A well-said, perhaps the briefest ever answer is: monad is just a monoid in the category of endofunctors.

monoid is defined as an algebraic structure (generally, a set) M with a binary operation (multiplication) • : M × M → M and an identity element (unit) η : 1 → M, following two axioms:

i. Associativity
∀ a, b, c ∈ M, (a • b) • c = a • (b • c)

ii. Identity
∃ e ∈ M ∀ a ∈ M, e • a = a • e = a

When specifying an endofunctor T : X → X (which is a functor that maps a category to itself) as the set M, the Cartesian product of two sets is just the composition of two endofunctors; what you get from here is a monad, with these two natural transformations:

1. The binary operation is just a functor composition μ : T × T → T
2. The identity element is just an identity endofunctor η : I → T

Satisfied the monoid axioms (i. & ii.), a monad can be seen as a monoid which is an endofunctor together with two natural transformations.

The name “monad” came from “monoid” and “triad”, which indicated that it is a triple (1 functor + 2 trasformations), monoidic algebraic structure.

In other words, monoid is a more general, abstract term. When applying it to the category of endofunctors, we have a monad.

Brian Beckman:

You can understand Monad without too much Category Theory.

Functional Programming = using functions to compose from small functions to very complex software (eg. Nuclear system, driverless car software…).

• Strong Types Safety: detect bugs at compile time.
• Data Protection thru Immutability: Share data safely in Concurrent / Parallel processing.
• Software ‘Componentisation’  ie Modularity : Each function always returns the same result, ease of software reliability testing.

Each “small” function is a Monoid.
f : a -> a (from input of type ‘a‘ , returns type ‘a’)
g: a -> a

compose h from f & g : (strong TYPING !!)
h = f。g : a -> a

[Note] : Object in Category, usually called  Type in Haskell, eg. ‘a’ = Integer)

You already know a Monoid (or Category in general) : eg Clock

1. Objects: 1 2 3 …12 (hours)
2. Arrow (Morphism): rule “+”:
• 7 + 10 = 17 mod 12 = 5
3. Law of Associativity:
x + (y + z) = (x + y) + z
4. Identity (or “Unit”):  (“12”):
x + 12 = 12 + x = x

More general than Monoid is a “Monoidal” Category where: (instead of only single object ‘a’, now more “a b c…”)
f : a -> b
g: b -> c
h = f。g : a -> c

Function under composition Associative rule and with an Identity => Monoid

Monad (M): a  way to manage  the side-effects (I/O, exception , SQL Database, etc) within the Functional Programming way like monoidal categories: ie associative composition, identity.

Remark: For the last 60 years in Software, there have been 2 camps:

1. Bottom-Up Design: from hardware foundation,  build performance-based languages: Fortran, C, C++, C#, Java…
2. Top-Down Design: from Mathematics foundation, build functional languages (Lambda-Calculus, Lisp, Algo, Smalltalk, Haskell…).
3. F# (Microsoft) is the middle-ground between 1 & 2.

Ref: What is a Monad ?

Monad = chaining operations with binding “>>=”

• Possible use: allows to write mini-language, parser…

# BM Category Theory 10: Monad & Monoid

Analogy

Function : compose “.“, Id

Imperative (with side effects eg. state, I/O, exception ) to Pure function by hiding or embellishment in Pure function but return “embellished” result.

10. 2 Monoid

Monoid in category of endo functors = Monad

Ref Book :

What is the significance of monoids in category theory? by Bartosz Milewski

# BM Category Theory 3 & 4 Monoid, Kleisli Category (Monad), Terminal /Initial Object, Product… Free Monoid

3.1 Monoid M (m, m)

Same meaning in Category as in Set: Only  1 object, Associative, Identity

Thin / Thick Category:

• “Thin” with only 1 arrow between 2 objects;
• “Thick” with many arrows between 2 objects.

Arrow : relation between 2 objects. We don’t care what an arrow actually is (may be total / partial order relations like = or $\leq$, or any relation), just treat arrow abstractly.

Note: Category Theory’s “Abstract Nonsense” is like Buddhism “空即色, 色即空” (Form = Emptiness).

Example of Monoid: String Concatenation: identity = Null string.

Strong Typing: function f calls function g, the type of the output of f must match with the type of the input of g.

Weak Typing:   no need to match type. eg. Monoid.

Category induces a Hom-Set: (Set of “Arrows”, aka Homomorphism 同态, which preserves structure after the “Arrow”)

• C (a, b) : a -> b
• C (a,a) for Monoid : a -> a

Morphism between a, b :
$\boxed{a \to (b, \text {string})}$

In brief, embellish the returned output b with string.

In the example of Kleisli Category, using Monoid (single object) with embellishing string => Monad (covered in later videos).

4.1 Monad (one definition) : Kleisli

Set and Category SET:

• Set has elements and function between sets.
• Category SET has NO element, only arrows (morphisms)

Universal Construction: define relations between Objects.

Set is rich with functions, except 1 case: There is no function from a non-empty set to Void (because a function must have an image in the co-domain, but Void has no element as the Image.)

Terminal Object : all other objects (each with UNIQUE arrow) point to it. [Denoted as () or Unit]

$\boxed { \forall a, \exists f :: a \to ()}$

$\boxed { \forall a, f :: a \to (), g :: a \to () \implies f = g }$ (Uniqueness of arrow)

Note: Terminal Object (T.O.) is the largest object. There is no T.O. in Natural numbers.

Initial Object: reverse of Terminal object. Outgoing arrow from Void.

absurd :: Void -> a

Category “Equality”:

• 2 objects are isomorphic (exists an inverse arrow), but we don’t say they are equal;
• Arrows can be equal by associativity :  $f \circ (g \circ h) = (f \circ g) \circ h$

Terminal Object (Similarly, Initial Object) is unique (up to ONLY ONE Isomorphism).

Assume both Ta and Tb are Terminal Objects. Prove Ta = Tb by showing there is only one isomorphism.

Note1: example of 2 isomorphisms between 2 Objects : (0, 1) and (T, F) =>

• 1st isomorphism : (1 ~ T), (0 ~ F)
• 2nd isomorphism: (1 ~ F), (0 ~ T)

Note2: There are outgoing arrows from T.O. => generalising.

4.2 Products

$C^{op}$ : also a Category with all arrows in C reversed.

Products (Cartesian) of 2 objects (a,b): by Universal Construction.

p and q are (good) projections of c.
p’ and q’ are (another) projections of c’.

c is the product of (a, b) if
there is a UNIQUE  isomorphism m,
$\boxed {m :: c' \to c }$
$\boxed {p' = p \circ m}$
$\boxed{q' = q \circ m}$

II 3.2 Free Monoid

Monoid : with a Set of 2 generators (a, b)

• Identity: e*a = a*e = a;
• Associativity

Free Monoid = {e, a, b, ab, ba, aab, abb, ….}, notice it is generated by the set {a, b} and thus not a Monoid (noted as Mon).

We can create a Functor U (aka  Forgetting function) between Monoid and its underlying Set:

$\boxed { U :: Mon \to Set }$

# Minggatu-Catalan Number

Minggatu-Catalan Number

First discovered by the Chinese Mathematician Minggatu 明安图 (清 康熙, 1730), later by the French (né Belgian) École Polytechnique mathematician Eugène Charles Catalan (1814 – 1894).

Proof: Consider ways of making sums with {0, 1, 2, 3, 4…} and +, ( , ).

0 = (0)
1 way for 0
$\boxed {C_{0} = 1}$

0+1 =(0+1) :
1 way for 1 $\boxed {C_{1} = 1}$

0 +1 +2 = ((0+(1+2)) = ((0+1)+2):
2 ways for 2 $\boxed {C_{2} = 2}$

0+1+2+3 = (0+(1+(2+3))) = (0+((1+2)+3))= ((0+1)+(2+3)) = ((0+(1+2))+3) = (((0+1)+2)+3) :
5 ways for 3 $\boxed {C_{3} = 5}$

0+1+2+3+…+n = ?
? ways for n $\boxed {C_{n} = ?}$

Try finding the pattern for $C_{4}$:

Let A represents {0, 1, 2, 3, 4}.
There are 4 cases:

Case 1:
($\underbrace{(A)}_{C_{0}}$ + $\underbrace{(A+ A+ A+ A)} _{C_{3}}$)

Case 2:
( $\underbrace{ (A+A ) }_{C_{1}}$ + $\underbrace{ (A+ A+ A) }_{C_{2}}$)

Case 3:
( $\underbrace{ (A+A+A ) }_{C_{2}}$ + $\underbrace{ ( A+ A) }_{C_{1}}$)

Case 4:
( $\underbrace{ (A+A+A+A) }_{C_{3}}$ + $\underbrace{ (A) }_{C_{0}}$ )

$C_{4} = C_{0}.C_{3} + C_{1}.C_{2} + C_{2}.C_{1}+ C_{3}.C_{0}$

Generalise:
$C_{n+1} = C_{0}.C_{n-0} + C_{1}.C_{n-1} + C_{2}.C_{n-2} + ...+ C_{k}.C_{n-k} + ...+ C_{n-0}.C_{0}$

Recurrence Sequence of $C_{n}$:
$\boxed{ \begin{cases} C_{0} =1, \\ C_{n+1} = \displaystyle\sum_{k=0}^{n}C_{k}C_{n-k} , & \text{( } n \geq {0} \text {)} \end{cases} }$

Step 1: Generating Function C(x) :
Let $C(x) =C_{0} + C_{1}x+ C_{2}x^2 + ...+ C_{n}x^n$

$\boxed {\displaystyle C(x) = \sum_{n=0}^{\infty} C_{n} x^n }$

$C(x)^{2} = (C_{0}C_{0})+ (C_{0}C_ {1} + C_{1}C_{0})x+ (C_{0}C_{2}+ C_{1}C_{1}+ C_{2}C_{0} ) x^2 + ...$

The coefficient for $x^{n}$: $\displaystyle \sum_{k=0}^{n}C_{k}C_{n-k}$

$\displaystyle C(x)^2 = \sum_{n=0}^{\infty} \underbrace{ \left (\sum_{k=0}^{n}C_{k}C_{n-k} \right)}_{\text {coeff of }x^{n}} x^n = \sum_{n=0}^{\infty} \underbrace{ C_{n+1}}_{\text {recurrence}} x^n$

$\displaystyle x.C(x)^{2} =x. \sum_{n=0}^{\infty} C_{n+1} x^n = \sum_{n=0}^{\infty} C_{n+1} x^{n +1}$
Change (n+1) to n by adjusting initial value from n= 0 to 1:
$\displaystyle x.C(x)^{2} = \sum_{n=1}^{\infty} C_{n} x^n$

Reset the initial value from n=1 to 0:
$\displaystyle x.C(x)^{2} = \sum_{n=0}^{\infty} C_{n} x^n - C_{0}$
We have seen
$C_{0}= 1$ and
$\displaystyle C(x) = \sum_{n=0}^{\infty} C_{n} x^n$

$\displaystyle x.C(x)^{2} - C(x) +1 = 0$
This is the quadratic equation on C (x):

$C(x)= \frac{1 \pm \sqrt{1-4x} }{2x}$

$2x.C(x)= 1 \pm \sqrt{1-4x}$

Case: $2x.C(x)= 1 + \sqrt{1-4x}$
When x $\rightarrow$ 0
$2x.C(x)= 0$
$1 + \sqrt{1-4x}= 1+1=2$
Impossible !

Case: $2x.C(x)= 1 - \sqrt{1-4x}$
When x $\rightarrow$ 0
$2x.C(x)= 0$
$1 - \sqrt{1-4x}= 1- 1=0$

Step 2: The closed form for the generating function C(x) :

$\boxed { C(x)= \frac{1 - \sqrt{1-4x} }{2x} }$

Step 3: The closed form for $C_{n}$

We expand $\sqrt{1-4x}$ into power series:

Let $\displaystyle \sqrt{1-4x}= K(x)= \sum_{k=0}^{\infty}K_{k}x^{k}$

$2xC(x)= 1 - \sqrt{1-4x}$
$\displaystyle 2xC(x)= 1 - \sum_{k=0}^{\infty}K_{k}x^{k}$

$\displaystyle 2x\sum_{k=0}^ {\infty}C_{k}x^{k} = 1 - \sum_{k=0}^{\infty}K_{k}x^{k}$

Move 2x inside the left sum:
$\displaystyle \sum_{k=0}^ {\infty}2C_{k}x^{k+1} = 1 - \sum_{k=0}^{\infty}K_{k}x^{k}$

Synchronise the indices k,
$\displaystyle \sum_{k=1}^ {\infty}2C_{k-1}x^{k} = 1 - K_{0} - \sum_{k=1}^{\infty}K_{k}x^{k}$

$\displaystyle \sum_{k=1}^ {\infty}2C_{k-1}x^{k} + \sum_{k=1}^{\infty}K_{k}x^{k} = 1 - K_{0}$

$\displaystyle \sum_{k=1}^ {\infty} (2C_{k-1} + K_{k})x^{k} = 1 - K_{0}$

We compare the coefficient of each term $x^{k}$ :

$x^{0}: 0 = 1- K_{0}$
$x^{1}: 2C_{0} + K_{1} = 0$
$x^{2}: 2C_{1} + K_{2} = 0$

$x^{n}: 2C_{n} + K_{n+1} = 0$

$\boxed{ \begin{cases} K_{0} =1, \\ C_{n} = \displaystyle - \frac {K_{n+1}}{2} , & \text{( } n \geq {0} \text {)} \end{cases} }$

We need to express $K_{n+1}$ in term of n only.

Go back to the definition of K(x):
$\displaystyle K(x)= \sum_{k=0}^{\infty}K_{k}x^{k}$

$\displaystyle K(x)= K_{0}+K_{1}x + K_{2}x^{2} + ...+ K_{n}x^{n}+...$

Differentiate:
$\displaystyle K'(x)=1.K_{1}x + 2K_{2}x^{1} + ...+ nK_{n}x^{n-1}+...$

$\displaystyle K'(0)= 1K_{1}$

Differentiate 2nd time:
$\displaystyle K''(x)=2.1K_{2} + ...+ n. (n-1)K_{n}x^{n-2}+...$

$\displaystyle K''(0)= 2.1K_{2}$

Continue …

$K^{(n)}(x)= n (n-1)(n-2)...2.1K_{n} + ...$

$K^{(n)}(0)= n! K_{n}$

$\boxed { K_{n} = \frac { K^{(n)}(0)}{n!} = \frac { K^{(n)}(0)}{n^{\underline {n}}} }$

Note: See blog on “Falling Factorial”: $x^{\underline {n}}$

Next step, we shall find $K^{(n)}(0)$ in term of n
Recall
$K(x) = (1-4x)^{\frac {1}{2}}$

$K'(x) = -2.(1-4x)^{-\frac {1}{2}}$
$K''(x) = -2.2(1-4x)^{-\frac {3}{2}}$
$K'''(x) = -2.4.3(1-4x)^{-\frac {5}{2}}$
$K''''(x) = -2.6.5.4(1-4x)^{-\frac {7}{2}}$

$K^{(n)}(x) = -2.(2n-2)^{\underline {n-1}}(1-4x)^{-\frac {2n-1}{2}}$

$K^{(n+1)}(x) = -2.(2n)^{\underline {n}}(1-4x)^{-\frac {2n+1}{2}}$

$\boxed { K^{(n+1)}(0) = -2.(2n)^{\underline {n}} }$

Also from above, we have seen:
$\boxed { K_{n} = \frac { K^{(n)}(0)}{n^{\underline {n}}} }$

Change n to (n+1):
$K_{n+1} = \frac { K^{(n+1)}(0)}{(n+1)^{\underline {n+1}}}$

$K_{n+1} = \frac { -2.(2n)^{\underline {n}} } {(n+1)^{\underline {n+1}}}$
From step 3 above, we also know:
$C_{n} = - \frac {K_{n+1}} {2}$

$C_{n} = \frac { (2n)^{\underline {n}} } {(n+1)^{\underline {n+1}}} = \frac {1}{n+1} \frac { (2n)^{\underline {n}} } {(n)^{\underline {n}}}$
since $(n+1)^{\underline {n+1}} = (n+1)!= (n+1).n! = (n+1). (n)^{\underline {n}}$
and
$\frac { (2n)^{\underline {n}}} {(n)^{\underline {n}}} = \binom {2n}{n}$ (by definition of Falling Factorial)

$\boxed {\displaystyle C_{n} = \frac {1}{n+1} \binom {2n}{n} }$ [QED]

Ref:

1. History:

3. Monoid

5. Ref:
Math Girls by Hiroshi Yuki et al.
http://www.amazon.co.uk/dp/0983951306/ref=cm_sw_r_udp_awd_UKSvtb1AXNWCE