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Stream: theory: category theory

Topic: Connected elements of the category of elements

Bruno Gavranović (Mar 15 2023 at 12:00):

Let $\mathcal{C}$ be a connected category (i.e. $\pi_0(\mathcal{C}) = \mathbf{1}$). Let $F : \mathcal{C} \to \mathbf{Set}$ be a functor.
Then is it correct that the category of connected components of the category of elements $\pi_0(\mathbf{El}(F))$ is equivalent to the terminal category $\mathbf{1}$?

It feels like there should be a straightforward way to prove this, but I'm getting confused about something here and can't pinpoint what

Dan Marsden (Mar 15 2023 at 12:03):

Bruno Gavranovic said:

Let $\mathcal{C}$ be a connected category (i.e. $\pi_0(\mathcal{C}) = \mathbf{1}$). Let $F : \mathcal{C} \to \mathbf{Set}$ be a functor.
Then is it correct that the connected components of the category of elements $\pi_0(\mathbf{El}(F))$ is equivalent to the terminal category $\mathbf{1}$?

It feels like there should be a straightforward way to prove this, but I'm getting confused about something here and can't pinpoint what

What if F is constantly the empty set?

Bruno Gavranović (Mar 15 2023 at 12:08):

Ah, right. Then $\mathbf{El}(F)$ is the empty category.

Dan Marsden (Mar 15 2023 at 12:11):

Bruno Gavranovic said:

Ah, right. Then $\mathbf{El}(F)$ is the empty category.

Yes, that was my thought, although I'm not sure if that answers your question, as I completely missed the category of connected components part, which is something I'm unfamiliar with :)

Martti Karvonen (Mar 15 2023 at 12:38):

If $F$ is constant at some set $X$, won't you get a connected component for each element of $X$?

Bruno Gavranović (Mar 15 2023 at 13:12):

If $F$ is constant at $X$, then a morphism $(M, x) \to (N, y)$ in $\mathbf{El}(F)$ is a map $r : M \to N$ in $\mathcal{M}$ such that $x = y$. Meaning that there's no way to connect $(M, x)$ and $(M, y)$ when $x \neq y$. So it sounds like you're right! Uh, it sounds like my conjecture is just flat out wrong :)

Matteo Capucci (he/him) (Mar 15 2023 at 14:27):

You can use functoriality of $\pi_0$ applied to the projection functor $El(F) \to C$ to understand the situation. The projection is sent to a function $\pi_0(El(F)) \to\pi_0(C) = 1$. This map doesn't constrain at all the $\pi_0$ of $El(F)$. Martti's example shows what goes wrong.

Hugo Bacard (Mar 15 2023 at 22:53):

Bruno Gavranović said:

Let $\mathcal{C}$ be a connected category (i.e. $\pi_0(\mathcal{C}) = \mathbf{1}$). Let $F : \mathcal{C} \to \mathbf{Set}$ be a functor.
Then is it correct that the category of connected components of the category of elements $\pi_0(\mathbf{El}(F))$ is equivalent to the terminal category $\mathbf{1}$?

It feels like there should be a straightforward way to prove this, but I'm getting confused about something here and can't pinpoint what

I know this is not exactly what you asked but I was wondering if your statement can be true if we have $\cal Cat$ instead of $Set$, that is $F: \cal C \to \cal Cat$ [then apply the Grothendieck construction] and further demanding $F$ to be ''pointwise connected'', i.e, $\pi_0(F(c))=1$ for every $c \in \cal C$?.

Jonas Frey (Mar 16 2023 at 00:47):

In general, the set of connected components of the cat of elements of a set valued functor is the colimit of the functor. If the index category is connected we speak of a connected colimit. An example are pushouts.