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Stream: learning: questions

Topic: From Groups to Categorical Algebra exercise; reflexive graph

David Sprayberry (Dec 23 2024 at 22:09):

In Dominique Bourn's From Groups to Categorical Algebra, there's Definition 1.4.1 (near the beginning, so visible from Amazon's preview sample) that in a category $\mathbb{E}$ a reflexive graph $R$ on $X$ is a pair of split epis $d_0$ and $d_1$ with a common section $s_0$:
image.png
Example 1.4.3 is $X\times X$ on $X$ where the split epis are the product projections $p_0^X$ and $p_1^X$ with common section the diagonal map $s_0^X$.
A morphism of reflexive graphs is then defined to be a pair of maps forming a square, one map from domain to domain, and the other from codomain to codomain, of the reflexive graphs, 'making the diagram commute'.
(*) I am not sure whether this means that just the squares indexed by 0 and 1 ought to commute or also the square formed with the sections.
Part of Exercise 1.4.5 asks to show that there is a unique morphism from the reflexive graph in Definition 1.4.1 to the reflexive graph in Example 1.4.3.
image.png
I haven't been able to show that such a pair $((g_0,g_1),f)$ must be unique.
Some things I've worked out in this exercise are as follows.
Since at least the squares indexed by 0 and 1 must commute, we have $f\circ d_0=g_0$ and $f\circ d_1=g_1$. Precomposing these equations by first going along $s_0$ gives $f\circ d_0 \circ s_0=g_0\circ s_0$ and $f\circ d_1 \circ s_0=g_1\circ s_0$. Using the property of split epis this reduces to $g_0\circ s_0=f=g_1\circ s_0$. Since in this case the section on the bottom of the diagram is just the diagonal map, the uncertainty I expressed in (*) does not make any difference here, as the square formed with the sections commutes anyways.
There's also that each $d_i$ is the coequalizer of $(1_R,s_0\circ d_i)$. Since $g_i=f\circ d_i$ does coequalize $(1_R,s_0\circ d_i)$, $f$ is unique for each $g_i$.
So although it looks as if given one component of the pair, the other is uniquely determined, I can't guarantee there isn't some other pair satisfying the same equations.
I'm hoping to get help solving this as the rest of the book depends on this unique morphism, which is given the name $d_R$.

Jean-Baptiste Vienney (Dec 23 2024 at 22:19):

I think tikzcd doesn't work here, so the best way to insert a commutative diagram is maybe to take a screenshot and then insert a picture here.

Jean-Baptiste Vienney (Dec 23 2024 at 22:20):

(By the way, I would like to study this book, it seems very cool.)

Jean-Baptiste Vienney (Dec 23 2024 at 22:30):

(Also, there is a button to edit your message in case you don’t know it exists)

David Sprayberry (Dec 23 2024 at 22:39):

Thanks your suggestions were helpful. I put images now. That's great hopefully we can collaborate. I haven't been able to move past this point. The book is very cool. The interspersed exercises and cryptic nature of the following material unless you have solved them makes it similar to Paul Halmos's Linear Algebra Problem Book except for category theory.

Jean-Baptiste Vienney (Dec 23 2024 at 23:11):

I don’t know if I will have the time to study the book soon but I’ll try to help you on your question as soon as I can.

Jean-Baptiste Vienney (Dec 23 2024 at 23:52):

« making the diagram commute » means that all the paths from a same object to another same object must be equal.

Jean-Baptiste Vienney (Dec 23 2024 at 23:52):

so all the squares must commute in this definition

David Sprayberry (Dec 24 2024 at 00:01):

You could be right that the square with the sections must commute as well. He can't have meant that the diagram is commutative in it's entirety, though, because it would imply that $d_0=d_1$. Also later on in Corollary 1.6.4, although it is in a different context, he mentions diagrams indexed by 0 and 1. I am not sure whether this hints that was his intended meaning here.

Jean-Baptiste Vienney (Dec 24 2024 at 00:03):

I think he means that all the squares in the diagram commute (probably).

Jean-Baptiste Vienney (Dec 24 2024 at 00:08):

Think about how a classical graph is a graph in this sense in the category of sets and it should become clearer what must be a morphism of graphs in any category

Jean-Baptiste Vienney (Dec 24 2024 at 00:10):

You can also try to do these exercises in the category of sets first

Jean-Baptiste Vienney (Dec 24 2024 at 00:36):

Let’s say we have two reflexive graphs (in the category of sets).

Jean-Baptiste Vienney (Dec 24 2024 at 00:36):

What is a reflexive graph in this case?

Jean-Baptiste Vienney (Dec 24 2024 at 00:37):

It is given by a set $G$ of edges

Jean-Baptiste Vienney (Dec 24 2024 at 00:37):

And a set $X$ of vertices

Jean-Baptiste Vienney (Dec 24 2024 at 00:38):

each edge has a domain and a codomain

Jean-Baptiste Vienney (Dec 24 2024 at 00:38):

and each vertex has an edge from itself to itself (the identity edge of the vertex)

Jean-Baptiste Vienney (Dec 24 2024 at 00:38):

Say we have two reflexive graphs $G_1,G_2$

Jean-Baptiste Vienney (Dec 24 2024 at 00:39):

What is a map from $G_1$ to $G_2$?

Jean-Baptiste Vienney (Dec 24 2024 at 00:39):

You want to map each edge to an edge and each vertex to a vertex, in a coherent way

Jean-Baptiste Vienney (Dec 24 2024 at 00:40):

Also you want to map each identity edge to an identity edge

Jean-Baptiste Vienney (Dec 24 2024 at 00:41):

At the end, you want three equations to be verified: one about the domains, one about the codomains and one about the identities edges

Jean-Baptiste Vienney (Dec 24 2024 at 00:46):

Namely, if I send the edge $e$ to the edge $f(e)$, the vertex $dom(e)$ to the vertex $g(dom(e))$ and the vertex $cod(e)$ to the vertex $g(cod(e)$, I want:

• $dom(f(e))=g(dom(e))$
• $cod(f(e))=g(cod(e))$

And for the identity vertices, I want that

• the identity $s(v)$ of the vertex $v$ is send to the identity of the vertex $g(v)$ (here the equation involve the sections):
  $s(g(v))=f(s(v))$

Jean-Baptiste Vienney (Dec 24 2024 at 00:48):

I didn’t use the same notations (I just tried to explain informally) but it must be clear now what the $3$ required equations in the text mean and that we want exactly these three equations to define a morphism of reflexive graphs (I hope).

David Sprayberry (Dec 24 2024 at 00:53):

Ok I understand now, thanks. So yes the square with the sections must commute.

Jean-Baptiste Vienney (Dec 24 2024 at 00:54):

I haven’t yet read all your questions by the way. I try to understand the text and your questions in parallel

Jean-Baptiste Vienney (Dec 24 2024 at 01:13):

The idea of a reflexive relation is that it is a reflexive graph but with at most on edge between any two vertices.

Jean-Baptiste Vienney (Dec 24 2024 at 01:15):

It is like a poset which is a category where there the arrows are just here to say that an object is smaller than another one. The arrows don’t have any content. The question is just to know whether there is one between two given objects.

Jean-Baptiste Vienney (Dec 24 2024 at 01:18):

The idea of the indiscrete relation on an object is that it relates any two elements of the object.

David Sprayberry (Dec 24 2024 at 01:31):

That makes sense, good point.

David Sprayberry (Dec 24 2024 at 02:34):

I think maybe a counterexample to the claimed uniqueness, using the example of the category of sets, is if in the diagram the reflexive graph $R$ on $X$ can be transformed by a symmetry, some permutation of the vertices that preserves the incidence with the edges, like the rigid motions in the plane in the dihedral groups. Then both this symmetry $f$ with its corresponding map $(g_0,g_1)$ from $R$ to $X\times X$ ensuring that incidence is preserved, and the identity symmetry $1_X$ with the map $(dom,cod)$ would work as morphisms into the indiscrete relation (or complete graph) on the vertices.

Jean-Baptiste Vienney (Dec 24 2024 at 03:25):

What is probably true is that there is a unique map from any reflexive graph on $X$ to the indiscrete reflexive relation on $X$ which is the identity on vertices, in any category.

Jean-Baptiste Vienney (Dec 24 2024 at 03:32):

The original statement is false in the category of sets for every set $X$ of cardinal $\ge 2$. You can map any reflexive graph all to the same vertex and same identity edge in the indiscrete reflexive relation. And you can do it for any choice of “receiving vertex”.

Jean-Baptiste Vienney (Dec 24 2024 at 03:35):

(This is a generalization to reflexive graphs of the notion of constant functor.)

David Sprayberry (Dec 24 2024 at 04:23):

Oh yeah, hadn't thought of that. Thanks for your help on this.