 ## Applied Category Theory

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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
``F A —F[f]→ F B |          | αₐ         α_b ↓          ↓G A —G[f]→ G B``
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:
``X —f→ Y|     |m₁    m₂↓     ↓U —g→ V``
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
``F A —F[f]→ F B |          | αₐ         α_b ↓          ↓G A —G[f]→ G B``
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 MNo 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
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