• Monoid is similar to a Group, but it does not have an inverse for each element.
• Monoid can be formed by \((M, \otimes)\) and satified following laws:
  • If \(\forall m \in M\), then there exists an identity \(I\) such as \(I \otimes m = m \otimes I = m\)
  • If \(\forall m_1, m_2, m_3 \in M\), then \((m_1 \otimes m_2) \otimes m_3 = m_1 \otimes (m_2 \otimes m_3)\)

• A class of objects
• A set of morphisms or a set of arrows

• Category is a set including objects and morphisms/arrow

  • Each object has identity morphism
  • The composition of morphisms are associative

  

• Example
• Morphism is NOT function in Category
• Reddit

     Objects are nature number: 0, 1, 2, 3…
  Morphisms: m → n are \( m \times n \) matrices
  Identity Morphism: 3 → 3 is 3 × 3 matrix
   \[
      \begin{pmatrix}
   1 & 0 & 0 \\
         0 & 1 & 0 \\
         0 & 0 & 1
         \end{pmatrix}
   \]
  Composition: matrix multiplication ∘

  Identity
  M(1, 1) : 1 → 1
  M(2, 2) : 2 → 2

  Morphism:
  M(1, 2) : 1 → 2
  M(2, 3) : 2 → 3

  Composition
     M(1, 2) ∘ M(2, 3) → M (1, 3)

     Associative
  M₁(1, 2) ∘ M₂(2, 3) ∘ M₃(3, 4) = M₄(1, 4)
  M₁(1, 2) ∘ (M₂(2, 3) ∘ M₃(3, 4)) = M₁(1, 2) ∘ M₂₃(2, 4) = M₄(1, 4)
  ⇒ M₁ ∘ M₂ ∘ M₃ ≡ M₁ ∘ (M₂ ∘ M₃)

1. A monad is a monoid in a category of endofunctor

2. What does it mean?

   • functors are as elements in category.
     • \(T \in C\)
   • The operation is functor composition
     • \(T \otimes T = T\)
   • Each functor \(T\) has an identity functor \(I\) such that
     • \(I \otimes T = T \otimes I = T\)
   • The operation satisfied Associativity law because monoid satisfied associativity law.
   • \(T \otimes T \otimes T = T \otimes (T \otimes T)\)

f(x) = x² + x³
A = B → B ∘ μ η
C  E   F = f(9) = 4

Matrix \[ A= \begin{bmatrix} \cos(\beta) & -\sin(\beta)\\ \sin(\beta) & \cos(\beta) \end{bmatrix} + B = \begin{bmatrix} \cos \beta & -\sin \beta \\ \sin \beta & \cos \beta \end{bmatrix} \] Matrix multiplication

1. Matrix addition
   1. matrix division

f::(Monad m)=> m a -> (a -> m b) -> m b

transpose::[[a]] -> [[a]]
transpose [] -> repeat []
transpose (x:cs) = zipWith(:) x $ transpose cs

• The definition of Functor
  • Given a category \(\mathcal{C}\) and \(\mathcal{D}\)
  • Functor map Object in \(\mathcal{C}\) to Object in \(\mathcal{D}\)
    • \(F(a) \in \mathcal{D}\) \(a \in \mathcal{C}\)
    • F(a -> b) -> F(a) -> F(b) where a ∈ \(\mathcal{C}\) b ∈ \(\mathcal{D}\)
  • Functor map morphism/arrow in \(\mathcal{C}\) to morphism/arrow in \(\mathcal{D}\)
  • \( F Id_a = Id_{F a} \)

• The definition of Functor in Haskell. Functor is the type class with two methods,

  class Functor f where
  fmap::(a -> b) -> f a -> f b
  

  • Other way to think about it is as following:
  • \( F(a \rightarrow b) \rightarrow F(a) \rightarrow F(b) \)
  • \(F\) is a functor and transform a function \(g:a \rightarrow b\) to a new function \(F(a) \rightarrow F(b)\)
  • If \(g\) function is from type Int to type \(String\) and the functor is Maybe, then the new function is from Maybe(Int) → Maybe(String)
  • The instance of Functor has to satisfy following two laws:

  fmap id = id
          fmap (f . g) = (fmap f) . (fmap g)
  

• What is the difference between Functor and Monad.
  • Monad is the subtype of Functor
• Vector and matrix forms a category?
  • Vector and Matrix => Category

1. What is the definition of Monad? Monad is Monoid over the endofunctor

   class Applicative m => Monad m where
     return:: a -> m a
             return = pure
     (>>=)::(Monad m) => m a -> (a -> m b) -> m b
   

2. When to use Monad?

1. Monad definition from "What we talk about when we are talking about Monads" Tomas Petricek Monad Definition in Haskell

   • A monad is an operator M on types, together triple functions:
     • map::(a → b) → (M a → M b)
     • unit:: a → M a
     • join:: M (M a) → M a
   • Satisfying following laws
   • join ∘ join = join ∘ map unit
     • join ∘ unit = id = join ∘ map join

2. A Monad over its category 𝓒 is a triple (T, 𝜂, 𝜇) where T : \(\mathscr{C}\) → \(\mathscr{C}\) is a functor,

   • Good explanation in Reddit What is Monad
   • 𝜇 : T² → T and 𝜂 : Id → T are nature transformations such that:
     1. μ_A ∘ T μ_A = μ_A ∘ μ_TA

        • T μ_A ⇒ Apply μ_A inside T

          fmap join [[[1, 2]], [[3]]]
                    ↑
                   μ_A
          
          => [[1, 2], [3]]
          

        • Apply μ_A inside first then Apply μ_A from outside = Apply μ_A from outside first then Apply μ_A
        • Take the layer from inside first and then take the top layer = take the layer from inside and then take the top layer
     2. μ_A ∘ η_TA = id_TA = μ_A ∘ T η_A
     3. join = μ
     4. unit = η
   • 1 and 2 are equvalent as following
   1. join ∘ join = join ∘ map join
      • join ∘ join ⇒ peel the layers off from outside.

      • join ∘ join ⇒ T ⊗ T ⊗ T → T

        let s = [[[1], [2]], [[3]]]
                -- join::[[[]]] -> [[]]
                -- join::[[]] -> []
                -- => (join . join)::[[[]]] -> []
        (join . join) s -- [1, 2, 3]
        

        • join ∘ map join ⇒ peel the layers off from inside.

        let s = [[[1], [2]], [[3]]]
                -- join::[[[]]] -> [[]]
                -- join::[[]] -> []
                -- => 
        (join . map join) [[[1], [2]], [[3]]] -- [1, 2, 3]
                -- map(\x -> join x) [[[1] [2]], [[3]]]
                -- map(\x -> join [[1], [2]] => [1, 2] ) –- peel from inside
                -- map(\x -> join [[3]] => [3])
                -- [[1, 2], [3]]
                -- join [[1, 2], [3]] => [1, 2, 3]
        

        1. join ∘ unit = join ∘ map unit

   Reference

3. What is the difference between Monad and Monoid? There are couple axioms for Monoid
   1. m₁ ⊗ m₂ ∈ M if m₁ and m₂ ∈ M
   2. id ⊗ m = m ⊗ id = m
   3. m1 ⊗ m2 ⊗ m3 = m1 ⊗ (m2 ⊗ m3)
4. The mathematic definition of Monad
   1. μ :: I → T
   2. η :: T ⊗ T → T
   3. T is the endofunctor which means from a category to itself. (T : C → C)
5. The domain and co-domain of η are both Functor
   • ⊗ is Functor composition
   • e.g. if T = m a then T ⊗ T = m (m a)
   • e.g. T = [] then T ⊗ T = [[]]

   • In Haskell, type constructor is like a Functor

     class Applicative f => Monad f where
       return :: a -> f a
       join f (f a) -> f a
       fmap f (a -> b) -> (f a -> f b) -- f = (a -> m b)
     
     -- definition in GHC
     class Applicative m => Monad m where
       return :: a -> m a
       (>>=)::m a -> (a -> m b) -> m b
     
             -- f = (a -> m b)
     m >>= f = join $ fmap (a -> m b) m b
     m >>= f = join $ fmap f $ m b
     m >>= f = join $ m (f b)
     m >>= f = join $ m (m b)
     

     Use join and fmap represents (>>=)

     -- f = (a -> m b)                          
     m >>= f = join $ fmap f m b       
     m >>= f = join $ fmap f $ m b              
             m >>= f = join $ m (f b)                   
     m >>= f = join $ m (m b)                   
     

     1. Maybe is Monad

     instance Monad Maybe where
       return Nothing = Nothing
       (>>=) (Just a) f = Just f a
     
       addMaybe::Maybe Int -> Maybe Int -> Maybe Int
       addMaybe Nothing _ = Nothing
       addMaybe _ Nothing = Nothing
       addMaybe (Just a) (Just b) = Just (a + b)
     
       -- other implementation
       addMaybe::Maybe Int -> Maybe Int -> Maybe Int
       addMaybe m1 m2 = do
               a <- m1
               b <- m2
               return (a + b)
     

1. How to use Applicative
2. What is Applicative
3. What is the difference between Monad and Applicative

class Functor f => Applicative f where
  pure:: a -> f a
 (<*>):: f (a -> b) -> f a -> f b

class Applicative f => Monad f where
  return:: a -> f a
  (>>=)::m a -> (a -> m b) -> m b