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Stream: theory: category theory

Topic: New category from an endofunctor

Jean-Baptiste Vienney (Aug 02 2024 at 18:11):

Let $\mathcal{C}$ be category and $F:\mathcal{C} \rightarrow \mathcal{C}$ an endofunctor.

Then, I can obtain a new category $\mathcal{C}_F$ as follows:

• $\mathrm{Ob}(\mathcal{C}_F)=\mathrm{Ob}(\mathcal{C})$,
• $\mathcal{C}_F[A,B]=\mathcal{C}[F(A),F(B)]$,
• if $v \in \mathcal{C}_F[B,C]$ and $u \in \mathcal{C}_F[A,B]$, then $v \circ^{\mathcal{C}_F} u = v \circ u$,
• $\mathrm{id}^{\mathcal{C}_F}_A = \mathrm{id}_{F(A)}$.

It seems to me that it is similar to the idea of a Kleisli category but without assuming than the endofunctor is a monad, and even simpler that the category of algebras over an endofunctor. Somehow, the simplest category you can create from an endofunctor.

Does it have a name?

Jean-Baptiste Vienney (Aug 02 2024 at 18:14):

Hmm I'm unsure if I didn't want to write instead:

• $\mathcal{C}^F[A,B]=\{F(f),~f \in \mathcal{C}[A,B]\}$

Kevin Carlson (Aug 02 2024 at 18:15):

The first version you posted is simply the factorization of $F$ into a bijective-on-objects followed by a fully-faithful functor. It doesn't have anything to do with $F$ being an endofunctor.

Kevin Carlson (Aug 02 2024 at 18:16):

The second version won't work, since images of functors are not always subcategories.

fosco (Aug 02 2024 at 18:16):

By construction, the functor $\mathcal{C} \to$ {your category} is bijective on objects. This category arises in a certain sense as a Kleisli category: it's related to the Kleisli category of the codensity monad of F

Jean-Baptiste Vienney (Aug 02 2024 at 18:17):

Kevin Carlson said:

The second version won't work, since images of functors are not always subcategories.

What do you mean? $F(f) \circ F(g)=F(f \circ g)$, where is the issue?

fosco (Aug 02 2024 at 18:17):

(is it the same thing? Don't remember now)

Kevin Carlson (Aug 02 2024 at 18:17):

$f$ and $g$ aren't necessarily composable just because $F(f)$ and $F(g)$ are!

Kevin Carlson (Aug 02 2024 at 18:18):

Consider the functor from the disjoint union of two walking arrows that glues one domain to one codomain. The resulting quotient is a 2-simplex but the long arrow of the 2-simplex isn't in the image.

fosco (Aug 02 2024 at 18:18):

Jean-Baptiste Vienney said:

Kevin Carlson said:

The second version won't work, since images of functors are not always subcategories.

What do you mean? $F(f) \circ F(g)=F(f \circ g)$, where is the issue?

The issue is that if $$f,g$ are composable then $Ff, Fg$ are; not viceversa. For example, take a constant functor onto a single object, $f,g$ in two disconnected components of the domain category...

Jean-Baptiste Vienney (Aug 02 2024 at 18:18):

I see what you mean but I wrote that a morphism from $A$ to $B$ in the new category if of the form $F(f)$ where $f$ is necessarily from $A$ to $B$.

Jean-Baptiste Vienney (Aug 02 2024 at 18:19):

I don't ask just that it is of the form $F(f):F(A) \rightarrow F(B)$ for any morphism $f$.

Nathanael Arkor (Aug 02 2024 at 18:20):

Jean-Baptiste Vienney said:

It seems to me that it is similar to the idea of a Kleisli category but without assuming than the endofunctor is a monad

As Kevin has pointed out, the original definition is the [[full image]] of $F$. However, there's a reason it looks similar to a Kleisli category: it is precisely the Kleisli category for $F$ viewed as an $F$- [[relative monad]].

Jean-Baptiste Vienney (Aug 02 2024 at 18:22):

Woh, very cool!

Jean-Baptiste Vienney (Aug 02 2024 at 18:23):

So an endofunctor $F$ is a always an $F$-relative monad?

Jean-Baptiste Vienney (Aug 02 2024 at 18:24):

Jean-Baptiste Vienney said:

I see what you mean but I wrote that a morphism from $A$ to $B$ in the new category if of the form $F(f)$ where $f$ is necessarily from $A$ to $B$.

Does it solve the issue, or am I still misunderstanding?

Jean-Baptiste Vienney (Aug 02 2024 at 18:25):

Ok, I see the issue.

Jean-Baptiste Vienney (Aug 02 2024 at 18:34):

fosco said:

Jean-Baptiste Vienney said:

Kevin Carlson said:

The second version won't work, since images of functors are not always subcategories.

What do you mean? $F(f) \circ F(g)=F(f \circ g)$, where is the issue?

The issue is that if $$f,g$ are composable then $Ff, Fg$ are; not viceversa. For example, take a constant functor onto a single object, $f,g$ in two disconnected components of the domain category...

Argh, no I still don't understand. I think this is a subcategory of the first one and I could even get a bigger one by writing $\mathcal{C}^F[A,B]=\{F(f):F(A) \rightarrow F(B),~f \in \mathcal{C}[X,Y],~X,Y \in \mathrm{Ob}(\mathcal{C})\}$. No need that $f$ and $g$ are composable since you just compose $F(f)$ and $F(g)$.

Jean-Baptiste Vienney (Aug 02 2024 at 18:36):

Though, I think these two subcategories are not very useful.

Kevin Carlson (Aug 02 2024 at 18:44):

But the whole point is that $F(f)\circ F(g)$ need not be of the form $F(h)$ for any $h$!

Kevin Carlson (Aug 02 2024 at 18:44):

You could take the subcategory generated by these morphisms, if you wanted. This is the image.

Jean-Baptiste Vienney (Aug 02 2024 at 18:46):

Kevin Carlson said:

But the whole point is that $F(f)\circ F(g)$ need not be of the form $F(h)$ for any $h$!

Ok, so the third category doesn't make sense. But in the second one you're doing $F(f) \circ F(g)$ where $f$ and $g$ are composable.

Jean-Baptiste Vienney (Aug 02 2024 at 18:47):

So that $F(f) \circ F(g)=F(f \circ g)$.

Jean-Baptiste Vienney (Aug 02 2024 at 18:47):

No, sorry.

Kevin Carlson (Aug 02 2024 at 18:48):

The second one being $C_F(A,B)=\{F(f), f\in C(A,B)\}$?

Jean-Baptiste Vienney (Aug 02 2024 at 18:48):

It works if you write this:
$\mathcal{C}_F[A,B]=\{F(f), f:A \rightarrow B, \text{ such that }F(f):F(A) \rightarrow F(B)\}$

Kevin Carlson (Aug 02 2024 at 18:49):

You can quotient $C$ by the congruence $f:A\to B\sim g:A\to B$ if $F(f)=F(g),$ sure. The "such that" statement you wrote doesn't seem meaningful.

Jean-Baptiste Vienney (Aug 02 2024 at 18:50):

Yes, it doesn't make sense.

Kevin Carlson (Aug 02 2024 at 18:50):

I guess this is the factorization of $F$ as a bijective-on-objects-and-full functor followed by a faithful functor.

Kevin Carlson (Aug 02 2024 at 18:50):

But for sure it again doesn't require $F$ to be an endofunctor.

Jean-Baptiste Vienney (Aug 02 2024 at 18:50):

Kevin Carlson said:

The second one being $C_F(A,B)=\{F(f), f\in C(A,B)\}$?

But so, this one works?

Jean-Baptiste Vienney (Aug 02 2024 at 18:51):

You have $F(f):F(A) \rightarrow F(B)$, $F(g):F(B) \rightarrow F(C)$ and you're just doing $F(g) \circ F(f)$.

Kevin Carlson (Aug 02 2024 at 18:51):

Yes, as I said, this is the (bijective on object+full , faithful) factorization

Jean-Baptiste Vienney (Aug 02 2024 at 18:52):

$$\mathcal{C}^F[A,B]=\{F(f):F(A) \rightarrow F(B),~f \in \mathcal{C}[X,Y],~X,Y \in \mathrm{Ob}(\mathcal{C})\}$$. No need that $$f$$ and $$g$$ are composable since you just compose $$F(f)$$ and $$F(g)$$.
`````
And I think this one also works.

Kevin Carlson (Aug 02 2024 at 18:53):

I can't read that one.

Kevin Carlson (Aug 02 2024 at 18:53):

Isn't that the same third one we saw doesn't work a minute ago?

Jean-Baptiste Vienney (Aug 02 2024 at 18:53):

Jean-Baptiste Vienney said:

fosco said:

Jean-Baptiste Vienney said:

Kevin Carlson said:

The second version won't work, since images of functors are not always subcategories.

What do you mean? $F(f) \circ F(g)=F(f \circ g)$, where is the issue?

The issue is that if $$f,g$ are composable then $Ff, Fg$ are; not viceversa. For example, take a constant functor onto a single object, $f,g$ in two disconnected components of the domain category...

$\mathcal{C}^F[A,B]=\{F(f):F(A) \rightarrow F(B),~f \in \mathcal{C}[X,Y],~X,Y \in \mathrm{Ob}(\mathcal{C})\}$.

Kevin Carlson (Aug 02 2024 at 18:54):

Yeah, that continues not to work for the same reason it didn't work last time you agreed it doesn't work.

Kevin Carlson (Aug 02 2024 at 18:54):

You can compose $F(f)\circ F(g)$ all you want but you won't get something of the form $F(h)$ in general.

Jean-Baptiste Vienney (Aug 02 2024 at 18:54):

Ok, I see.

Jean-Baptiste Vienney (Aug 02 2024 at 18:55):

For my defense, the second version I wrote at the beginning works just fine and then I tried to modify it in ways which don't work anymore because you told me that this second version didn't work.

Kevin Carlson (Aug 02 2024 at 18:56):

Yeah, you're right, I misread your second version at first.

Eric M Downes (Aug 02 2024 at 19:20):

The way this issue ($\sf C\in Cat_0\not\implies\mathit{F}C\in Cat_0$) for $F:\mathsf{C\to D}$ is fixed in forming a [[category algebra]] (where $\mathcal{F}\mathsf{C}_0=\set*$) is to have $Ff\circ Fg:=0$ a sink morphism $0=0\circ h=h\circ 0~\forall h\in \mathsf{D}_1$, whenever $dom(f)\neq cod(g)$

Is there a reason to not do something like that here? You end up with a unique sink for every homset, s.t. $0_{xy}\circ 0_{yz}=0_{xz}$. It seems like the "smallest" category generated by $\mathcal{F}\mathsf{C}$ in some sense. I've probably just described some kind of named closure. :)

Jean-Baptiste Vienney (Aug 02 2024 at 20:04):

Yeah, I want to work with categories, not algebras. And your category algebras don't look much like categories, unless you make them into categories with a single object.

Eric M Downes (Aug 02 2024 at 21:02):

I apologize for the sloppiness with the codomain. It's fine to not want to use it, and it is algebraic in spirit, but what I have described is a category (Ed: when $\sf D$ is a [[gaunt category]] ). Or rather a natural transformation taking $F:{\sf C\to D}\implies F^{(0)}:{\sf C\to D^{(0)}}$, where the codomain of $F^{(0)}$ has $\sf D$ as a faithfully included wide subcategory, and has a unique sink morphism adjoined for every homset of $\sf D$. (The sinks form their own thin/posetal subcategory, excluded only the identities.) I'll do some digging and see if there is a name for it.

But it seems like you found a way to make what you're doing work. So I won't distract you unless you want to discuss it more.

Jean-Baptiste Vienney (Aug 02 2024 at 21:38):

Ok, so you say that if $F:\mathcal{C} \rightarrow \mathcal{D}$ is a functor, then you can define a category $\mathcal{C}^F$ as follows:

• $\mathrm{Ob}(\mathcal{C}^F):=\mathrm{Ob}(\mathcal{C})$
• $\mathcal{C}^F[A,B]:=\{F(f):F(A) \rightarrow F(B), f \in \mathcal{C}[X,Y],~X,Y \in \mathcal{C}\} \sqcup \{\mathrm{undefined}_{A,B}:F(A) \rightarrow F(B)\}$

• If $v \in \mathcal{C}^F[B,C]$ and $u \in \mathcal{C}^F[A,B]$, then:
  (a) $v \circ^{\mathcal{C}^F} u := v \circ u$ if $v=F(g)$ for some $g$ and $u=F(f)$ for some $f$ such that the codomain of $f$ is equal to the domain of $g$,
  (b) $v \circ^{\mathcal{C}^F} u:=\mathrm{undefined}_{A,C}$ if there doesn't exist any $f,g$ satisfying $v=F(g)$ and $u=F(f)$, and such that the codomain of $f$ is equal to the domain of $g$,
  (c) $v \circ^{\mathcal{C}^F} \mathrm{undefined}_{A,B} := \mathrm{undefined}_{A,C}$
  (d) $\mathrm{undefined}_{B,C} \circ^{\mathcal{C}^F} u:=\mathrm{undefined}_{A,C}$

• $\mathrm{id}_{A}^{\mathcal{C}_F}=\mathrm{id}_{F(A)}$

Fine, that's interesting.

Jean-Baptiste Vienney (Aug 02 2024 at 21:48):

The issue is that I believe this is not necessarily a category because I think the identity will not always be unital.

Jean-Baptiste Vienney (Aug 02 2024 at 21:50):

I mean: it looks like sometimes postcomposing a morphism $u$ (or precomposing a morphism $v$) by the identity could return $\mathrm{undefined}_{A,C}$ instead of $u$ (instead of $v$).

Rémy Tuyéras (Aug 02 2024 at 22:28):

Note sure whether this will help @Jean-Baptiste Vienney , but your two constructions seem to be related as follows.

Your functor $F:\mathcal{C} \to \mathcal{C}$ defines a natural transformation:

$F_0:\mathsf{Obj}(\mathcal{C}) \to \mathsf{Obj}(\mathcal{C})$
$F_1:\mathsf{Arr}(\mathcal{C}) \to \mathsf{Arr}(\mathcal{C})$

compatible with identities, source, targets, compositions, etc.

It seems that your first construction is pulling back the structure of $\mathcal{C}$ along $F_0:\mathsf{Obj}(\mathcal{C}) \to \mathsf{Obj}(\mathcal{C})$. This gives you a new category $\mathcal{C}_F$ that gives you the following factorization of $F$:

$\mathcal{C} \mathop{\to}\limits^{a} \mathcal{C}_F \mathop{\to}\limits^{b} \mathcal{C}$

The morphism $a$ should be induced by universal maps for the cones formed by $F$ and the structure of $\mathcal{C}$. You can now use image factorizations on the components of morphism $a$ to form the following factorization:

$\mathcal{C} \mathop{\to}\limits^{a_0} \mathcal{C}^F \mathop{\to}\limits^{a_1} \mathcal{C}_F$

The two constructions seem to be related to the concept of weak [[Cartesian morphisms]], where your fibration would be $\mathcal{C} \mapsto \mathsf{Obj}(\mathcal{C})$ and an obvious Cartesian morphism would be $b:\mathcal{C}_F \to \mathcal{C}$. I am wondering whether the arrow $b \circ a_1:\mathcal{C}^F \to \mathcal{C}$ is Cartesian is some sense too.

Jean-Baptiste Vienney (Aug 02 2024 at 22:40):

Is $a$ doing $a(f)=F(f)$?

Jean-Baptiste Vienney (Aug 02 2024 at 22:41):

I don't understand what you mean by "universal maps for the cones ..."

Rémy Tuyéras (Aug 02 2024 at 22:41):

That would be $b$ that has this property

Jean-Baptiste Vienney (Aug 02 2024 at 22:42):

I think $b$ is doing $b(f)=f$, no?

Rémy Tuyéras (Aug 02 2024 at 22:47):

For example, you construct source and target of $\mathcal{C}_F$ by pulling back the map $(s,t):\mathsf{Arr}(\mathcal{C}) \to \mathsf{Obj}(\mathcal{C}) \times \mathsf{Obj}(\mathcal{C})$ along $F_0 \times F_0$. The pullback gives you a source and target

$(s',t'):\mathsf{Arr}(\mathcal{C}_F) \to \mathsf{Obj}(\mathcal{C})\times \mathsf{Obj}(\mathcal{C})$

where $\mathsf{Arr}(\mathcal{C}_F) = \{(a,f,b)~|~(a,b) \textrm{ in } \mathsf{Obj}(\mathcal{C})\textrm{ and }f:F(a) \to F(b)\}$

Rémy Tuyéras (Aug 02 2024 at 22:48):

The pullback square defines $b$, and if you look at the map at the bottom of that square, it should be $F_0\times F_0$ (representing the mapping $F_0$ on source and target separately)

Rémy Tuyéras (Aug 02 2024 at 22:49):

Ah, my bad, I realized I misread your message -- yes, $a(f) = F_1(f)$

Rémy Tuyéras (Aug 02 2024 at 22:50):

but $b(X) = F_0(X)$

Rémy Tuyéras (Aug 02 2024 at 22:53):

Jean-Baptiste Vienney said:

I don't understand what you mean by "universal maps for the cones ..."

Basically, the universal maps induced by the pullbacks to form $a$

Eric M Downes (Aug 02 2024 at 23:57):

Jean-Baptiste Vienney said:

I mean: it looks like sometimes postcomposing a morphism $u$ (or precomposing a morphism $v$) by the identity could return $\mathrm{undefined}_{A,C}$ instead of $u$ (instead of $v$).

Thank you for catching that; the problem comes if $F$ collapses two objects either of which has automorphisms, without collapsing the autos themselves. I’m looking back at my notes and I misremembered some stuff. But it’s getting late here. I will try again tomorrow.

Eric M Downes (Aug 03 2024 at 03:02):

Okay, couldn’t sleep! What I was suggesting only works if the codomain of $F$ is a [[gaunt category]]; all isos are identities. This is quite a severe restriction for me to forget! Sorry for wasting your time, and thanks for the sanity check; my memory is just shot lately.

John Baez (Aug 03 2024 at 07:56):

I hadn't known the term "gaunt category". I enjoyed this jargonesque definition: a gaunt category is a skeletal category whose core is thin.