| 19 Aug 2020 |
 Joel Sjögren | ... where n is a comonad, e.g. n a = d * a | 06:55:27 |
| Joel Sjögren | I 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 |
| Joel Sjögren | Given 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 Sjögren | It 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 |
| 25 Aug 2020 |
|  robin changed their display name from hekate to robin. | 17:55:00 |
| 21 Aug 2020 |
 Tim | hm, that's an interesting factorisation | 23:56:49 |
| Tim | sort of pulling back the morphisms from D along F | 23:57:11 |
| Tim | no conditions on F? | 23:57:37 |
| 22 Aug 2020 |
| Joel Sjögren | Yes, it is a pullback to create the apex of a span from another. | 08:17:29 |
| Joel Sjögren | No conditions | 08:20:55 |
| 23 Aug 2020 |
| Joel Sjögren | I just found it described here more elaborately https://arxiv.org/pdf/math/0608420.pdf | 12:56:06 |
| Joel Sjögren | page 18 | 12:56:37 |
| Tim | oh of course! i've read these notes so many times, can't believe i forgot about them | 14:17:03 |
| 26 Aug 2020 |
|  Flonkmaster3000 joined the room. | 14:17:37 |
| 27 Aug 2020 |
| 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? | 18:49:52 |
| 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 Sjögren | It sounds like a complete argument to me. | 19:00:42 |
| Flonkmaster3000 | Hah the few uni maths courses I had had me always spell out the maths, but I have been writing prose for most of these exercises | 19:02:26 |
| Tim | sometimes prose is a much nicer way of writing about maths! | 19:29:36 |
| Tim | and 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 |
| Flonkmaster3000 | Thanks 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 |
|  unrooted joined the room. | 13:45:08 |
| 9 Sep 2020 |
|  diamondtoken joined the room. | 00:18:37 |
| 10 Sep 2020 |
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