionally, examples with $\lambda$ and $\kappa$ regular. I think the $\mathbb R$ example generalizes to $2^\mu$ with the lex order, where $\mu$ is an ordinal satisfying $|2^{2^{<\mu}}|=|2^\mu|,$ for example $\mu=\beth_\alpha$ with $\alpha$ a limit ordinal. Here $2^{<\mu}$ means the functions $\mu\to 2$ with bounded support.)

Harry West
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