 | Applied Category Theory | 414 Members |
| 33 Servers |
| Applied Category Theory | 414 Members |
| 33 Servers |
| Sender | Message | Time |
|---|---|---|
| 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 needBut 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 morphismsX —f→ Y and U —g→ V, a morphism m: f → g in M from f to g is any pair m = (m₁, m₂) of morphismsm₁: X → U, m₂: Y → V in D such that the following square commutes: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, definen ∘ 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 defineα[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 defineα[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 needBut 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), wherem₁: 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 | |