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Applied Category Theory

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19 Sep 2020
@telegram_220807826:t2bot.ioV 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
@telegram_220807826:t2bot.ioV 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
@telegram_220807826:t2bot.ioV 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
@telegram_220807826:t2bot.ioV 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
@telegram_220807826:t2bot.ioV 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
@telegram_220807826:t2bot.ioV M * (This is an object of M, because αₓ is a morphism in D) 12:48:08
@telegram_220807826:t2bot.ioV 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
@telegram_220807826:t2bot.ioV 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
@telegram_220807826:t2bot.ioV 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
@telegram_220807826:t2bot.ioV 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
@telegram_220807826:t2bot.ioV M * (This is an object of M, because αₐ is a morphism in D) 12:59:12
@telegram_220807826:t2bot.ioV 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
@telegram_220807826:t2bot.ioV 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
@telegram_220807826:t2bot.ioV 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
@telegram_220807826:t2bot.ioV 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
@telegram_220807826:t2bot.ioV 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
@telegram_220807826:t2bot.ioV 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
@telegram_220807826:t2bot.ioV M So they're both functors from C to D. 13:07:31
@telegram_220807826:t2bot.ioV 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
@telegram_220807826:t2bot.ioV M * So they're both functors from C to D. 13:08:44
@telegram_13622077:t2bot.ioRoberto 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
@telegram_220807826:t2bot.ioV 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
@telegram_15404434:t2bot.ioFran Gómez García changed their display name from Francisco Gómez García to Fran Gómez García.15:24:36
@telegram_623687309:t2bot.ioMatteo @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
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21 Sep 2020
@juan:matrix.valenciawireless.orgjuan 17:19:46
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26 Sep 2020
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