arthur
arthur-i.bsky.social
arthur
@arthur-i.bsky.social
me too!! I just wish my brain could do both at once, but the past week has shown me that math will just take over completely whether I want it to or not haha
July 8, 2024 at 7:15 PM
lmaooo nice
July 8, 2024 at 7:13 PM
I mean the gator absolutely popped off especially on twitter so I'm pretty happy but this is still very difficult
July 8, 2024 at 4:59 PM
I've even made a to do list for stuff to try to prove/construct/look into like this would've consumed me for a while
July 4, 2024 at 7:43 PM
I only vaguely understand how inaccessibles solve this problem so honestly I'm just assume people have it figured out and continue to use CAT guilt-free
July 3, 2024 at 5:52 PM
lmao I'm glad, I haven't even worked with those yet and I already hate the name
July 3, 2024 at 4:21 PM
so fucking true!!!!!!
July 3, 2024 at 5:19 AM
I think once the gay furries get into category theory it'll usher in a new golden age of math
July 2, 2024 at 9:26 PM
omg I'm exactly the same lol, my main motto in life is that I can't do more than two things without severely underperforming on all of them as a result, and even two is stretching it. but I can't just let my interests go so unfortunately I guess I'll just suffer !
July 2, 2024 at 9:22 PM
true! though personally the directions of the 1-morphisms and 2-morphisms used for // feel very natural to me at least visually lol so it's well-deserved bias
July 1, 2024 at 4:19 PM
fair enough!
June 30, 2024 at 12:18 AM
oh wow this is a much better way of showing it haha
June 30, 2024 at 12:05 AM
me too tbh lol
June 29, 2024 at 11:28 PM
since Ab(Z,-) and Set(*,G(-)) were isomorphic in the first place, the products would be isomorphic too, and this would be natural in X. anyway I'm rambling, this is extremely clunky and overcomplicated but I'm glad I tried it at least lol I'm going to bed
June 29, 2024 at 11:15 PM
this gives you two functors Psh(1) ~= Set -> (Set^Ab)^op (i.e. Set^op x Ab -> Set), which basically take a set and return a product of Ab(Z,-)s or Set(*,G(-))s. if X is the coproduct of a diagram of *s, then the first functor would send X to ~=Ab(F(X),-) and the second would send it to ~=Set(X,G(-))
June 29, 2024 at 11:15 PM
basically I was trying to see if you can use my definition of F to extend this to a natural isomorphism Ab(F(-),-) -> Set(-,G(-)). you can do a funny yoneda extension on both sides using a functor 1 -> (Set^Ab)^op sending the single object to Ab(Z,-) or Set(*,G(-))
June 29, 2024 at 11:15 PM
this is probably the standard way of constructing left adjoints of functors to Set lol
June 27, 2024 at 5:31 PM
as soon as I realized sets are just small coproducts of singleton sets I was like Oh hold on a minute.........
June 27, 2024 at 5:31 PM