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What seems implicit in this talk on youtube, is the claim that every directed multigraph (with loops) can be identified with a finite category and vice versa, if we consider the paths of the directed multigraph rather than just the edges (= 1-edge paths).

Is this true?

If it is true, why is there not more cross-fertilization between category theory and graph theory?

EDIT: The proposed identification:

The nodes are the objects of the finite category.

The identity morphisms are implicit (i.e. do not need to be depicted), but if they are to be depicted, then they are just a loop starting and ending at the corresponding vertex.

As stewbasic pointed out in the commented below, we need to stipulate that each edge in a path can only be traversed once. Each path should have a well-defined source and sink, so that the domain and codomain of our corresponding morphism are well-defined. We do not want to include arbitrary cycles.

Anyway, the morphisms are the paths of the directed multigraph, or the edges of the corresponding transitive closure of said multigraph (we want to interpret the arrows such that the associative rule holds, i.e. the composition of any two morphisms is again a morphism).

My guess would be that finite categories are uninteresting to the typical category theorist, not to mention 1-categories in general, and directed multigraphs are probably too pathological for the average graph theorist to care about.

Note that for this identification to work, we have to assume that the identity morphisms are implicit in the directed multigraph, but that really isn't a problem, since when composed with any other morphism they "disappear", at least in the sense that neglecting to draw the loops corresponding to the identity morphisms on the directed multigraph would not affect whether the transitive closure represents the corresponding finite category or not.

I assumed that nLab would have something to say about this, but no such luck: https://ncatlab.org/nlab/show/category+of+simple+graphs

From this page there is a clear correspondence between the transitive closure of a directed multigraph (or considering the paths of the multigraph instead of the edges) and posets, since we can define the nodes to be reflexive, antisymmetry follows from the multigraph being directed (from the nLab page it follows that symmetry for the relation implied by the multigraph holds if it is undirected), and transitivity follows from taking the transitive closure/considering paths instead of edges.

So if the above correspondence is correct, then in some sense finite categories, directed multigraphs, and posets all "coincide", essentially since they all describe the same type of relation.

I know that every monoid can be represented by a category with one object and the morphisms are just the elements of the monoid (the invertible elements being the isomorphisms).

EDIT: I think this question might be a duplicate, so I am voting to close this question. The original question did not come up in the suggestions or similar questions while writing this one, so thanks must go to Musa Al-hassy for finding and pointing out the similarity:

Is there a monoid structure on the set of paths of a graph?

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Chill2Macht
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    Can a path include an edge more than once? eg if the graph is a single vertex with a single loop edge, I can follow the edge n times for any n>=0; are these considered distinct paths? If so, the resulting category wouldn't be finite. If not, how is composition defined? – stewbasic Jun 17 '16 at 03:30

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The path category on a graph is like a free algebra; in fact, ignoring size issues, every category can be presented as the path category modulo a congruence relation of a suitable type on the arrows.

(The morphisms in the path category are all paths; including the ones that contain cycles and repeated edges. This includes empty paths, with the caveat that an empty path still remembers where it starts and stops)

I'm not really sure what cross-fertilization would happen; that the underlying graph of a category has an edge set defining a transitive relation makes it rather degenerate from the perspective of graph theory. e.g. the shortest distance between any two objects is either $0$, $1$ or $\infty$, and I don't think I've ever seen anyone ask for an object coloring of a category.

Some things do come up, but they're more topological in nature, like whether a category is connected.

The underlying graph of a category is definitely not enough to determine the category itself.

For example, the category with one object and one nontrivial involution ($x^2 = 1$) has the same underlying graph as the category with one object and one nontrivial idempotent ($x^2=x$).

An important special case where the underlying graph does determine the category is when it is a preorder — since each homset has at most one element, it's easy to identify the product of any two composable morphisms.

  • Wait so automorphisms of finite categories can't be depicted by node/arrow diagrams? I.e. the representation only goes in one direction (such diagrams can always be used to represent a finite category), but not every finite category can be represented accurately with such a graph? – Chill2Macht Jun 17 '16 at 04:01
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    @William: Right, if I understand your question. A category is an algebra of sorts; simply knowing its elements is (usually) not enough to determine its operations. (and presenting the category in this fashion requires generators and relations; just generators usually isn't enough) –  Jun 17 '16 at 04:09
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    @William: ... although in the special case that every homset is either empty or has a unique element, knowing the elements is enough to determine the composition. Partial orders (and preorders and lattices) are important in category theotry –  Jun 17 '16 at 04:12
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You haven't made explicit just how you're making the identification, but I'm pretty sure the answer is no. Let me make sure I have this right: You're saying that you take a finite directed multigraph -- meaning, I take it, a directed graph with multiple edges allowed in each direction, as well as self-loops (in multiples, of course) -- and then taking the category of paths in this graph. (The objects are the vertices, morphisms are directed paths, identity morphism is null path, composition is concatenation.) And then you're asking if every finite category (in the sense of having finitely many objects) is obtained this way.

The answer is no, they aren't, because in such a category, it is impossible for two non-identity morphisms to compose to the identity. I.e., there are no nontrivial isomorphisms, not even nontrivial automorphisms.

Harry Altman
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  • Wait so are you saying that every morphism from an object to itself has to be the identity morphism? I thought that was only true if the object was terminal? – Chill2Macht Jun 17 '16 at 03:37
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    No, there can be trivial endomorphisms, just not trivial automorphisms. And having no nontrival automorphisms can happen with plenty of non-terminal objects. In these categories it happens with every object. In, say, the category of rings, it happens with, e.g., $\mathbb{Z}$ (that one actually has no nontrivial endomorphisms!). – Harry Altman Jun 17 '16 at 03:45
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    What do you mean by "trivial" here? –  Jun 17 '16 at 03:45
  • I assume he means any automorphism which is "trivial" must be the corresponding identity morphism. However, I don't understand why we can't have non-trivial automorphisms -- why can't we just draw a loop around a node and say that "this morphism is not the identity morphism"? I guess defining transitive closure becomes trickier, because then for every morphism mapping to that node you have to add the morphism created by left-composing the automorphism, and for every morphism leaving that node you have to right-compose. I was just saying that we don't have to worry about it for identity morphm. – Chill2Macht Jun 17 '16 at 03:49
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    @William: An automorphism is invertible; in particular you can multiply it by something to get the identity. That can't happen in a path category (unless the arrow is itself the identity); the identity is the only endomorphism that is an automorphism. –  Jun 17 '16 at 03:54
  • @Hurkyl sorry I don't understand this claim -- any monoid with an invertible element has an automorphism (as a category with one object), and every such monoid can be represented as a directed multigraph with loops -- so why is it that this can't happen in a path category? – Chill2Macht Jun 17 '16 at 12:52
  • @William: And the path category generated by a graph with one object is the free monoid generated by the loops; free monoids don't have invertible elements. More generally, every path has a nonnegative length, and composing paths adds their lengths. –  Jun 17 '16 at 13:11
  • @Hurkyl but groups have invertible elements and they are monoids? Also I don't see why we would need nonnegative length, or why a loop of length two couldn't compose to the identity. If the natural numbers can be represented by such a graph (with infinite loops), I don't understand what bars the integers from also being represented – Chill2Macht Jun 17 '16 at 13:12
  • @William: Because that's the definition of the path category generated by a graph. It's the category whose morphisms are the paths in the graph, with composition given by concatenating paths. –  Jun 17 '16 at 13:14
  • @Hurkyl Yeah so why can't you concatenate two loops and say that they are the identity morphism? – Chill2Macht Jun 17 '16 at 14:32
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    Just as with the free group construction, the multiplication is concatenation with possible cancellation, so too we can do that with the composition in the path category. Then we can get non-trivial (non-endo) isomorphisms. See here for details: http://math.stackexchange.com/questions/1826717/is-there-a-monoid-structure-on-the-set-of-paths-of-a-graph – Musa Al-hassy Jun 17 '16 at 18:39
  • @MusaAl-hassy Oh wow thank you so much! I was surprised this question hadn't already been asked for, but it turns out it was! The answers on that post are also very illuminating as well. I really appreciate you taking the time to find, notice, and point out this similarity to me -- thank you again so much! – Chill2Macht Jun 17 '16 at 22:51