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Category Theory

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

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19 Aug 2020
@joel135:matrix.orgJoel Sjögren... where n is a comonad, e.g. n a = d * a06:55:27
@joel135:matrix.orgJoel SjögrenI am finding the introduction to "Combining a monad and a comonad" informative. Its reference [2] has a generalized strength like that.07:04:55
21 Aug 2020
@joel135:matrix.orgJoel 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
@joel135:matrix.orgJoel SjögrenIt seems like the factorization is initial among C -> E' -> D with E' -> D fully faithful.07:13:39
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25 Aug 2020
@hekate:matrix.arcana.loverobin changed their display name from hekate to robin.17:55:00
21 Aug 2020
@thosgood:matrix.orgTimhm, that's an interesting factorisation23:56:49
@thosgood:matrix.orgTimsort of pulling back the morphisms from D along F23:57:11
@thosgood:matrix.orgTimno conditions on F?23:57:37
22 Aug 2020
@joel135:matrix.orgJoel SjögrenYes, it is a pullback to create the apex of a span from another.08:17:29
@joel135:matrix.orgJoel SjögrenNo conditions08:20:55
23 Aug 2020
@joel135:matrix.orgJoel SjögrenI just found it described here more elaborately https://arxiv.org/pdf/math/0608420.pdf12:56:06
@joel135:matrix.orgJoel Sjögrenpage 1812:56:37
@thosgood:matrix.orgTimoh of course! i've read these notes so many times, can't believe i forgot about them14:17:03
26 Aug 2020
@flonk_:matrix.orgFlonkmaster3000 joined the room.14:17:37
27 Aug 2020
@flonk_:matrix.orgFlonkmaster3000So 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
@flonk_:matrix.orgFlonkmaster3000* 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
@flonk_:matrix.orgFlonkmaster3000* 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
@flonk_:matrix.orgFlonkmaster3000 * 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
@flonk_:matrix.orgFlonkmaster3000 * 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
@joel135:matrix.orgJoel SjögrenIt sounds like a complete argument to me.19:00:42
@flonk_:matrix.orgFlonkmaster3000Hah 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
@thosgood:matrix.orgTimsometimes prose is a much nicer way of writing about maths!19:29:36
@thosgood:matrix.orgTimand 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
@flonk_:matrix.orgFlonkmaster3000Thanks I guess :) If you guys are happy with the argument then I am too!19:31:52
4 Sep 2020
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5 Sep 2020
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8 Sep 2020
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9 Sep 2020
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10 Sep 2020
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