 ## Category Theory

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Enriched categories, topoi, abelian categories, monoidal categories, homological algebra || part of the +mathematics:matrix.org community6 Servers

SenderMessageTime
19 Aug 2020 Joel Sjögren... where n is a comonad, e.g. n a = d * a06:55:27 Joel SjögrenI am finding the introduction to "Combining a monad and a comonad" informative. Its reference  has a generalized strength like that.07:04:55
21 Aug 2020 Joel SjögrenGiven a functor F : C -> D consider the factorization C -> E -> D where ob(E) = ob(C) but homE(c1, c2) = homD(Fc1, Fc2). Can this be characterized by a universal property? Has anyone used it?07:02:30 Joel SjögrenIt seems like the factorization is initial among C -> E' -> D with E' -> D fully faithful.07:13:39 zigong joined the room.16:15:59
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21 Aug 2020 Timhm, that's an interesting factorisation23:56:49 Timsort of pulling back the morphisms from D along F23:57:11 Timno conditions on F?23:57:37
22 Aug 2020 Joel SjögrenYes, it is a pullback to create the apex of a span from another.08:17:29 Joel SjögrenNo conditions08:20:55
23 Aug 2020 Joel SjögrenI just found it described here more elaborately https://arxiv.org/pdf/math/0608420.pdf12:56:06 Joel Sjögrenpage 1812:56:37 Timoh of course! i've read these notes so many times, can't believe i forgot about them14:17:03
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27 Aug 2020 Flonkmaster3000So the book I'm working through asks: Given P, Q preorders, show that the functor category Q^P is a preorder. It seems pretty obvious, if F, G are functors P -> Q then, since Q is a preorder, there is at most one arrow F a -> G a. And a natural transformation can only choose that one arrow as it's component arrow at a. So there is at most one natural transformation between any F -> G which means Q^P is a preorder. But I suck at maths and I have a hard time writing down the equations. Can someone spell it out?18:49:52 Flonkmaster3000* So the book I'm working through asks: Given P, Q preorders, show that the functor category Q^P is a preorder. It seems pretty obvious, if F, G are functors P -> Q then, since Q is a preorder, there is at most one arrow F a -> G a. And a natural transformation can only choose that one arrow as it's component arrow at a. But I suck at maths and I have a hard time writing down the equations. Can someone spell it out?18:50:01 Flonkmaster3000* So the book I'm working through asks: Given P, Q preorders, show that the functor category Q^P is a preorder. It seems pretty obvious, if F, G are functors P -> Q then, since Q is a preorder, there is at most one arrow F a -> G a. And a natural transformation can only choose that one arrow as it's component arrow at a. But I suck at maths and I have a hard time writing down the equations. Can someone spell it out?18:50:09 Flonkmaster3000 * So the book I'm working through asks: Given P, Q preorders, show that the functor category Q^P is a preorder. It seems pretty obvious, if F, G are functors P -> Q then, since Q is a preorder, there is at most one arrow F a -> G a. And a natural transformation can only choose that one arrow as it's component arrow at a. So there only is one natural transformation between any F -> G which means Q^P is a preorder. But I suck at maths and I have a hard time writing down the equations. Can someone spell it out?18:59:19 Flonkmaster3000 * So the book I'm working through asks: Given P, Q preorders, show that the functor category Q^P is a preorder. It seems pretty obvious, if F, G are functors P -> Q then, since Q is a preorder, there is at most one arrow F a -> G a. And a natural transformation can only choose that one arrow as it's component arrow at a. So there is at most one natural transformation between any F -> G which means Q^P is a preorder. But I suck at maths and I have a hard time writing down the equations. Can someone spell it out?18:59:52 Joel SjögrenIt sounds like a complete argument to me.19:00:42 Flonkmaster3000Hah the few uni maths courses I had had me always spell out the maths, but I have been writing prose for most of these exercises19:02:26 Timsometimes prose is a much nicer way of writing about maths!19:29:36 Timand yes, I honestly don't think I'd change anything about the way you've written that proof — it's really slick!19:30:47 Flonkmaster3000Thanks I guess :) If you guys are happy with the argument then I am too!19:31:52
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