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 |

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9 Sep 2020 |

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10 Sep 2020 |

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