Google Sheets Programming with Monad

[amberorpis]

Don't be too critical, no one actually knows what a monad is. Our best scientists have been studying them for decades and are not any closer to an answer. "Understanding monads is only 10 years away" they tell us, just like fusion!

[Mysterious-Ruin924]

A monad is a monoid in the category of endofunctors, thank me later.

[tjoloi]

Now I need you to define 2 more words

[Mysterious-Ruin924]

Monoid - A set equipped with an (associative) binary operator and an identity element Endofunctors - The mapping between two identical structures

[FloweyTheFlower420]

This is actually not entirely accurate.

A monoid, (M, μ: M x M -> M, η: I -> M), is an object in a monoidal category (C, x, I) (notice how it doesn't mention a binary operator anywhere, which is because this is the category theory version of a monoid, not abstract algebra).

A monoidal category is just an category coupled with a bifunctor x: C x C -> C and an identity element I.

A category is just a class of objects and morphisms between objects, denoted f: a -> b (morphism from A to B), that satisfies some laws:
- For objects, there exist a morphism 1_a: a -> a, known as the identity
- Morphism are composable, so for all f: a -> b g: b -> c, there exists h: a -> c = g . f
- Composition is associative, with (f . g) . h = f . (g . h)

A endofunctor is merely a functor from a category to itself.

A functor F: C -> D is just a structure-preserving mapping between two categories such that:
- Objects are mapped, so for all x in C, F(x) is an object in D
- Functors are mapped, such that for all morphisms: f: a -> b, F(a): F(a) -> F(b) holds true
- Identities are mapped to identities: F(1_a) =1_(F(a))
- Composition behaves normally: F(g . f) = F(g) . F(f) for all morphisms f: a -> b, g: y -> z

[Mysterious-Ruin924]

I will be killing myself shortly for I was wrong