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Free categories #666
Free categories #666
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If Node and Edge are both sets then it should be a set. Unless I'm confused this is Lemma 4.2 in https://arxiv.org/abs/2112.06609 |
Thanks, good catch. I've tried to formalise this but I'm running into universe levels. The |
Hmm, can you get it to work if everything is at the same level (so not separate levels for nodes and edges)? Might be a good start. Most (all?) examples used for defining (co)limits should have the nodes and graphs in level 0 anyway |
@mortberg I've now completed this for the case where nodes and edges have the same universe levels. I still get a problem with the definition of However, I was wondering if it could be fixed by defining something like data tlift (A : Type ℓv) : Type (ℓ-max ℓv ℓe) where
↑↑ : A → tlift A and then using |
How does one get the CI to run for this PR? |
@mortberg from memory, the code doesn't type check yet - see my comment dated 29 Dec 2021. |
Yes, there is a type called |
I've tried to define the free category over a directed graph. This should make it much easier to define many finite categories (see
Categories.Instances.Cospan
for what we currently have to do to define the free category over● → ● ← ●
).I defined a notion of path in a graph to serve as the homset. The issue I'm having is proving that
Path v w
is a set. I'm guessing it should be true whenNode G
andEdge G
are both sets, but don't see a good way to prove this. One idea I had was to convert paths to lists, then useisOfHLevelRetract
and the fact thatisSet A → isSet (List A)
. However, this path-to-list function doesn't have an inverse (for some lists).Ideas and contributions appreciated.