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Stream: learning: questions

Topic: joint equalizer

Jan Pax (Oct 31 2023 at 20:42):

How do we utilize $t\cdot t=t$ to obtain a factorization $\tau$ of $t$ throguh $d$ ? Snímek-obrazovky-2023-10-31-212130.png

Morgan Rogers (he/him) (Oct 31 2023 at 20:58):

I think there is an error in the question, please could you edit it?

Jan Pax (Oct 31 2023 at 20:59):

Thank you for noticing my question. Could you specify that error for me ?

Morgan Rogers (he/him) (Oct 31 2023 at 21:00):

I mean grammatically, rather than mathematically, it doesn't currently make sense.

Jan Pax (Oct 31 2023 at 21:02):

My English is poor, but I have done my best to make it clear. What's wrong with it ?

Jan Pax (Oct 31 2023 at 21:03):

Is it better now ?

David Egolf (Oct 31 2023 at 21:58):

It looks like we have two categories: $\mathcal{A}$ and $RGraph$. The data of an object in $\mathcal{A}$ is a tuple $(G_e, t, s)$ where $G_e$ is a set and $t,s: G_e \to G_e$ are functions. Further, the conditions $s \circ t = t$ and $t \circ s = s$ must be satisfied. The goal, I think, is to create a reflexive graph from any such tuple.

The data of a reflexive graph, an object in $RGraph$, is a tuple $(G_e, G_v, \tau, \sigma, d)$. Here $G_e$ and $G_v$ are I think sets, $\tau: G_e \to G_v$ tells us the target of an edge, and $\sigma: G_e \to G_v$ tells us the source of an edge, and $d: G_v \to G_e$ assigns to each vertex the loop from that vertex to itself. That fact that, for each vertex $v$, the output $d(v)$ of $d$ is a loop from $v$ to $v$ is expressed by the condition that $\tau \circ d = s \circ d = 1_{G_v}$.

To make a reflexive graph from the data $(G_e, t, s)$, we need to come up with a set of vertices. To do this, we take the "joint equalizer" of the below diagram in $\mathsf{Set}$, which I assume is just the limit of this diagram:
diagram

I believe that the limit of this should be the subset of $G_e$ containing elements $e$ so that $s(e)=t(e)=e$. So, our set of vertices $G_v$ I think is the subset of elements $e$ from $G_e$ satisfying $s(e)=t(e)=e$. Then we can define the function $d: G_v \to G_e$ as the inclusion of these vertices into $G_e$. (I think this is the key morphism that makes $G_v$ a limit of this diagram).

We then want to determine a $\tau: G_e \to G_v$, which will be the "target" function for our reflexive graph. It is at this point that I get stuck! I think this is what the question above was asking about, as well. The book referenced in the screenshot says this:

This [joint equalizer diagram] yields the canonical factorizations $\tau : G_e \to G_v$ of $t$ through $d$, and $\sigma: G_e \to G_v$ of $s$ though $d$ (such factorizations exist because $t = s \circ t = t \circ s \circ t = t \circ t$ and analogously for s).

But I don't follow what the author is saying here....

I think maybe the author is indicating there is a function $\tau$ that makes this diagram commute:
factoring t through d

David Egolf (Oct 31 2023 at 22:09):

Ah, I think I get it! We use the universal property of a limit to get our $\tau$:
universal property

We need $s \circ t =t$ and $t \circ t = t$ for the "cone" with tip $G_e$ to actually be a cone. But both of these are true equations, so this is a real cone, and so we can use this setup to uniquely specify $\tau$.

To specify $\sigma$, I think one just has to use an analogous cone with tip $G_e$ with "legs" made from $s$ instead of $t$.

Morgan Rogers (he/him) (Nov 01 2023 at 09:01):

Yes @Jan Pax , your question makes sense now. I'm sorry for making a comment on your English, but I didn't want your question to be ignored because it wasn't understood :sweat_smile:
Thanks for the thorough response @David Egolf