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In the spirit of "no question is dumb"
Is there a way to map from a category C with ob(C) and Mor(C), to a category D where ob(D)=Mor(C) and Mor(D)=ob(C)?
I don't think there is a generic way, imagine the category π,
You'd have three objects in the resulting category, and , but only two arrows, and , but you need at least three arrows, because each object needs an identity.
That said, your question makes me think of string diagrams, where you visually flip the roles of what's a point, a line, and a region, so maybe there's an interesting duality here if you're careful.
Do you want a functor?
@Alex Kreitzberg Yes this is my understanding. Can't get the simplest example to port over, but I wondered if higher categories or other advanced concepts (things I have not played with yet) had anything to offer.
@Jean-Baptiste Vienney Would be nice, but by definitions not sure it can work out unless something creative is done.
Basically the answer to the stated question is "no". But there are sneaky things you can do. A "category internal to Vect" is a category with a vector space of objects and a vector space of morphisms, where everything in sight is linear.
A nontrivial little fact is that a category internal to Vect is equivalent to a pair of vector spaces and a linear map between them:
where is the vector space of objects and is the vector space, not of all morphisms, but of morphisms for some object .
Then by taking the dual vector spaces we get
which by that little fact is equivalent to another category internal to Vect! The space of morphisms in this new category is the dual of the space of objects in the original category.
John Baez said:
Basically the answer to the stated question is "no".
But why? You can just map every object to a single chosen object and every morphism to the corresponding identity.
(Taking into acounts that morphisms in the second category are objects of the first and objects of the second are morphisms of the first but it should not change anything to the existence of these functors.)
But first you have to build such a category .
But is this category given in the data of the problem or are you allowed to build any one?
@Jean-Baptiste Vienney Trying to "preserve" everything from the first category, if at all possible. Collapsing like this isn't ideal (although tbf that may not have been clear from how it the question was posed).
@John Baez Thank you, this is the kind of insight I was hoping for :thumbs_up:
Jean-Baptiste Vienney said:
John Baez said:
Basically the answer to the stated question is "no".
But why?
As far as I can tell, @sean was asking if there's a way to take an arbitrary category C with object set ob(C) and morphism set mor(C) and construct a category D with object set ob(D)=mor(C) and morphism set mor(D)=ob(C).
We cannot. Let C be a category with 1 object and 2 morphisms. Then D would need to have 2 objects and 1 morphism. But there is no category with 2 objects and 1 morphism.
One might consider PoincarΓ© dual string diagrams an attempt to reverse dimensions:
image.png
But of course these diagrams aren't categories of any kind. Which raises a question: what kinds of structured entity are these dual string diagrams?
It also makes me think of hypergraph.
If I remember correctly a hypergraph is given by:
In that case, swapping vertices and edges amount to taking the transpose of the incidence relation.
I remember that there is a notion of hypergraph category but I donβt know much about them.
I know too much about hypergraph categories. :weary: But I've never tried to imagine the 'dual' of a hypergraph category. It's an interesting idea, but don't see how it would make sense, since we know how to compose morphisms, but not objects.
Petri nets are very similar to hypergraphs, except that each hyperedge is oriented, going from some finite set of vertices to some other finite set of vertices. So we have two relations, saying which edges go in to each vertex and saying which edges go out of each vertex.
People have observed that you can dualize any Petri net and get a new Petri net, but nobody has ever done anything with this idea, as far as I know! I have wondered about it.