## Applied Category Theory | 414 Members | |

33 Servers |

Sender | Message | Time |
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19 Sep 2020 | ||

V M | (Indeed, `F[f]: F A → F B` and `G[f]: G A → G B` , as we require in the definition of our `M` -morphism). | 12:39:40 |

V M | But in order for this to be a morphism, it needs to satisfy that commutativity condition. So we need to commute.But that's exactly the naturality square, so it's satisfied! | 12:41:53 |

V M | Hold on, I messed up the definition a bit. `M` should be the category of morphisms of `D` (not `C` ). I'll edit that.Fixed! | 12:46:12 |

V M | * Let `F, G: C → D` be two functors from a category `C` to a category `D` .Let `M` be a category of morphisms of `D` : That is, every object of `M` is a morphism `X —f→ Y` in `D` , and given two such morphisms`X —f→ Y` and `U —g→ V` , a morphism `m: f → g` in `M` from `f` to `g` is any pair `m = (m₁, m₂)` of morphisms`m₁: X → U` , `m₂: Y → V` in `D` such that the following square commutes: So to be clear, a single morphism `m` in `M` is actually a pair of morphisms `(m₁, m₂)` of our original category `D` , satisfying that commutativity condition. And composition of morphisms in `M` is defined in the obvious way. Given `m: f → g` and `n: g → h` , define`n ∘ m = (n₁ ∘ m₁, n₂ ∘ m₂)` [where `mᵢ` and `nᵢ` are the two "components" of `m` and `n` as in our definition].You can verify that this `M` is a category. | 12:47:20 |

V M | * Since `α` is a natural transformation, by definition for each object `X` of `C` , there is a morphism `αₓ: F X → G X` in `D` .So, let the action of the functor `α` on objects be as follows:For all `X ∈ Ob C` , `α(X) = αₓ` . | 12:48:00 |

V M | * (This is an object of `M` , because `αₓ` is a morphism in `D` ) | 12:48:08 |

V M | * What about the action of `α` on morphisms?For any `f: X → Y` in `C` , we need to define `α[f]: α(X) → α(Y)` in `M` .But remember, a (single) morphism in `M` is a pair of morphisms of `D` . So we define`α[f] = (F[f], G[f])` . | 12:48:39 |

V M | * Hold on, I messed up the definition a bit. `M` should be the category of morphisms of `D` (not `C` ). I'll edit that.Fixed! | 12:49:12 |

V M | Hm… I don't remember if I checked this, but I think the converse is true too: Any functor from `C` to `M` must be some natural transformation between two parallel functors in `C` .If that's true, we can say natural transformations in `C` are exactly functors from `C` to `M` . | 12:52:43 |

V M | * Since `α` is a natural transformation, by definition for each object `A` of `C` , there is a morphism `αₐ: F A → G A` in `D` .So, let the action of the functor `α` on objects be as follows:For all `A ∈ Ob C` , `α(A) = αₐ` . | 12:59:00 |

V M | * (This is an object of `M` , because `αₐ` is a morphism in `D` ) | 12:59:12 |

V M | * What about the action of `α` on morphisms?For any `f: A → B` in `C` , we need to define `α[f]: α(A) → α(B)` in `M` .But remember, a (single) morphism in `M` is a pair of morphisms of `D` . So we define`α[f] = (F[f], G[f])` . | 12:59:30 |

V M | * (Indeed, `F[f]: F A → F B` and `G[f]: G A → G B` , as we require in the definition of our `M` -morphism). | 12:59:44 |

V M | * But in order for this to be a morphism, it needs to satisfy that commutativity condition. So we need to commute.But that's exactly the naturality square, so it's satisfied! | 13:00:03 |

V M | Let `C` be a category, and `M` be the category of morphisms in any category `D` , as we've defined.[Note: I think there are a few different categories called "category of morphisms of a category" so don't take this as the definition]Now, consider any functor from `C` to `M` , and let's label it `α` (though we don't as yet know if it's a natural transformation).Now firstly, for every `A ∈ Ob C` , `α(A)` should be an object in `M` , which means it's actually a morphism in `D` :`α(A) = X —f→ Y` ,for some objects `X` and `Y` in `D` .So we immediately get two mappings from `C` to `D` , namely:`F, G: C → D` , `F A = X` , `G A = Y` (for `X` and `Y` as given above).Or if you wish: `F A = dom α(A)` , `G A = cod α(A)` (resp. the domain and codomain of the object `α(A)` of `M` , which is a morphism of `D` ). | 13:02:24 |

V M | Also, given a morphism `f: A → B` in `C` , we get a morphism `α[f]: α(A) → α(B)` in `M` .Which means we get `α[f] = (m₁, m₂)` , where`m₁: dom α(A) → dom α(B)` and `m₂: cod α(A) → cod α(B)` are morphisms in `D` .That is, `m₁: F A → F B` and `m₂: G A → G B` (by our definition of the mappings `F` and `G` on objects).So now define `F[f] = m₁` , and `G[f] = m₂` , which gives the action of `F` and `G` on morphisms. | 13:05:51 |

V M | Since `α` is a functor, it preserves compositions and identities. And our definition of composition of `M` -morphisms is as component-wise composition (since each `M` -morphism is a pair of `D` -morphisms).So this implies (I can more or less see that working out…) that `F` and `G` as defined above also preserve compositions and identities. | 13:07:29 |

V M | So they're both functors from `C` to `D` . | 13:07:31 |

V M | And then `α` is a natural transformation from `F` to `G` , if we define its components as:For each `A ∈ Ob C` , let `αₐ = α(A)` . | 13:08:08 |

V M | * So they're both functors from `C` to `D` . | 13:08:44 |

Roberto Abdelkader Martínez Pérez | @MVVVVVVVVVVVV wow, I was expecting a yes/no answer and maybe a link but you gave me a full explanation. Thank you very much! | 14:08:14 |

V M | No problem. It's something I figured out a while ago, but I'd completely forgotten about it, so I'm glad I got reminded of it, thanks to your question. | 14:31:40 |

Fran Gómez García changed their display name from Francisco Gómez García to Fran Gómez García. | 15:24:36 | |

Matteo | @MVinay great write up! I used to know this when I first learn CT but then I forgot it and couldn't come up with it anymore 😂 so it's cool to see it again | 16:52:09 |

wendy joined the room. | 17:08:31 | |

juan joined the room. | 23:49:00 | |

21 Sep 2020 | ||

juan | 17:19:46 | |

darkharmony9999 joined the room. | 19:22:28 | |

gabe joined the room. | 22:49:46 | |

26 Sep 2020 | ||

Yosbi J. Gollés joined the room. | 21:02:34 |

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