Theorem 1. let $Q$ be a thin quiver, then every subobject of $Q$ is thin.
Proof. Thinness is a non-existence condition stating that there do not exist morphisms $m: A \to B$ and $n : A \to B$. Non-existence conditions are subset closed. $\square$
It is rather trivial that thin quivers are subobject closed, and that the subobjects of thin quivers are again thin. Of course, it also follows that any subcategory of a thin category is again thin. Far more interesting is the case of congruences of quivers that have thin quotients.
Lemma 1. let $Q$ be a thin quiver. Then $Q$ has no congruences that collapse two morphisms $m: A \to B$ and $n : C \to D$ without also equating two objects. $Q$ has a unique object-preserving partition.
Proof. $Q$ is a thin quiver it therefore follows that for the two morphisms $m$ and $n$ we must have either $A \not= C$ or $B \not= D$. Suppose that $A \not= C$ then we would have to equate $A$ and $C$ under the congruence so that would produce an equal pair of objects. Likewise, if $B \not= D$ then we would have to equate $B$ and $D$ also producing an equal pair of objects. So no two non-parallel morphisms can be equated under a thin quiver congruence. $\square$
This is the base case, which is that there is a unique congruence for the object partition that preserves all objects. Our goal is to show that there is a unique quiver congruence associated to every object partition.
Lemma 2. let $Q$ be a quiver, then the equivalence minimal congruence that turns $Q$ into a thin quiver $Thin(Q)$ is the congruence that equates all parallel pairs of morphisms in $Q$ and no objects. All other morphisms $f: Q \to T$ from $Q$ to a thin quiver factor through $Thin(Q)$.
Proof. let $Q$ be a quiver. Then in order for $Q$ to be thin there must not be any pairs of morphisms $m$ and $n$ with the property that $s(m) = s(n)$ and $t(m) = t(n)$. Thusly, to make $Q$ thin we can equate all such pairs of morphisms $m$ and $n$ to get the congruence $Thin(Q)$. This has as a quotient a thin quiver because it can not have any pairs $m$ and $n$ with $s(m) = s(n)$ and $t(m) = t(n)$ for if it did it would contradict the fact that we collapsed them. Then to see minimality, consider that any other congruence $C$ of $Q$ must also collapse all such pairs $m$ and $n$. It follows that $Thin(Q)$ is the minimal thin congruence. $\square$
With these two lemmas we now have have the means we need to prove our main theorem, which is that the thin congruences of a quiver are fully determined by object partitions.
Theorem 2. Let $Q$ be a quiver and $B$ a partition of its objects. Then there is a unique quiver congruence of $Q$ with $B$ its object partition that has a thin quotient.
Proof. there is always a unique quiver congruence associated to any object partition $B$ of a quiver $Q$: the congruence that collapses all objects in $B$ but no morphisms. We can always form this congruence by the pair $(id,B)$. Then the quotient of this congruence need not be thin. So by lemma 2 we can form a partition $A$ of the morphism set $E$ by equating all parallel edges which turns $\frac{Q}{(id,B)}$ into a thin quiver. This forms the following commutative diagram:
Then also by lemma one, we see that every morphism from $Q$ to a thin quiver $T$ that goes through $B$ must filter through $\frac{Q}{(A,B)}$. However, by lemma one we know that no such further congruence can exist without further equating the objects of $B$. So $\frac{Q}{(A,B)}$ is the only thin quiver produced by a congruence with object partition $B$. It follows that thin congruences are fully determined by their object partitions. $\square$
If by theorem 2, we have that each object partition is uniquely associated to a thin congruence, then there is a one to one mapping $U : Con(Ob(Q)) \to Con(Q)$ that maps any object partition into a thin congruence.
Corollary. let $Q$ be a quiver then the suborder of $Con(Q)$ consisting of all thin congruences is isomorphic to the partition lattice $Con(Ob(Q))$ of partitions of the object set $Ob(Q)$.
It follows from this that the lattice of thin congruences of $Q$ is simply a partition lattice. We can further consider the thin mapping we defined in lemma 2 to be a closure operation on $Con(Q)$.
Corollary. $Thin: Con(Q) \to Con(Q)$ is a closure operation on the lattice of congruences of a quiver $Con(Q)$ that maps any congruence to its thin component.
The relevance of this result is that we can understand the epi-mono factorisations in the full subcategory of the topos $Quiv$ of thin quivers. We see from this that every morphism of directed graphs is determined by a vertex partition on the one hand a subset of vertices and edges on the other. The interesting thing about this is that it produces a dichotomy between congruences and subobjects, where only the later requires both object and morphism components.
Of particular interest is the application of this theory to considering subobjects and congruences of binary operations, which can simply be considered to be quivers with an algebraic function operation adjoined to them in the topos $Sets^{T_{2,3}}$ of ternary quivers. This leads into a more advanced topos theoretic theory of algebraic operations and their subobjects and congruences, which will be helpful in defining the topos theoretic foundations of algebra.
We see that topos theory provides the best foundation for combinatorics because topoi like $Quiv$ let you reason logically about graphs, digraphs, etc. $Sets^{[1,2]}$ lets you reason about hypergraphs. In order to make it the best foundation for abstract algebra we need to consider other topoi like $Sets^{T_{2,3}}$. In order to use topoi in geometry, as was originally intended, we can consider topoi of sheaves $Sh(X)$ of topological spaces. Every branch has its own topoi available for further study, which makes topos theory such an exciting field for further research and analysis.
References:
Quiver in nlab
Topoi: The Categorial Analysis of Logic
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