`(S,⋆,z)`

consisting of an associative binary operation `⋆`

over some set `S`, which is closed under `⋆`

, and which contains an identity element `z` for `⋆`

. That is:
for all s1, s2, s3 in S: (i) s1⋆s2 etc are also in S (ii) (s1⋆s2)⋆s3 = s1⋆(s2⋆s3) (iii) z⋆s1 = s1 = s1⋆zSome examples of monoids are: * finite strings of an alphabet `A`, with

`⋆`

being concatenation and `z` being the empty string
* all functions `X→X`

over a set `X`, with `⋆`

being composition and `z` being the identity function over `X`
* the natural numbers with `⋆`

being plus and `z` being 0 (in particular, this is a **commutative monoid**). If we use the integers, or the naturals mod n, instead of the naturals, then every element will have an inverse and so we have not merely a monoid but a **group**.
* if we let `⋆`

be multiplication and `z` be 1, we get different monoids over the same sets as in the previous item.
Categories
----------
A **category** is a generalization of a monoid. A category consists of a class of **elements**, and a class of **morphisms** between those elements. Morphisms are sometimes also called maps or arrows. They are something like functions (and as we'll see below, given a set of functions they'll determine a category). However, a single morphism only maps between a single source element and a single target element. Also, there can be multiple distinct morphisms between the same source and target, so the identity of a morphism goes beyond its "extension."
When a morphism `f` in category `f:C1→C2`

.
To have a category, the elements and morphisms have to satisfy some constraints:
(i) the class of morphisms has to be closed under composition: where f:C1→C2 and g:C2→C3, g ∘ f is also a morphism of the category, which maps C1→C3. (ii) composition of morphisms has to be associative (iii) every element E of the category has to have an identity morphism 1These parallel the constraints for monoids. Note that there can be multiple distinct morphisms between an element `E` and itself; they need not all be identity morphisms. Indeed from (iii) it follows that each element can have only a single identity morphism. A good intuitive picture of a category is as a generalized directed graph, where the category elements are the graph's nodes, and there can be multiple directed edges between a given pair of nodes, and nodes can also have multiple directed edges to themselves. Morphisms correspond to directed paths of length ≥ 0 in the graph. Some examples of categories are: * Categories whose elements are sets and whose morphisms are functions between those sets. Here the source and target of a function are its domain and range, so distinct functions sharing a domain and range (e.g., `sin` and `cos`) are distinct morphisms between the same source and target elements. The identity morphism for any element/set is just the identity function for that set. * any monoid_{E}, which is such that for every morphism f:C1→C2: 1_{C2}∘ f = f = f ∘ 1_{C1}

`(S,⋆,z)`

generates a category with a single element `x`; this `x` need not have any relation to `S`. The members of `S` play the role of *morphisms* of this category, rather than its elements. All of these morphisms are understood to map `x` to itself. The result of composing the morphism consisting of `s1` with the morphism `s2` is the morphism `s3`, where `s3=s1⋆s2`

. The identity morphism for the (single) category element `x` is the monoid's identity `z`.
* a **preorder** is a structure `(S, ≤)`

consisting of a reflexive, transitive, binary relation on a set `S`. It need not be connected (that is, there may be members `s1`,`s2` of `S` such that neither `s1≤s2`

nor `s2≤s1`

). It need not be anti-symmetric (that is, there may be members `s1`,`s2` of `S` such that `s1≤s2`

and `s2≤s1`

but `s1` and `s2` are not identical). Some examples:
* sentences ordered by logical implication ("p and p" implies and is implied by "p", but these sentences are not identical; so this illustrates a pre-order without anti-symmetry)
* sets ordered by size (this illustrates it too)
Any pre-order `(S,≤)`

generates a category whose elements are the members of `S` and which has only a single morphism between any two elements `s1` and `s2`, iff `s1≤s2`

.
Functors
--------
A **functor** is a "homomorphism", that is, a structure-preserving mapping, between categories. In particular, a functor `F` from category (i) associate with every element C1 ofA functor that maps a category to itself is called an **endofunctor**. The (endo)functor that maps every element and morphism ofCan element F(C1) ofD(ii) associate with every morphism f:C1→C2 ofCa morphism F(f):F(C1)→F(C2) ofD(iii) "preserve identity", that is, for every element C1 ofC: F of C1's identity morphism inCmust be the identity morphism of F(C1) inD: F(1_{C1}) = 1_{F(C1)}. (iv) "distribute over composition", that is for any morphisms f and g inC: F(g ∘ f) = F(g) ∘ F(f)

`η[C1]:G(C1)→H(C1)`

in `η[C1]`

has as source `C1`'s image under `G` in for every morphism f:C1→C2 inThat is, the morphism via `G(f)` from `G(C1)` to `G(C2)`, and then viaC: η[C2] ∘ G(f) = H(f) ∘ η[C1]

`η[C2]`

to `H(C2)`, is identical to the morphism from `G(C1)` via `η[C1]`

to `H(C1)`, and then via `H(f)` from `H(C1)` to `H(C2)`.
How natural transformations compose:
Consider four categories -ThenB-+ +---C--+ +----D-----+ +--E-- | | | | | | F: ------> G: ------> K: ------> | | | | | η | | | ψ | | | | v | | v | | H: ------> L: ------> | | | | | φ | | | | | | v | | | | J: ------> | | -----+ +--------+ +------------+ +-------

`(η F)`

is a natural transformation from the (composite) functor `GF` to the composite functor `HF`, such that where `B1` is an element of category `(η F)[B1] = η[F(B1)]`

---that is, the morphism in `η`

assigns to the element `F(B1)` of `(K η)`

is a natural transformation from the (composite) functor `KG` to the (composite) functor `KH`, such that where `C1` is an element of category `(K η)[C1] = K(η[C1])`

---that is, the morphism in `η[C1]`

of `(φ -v- η)`

is a natural transformation from `G` to `J`; this is known as a "vertical composition". For any morphism `f:C1→C2`

in φ[C2] ∘ H(f) ∘ η[C1] = φ[C2] ∘ H(f) ∘ η[C1]by naturalness of

`φ`

, is:
φ[C2] ∘ H(f) ∘ η[C1] = J(f) ∘ φ[C1] ∘ η[C1]by naturalness of

`η`

, is:
φ[C2] ∘ η[C2] ∘ G(f) = J(f) ∘ φ[C1] ∘ η[C1]Hence, we can define

`(φ -v- η)[\_]`

as: `φ[\_] ∘ η[\_]`

and rely on it to satisfy the constraints for a natural transformation from `G` to `J`:
(φ -v- η)[C2] ∘ G(f) = J(f) ∘ (φ -v- η)[C1]An observation we'll rely on later: given the definitions of vertical composition and of how natural transformations compose with functors, it follows that:

((φ -v- η) F) = ((φ F) -v- (η F))I'll assert without proving that vertical composition is associative and has an identity, which we'll call "the identity transformation."

`(ψ -h- η)`

is natural transformation from the (composite) functor `KG` to the (composite) functor `LH`; this is known as a "horizontal composition." It's trickier to define, but we won't be using it here. For reference:
(φ -h- η)[C1] = L(η[C1]) ∘ ψ[G(C1)] = ψ[H(C1)] ∘ K(η[C1])Horizontal composition is also associative, and has the same identity as vertical composition. Monads ------ In earlier days, these were also called "triples." A **monad** is a structure consisting of an (endo)functor `M` from some category

`φ`

, each being between some arbitrary endofunctor `F` on `φ`

assigns `C1` a morphism from element `F(C1)` to element `MF'(C1)`, satisfying the constraints detailed in the previous section. For different members of `T`, the relevant functors may differ; that is, `φ`

is a transformation from functor `F` to `MF'`, `γ`

is a transformation from functor `G` to `MG'`, and none of `F`, `F'`, `G`, `G'` need be the same.
One of the members of `T` will be designated the `unit` transformation for `M`, and it will be a transformation from the identity functor `1C` for `φ`

and `γ`

be members of `T`, that is they are natural transformations from `F` to `MF'` and from `G` to `MG'`, respectively. Let them be such that `F' = G`. Now `(M γ)`

will also be a natural transformation, formed by composing the functor `M` with the natural transformation `γ`

. Similarly, `(join G')` will be a natural transformation, formed by composing the natural transformation `join` with the functor `G'`; it will transform the functor `MMG'` to the functor `MG'`. Now take the vertical composition of the three natural transformations `(join G')`, `(M γ)`

, and `φ`

, and abbreviate it as follows. Since composition is associative I don't specify the order of composition on the rhs.
γ <=< φ =def. ((join G') -v- (M γ) -v- φ)In other words, `<=<` is a binary operator that takes us from two members

`φ`

and `γ`

of `T` to a composite natural transformation. (In functional programming, at least, this is called the "Kleisli composition operator". Sometimes it's written `φ >=> γ`

where that's the same as `γ <=< φ`

.)
`φ`

is a transformation from `F` to `MF'`, where the latter = `MG`; `(M γ)`

is a transformation from `MG` to `MMG'`; and `(join G')` is a transformation from `MMG'` to `MG'`. So the composite `γ <=< φ`

will be a transformation from `F` to `MG'`, and so also eligible to be a member of `T`.
Now we can specify the "monad laws" governing a monad as follows:
(T, <=<, unit) constitute a monoidThat's it. Well, there may be a wrinkle here. I don't know whether the definition of a monoid requires the operation to be defined for every pair in its set. In the present case,

`γ <=< φ`

isn't fully defined on `T`, but only when `φ`

is a transformation to some `MF'` and `γ`

is a transformation from `F'`. But wherever `<=<` is defined, the monoid laws must hold:
(i) γ <=< φ is also in T (ii) (ρ <=< γ) <=< φ = ρ <=< (γ <=< φ) (iii.1) unit <=< φ = φ (here φ has to be a natural transformation to M(1C)) (iii.2) ρ = ρ <=< unit (here ρ has to be a natural transformation from 1C)If

`φ`

is a natural transformation from `F` to `M(1C)` and `γ`

is `(φ G')`

, that is, a natural transformation from `FG'` to `MG'`, then we can extend (iii.1) as follows:
γ = (φ G') = ((unit <=< φ) G') = (((join 1C) -v- (M unit) -v- φ) G') = (((join 1C) G') -v- ((M unit) G') -v- (φ G')) = ((join (1C G')) -v- (M (unit G')) -v- γ) = ((join G') -v- (M (unit G')) -v- γ) since (unit G') is a natural transformation to MG', this satisfies the definition for <=<: = (unit G') <=< γwhere as we said

`γ`

is a natural transformation from some `FG'` to `MG'`.
Similarly, if `ρ`

is a natural transformation from `1C` to `MR'`, and `γ`

is `(ρ G)`

, that is, a natural transformation from `G` to `MR'G`, then we can extend (iii.2) as follows:
γ = (ρ G) = ((ρ <=< unit) G) = (((join R') -v- (M ρ) -v- unit) G) = (((join R') G) -v- ((M ρ) G) -v- (unit G)) = ((join (R'G)) -v- (M (ρ G)) -v- (unit G)) since γ = (ρ G) is a natural transformation to MR'G, this satisfies the definition <=<: = γ <=< (unit G)where as we said

`γ`

is a natural transformation from `G` to some `MR'G`.
Summarizing then, the monad laws can be expressed as:
For all γ, φ in T for which ρ <=< γ and γ <=< φ are defined: (i) γ <=< φ is also in T (ii) (ρ <=< γ) <=< φ = ρ <=< (γ <=< φ) (iii.1) (unit G') <=< γ = γ when γ is a natural transformation from some FG' to MG' (iii.2) γ = γ <=< (unit G) when γ is a natural transformation from G to some MR'GThe standard category-theory presentation of the monad laws ----------------------------------------------------------- In category theory, the monad laws are usually stated in terms of `unit` and `join` instead of `unit` and `<=<`. Let's remind ourselves of some principles: * composition of morphisms, functors, and natural compositions is associative * functors "distribute over composition", that is for any morphisms `f` and `g` in `F`'s source category:

`F(g ∘ f) = F(g) ∘ F(f)`

* if `η`

is a natural transformation from `F` to `G`, then for every `f:C1→C2`

in `F` and `G`'s source category `η[C2] ∘ F(f) = G(f) ∘ η[C1]`

.
Let's use the definitions of naturalness, and of composition of natural transformations, to establish two lemmas.
Recall that join is a natural transformation from the (composite) functor `MM` to `M`. So for elements `C1` in `f:C1→C2`

in (1) join[C2] ∘ MM(f) = M(f) ∘ join[C1]Next, consider the composite transformation

`((join MG') -v- (MM γ))`

.
* `γ`

is a transformation from `G` to `MG'`, and assigns elements `C1` in `γ\*: G(C1) → MG'(C1)`

. `(MM γ)`

is a transformation that instead assigns `C1` the morphism `MM(γ\*)`

.
* `(join MG')` is a transformation from `MMMG'` to `MMG'` that assigns `C1` the morphism `join[MG'(C1)]`.
Composing them:
(2) ((join MG') -v- (MM γ)) assigns to C1 the morphism join[MG'(C1)] ∘ MM(γ*).Next, consider the composite transformation

`((M γ) -v- (join G))`

.
(3) This assigns to C1 the morphism M(γ*) ∘ join[G(C1)].So for every element `C1` of

((join MG') -v- (MM γ))[C1], by (2) is: join[MG'(C1)] ∘ MM(γ*), which by (1), with f=γ*: G(C1)→MG'(C1) is: M(γ*) ∘ join[G(C1)], which by 3 is: ((M γ) -v- (join G))[C1]So our **(lemma 1)** is:

((join MG') -v- (MM γ)) = ((M γ) -v- (join G)), where γ is a transformation from G to MG'.Next recall that unit is a natural transformation from `1C` to `M`. So for elements `C1` in

`f:a→b`

in (4) unit[b] ∘ f = M(f) ∘ unit[a]Next consider the composite transformation

`((M γ) -v- (unit G))`

:
(5) This assigns to C1 the morphism M(γ*) ∘ unit[G(C1)].Next consider the composite transformation

`((unit MG') -v- γ)`

.
(6) This assigns to C1 the morphism unit[MG'(C1)] ∘ γ*.So for every element C1 of

((M γ) -v- (unit G))[C1], by (5) = M(γ*) ∘ unit[G(C1)], which by (4), with f=γ*: G(C1)→MG'(C1) is: unit[MG'(C1)] ∘ γ*, which by (6) = ((unit MG') -v- γ)[C1]So our **(lemma 2)** is:

(((M γ) -v- (unit G)) = ((unit MG') -v- γ)), where γ is a transformation from G to MG'.Finally, we substitute

`((join G') -v- (M γ) -v- φ)`

for `γ <=< φ`

in the monad laws. For simplicity, I'll omit the "-v-".
for all φ,γ,ρ in T, where φ is a transformation from F to MF', γ is a transformation from G to MG', R is a transformation from R to MR', and F'=G and G'=R: (i) γ <=< φ etc are also in T ==> (i') ((join G') (M γ) φ) etc are also in T (ii) (ρ <=< γ) <=< φ = ρ <=< (γ <=< φ) ==> (ρ <=< γ) is a transformation from G to MR', so: (ρ <=< γ) <=< φ becomes: (join R') (M (ρ <=< γ)) φ which is: (join R') (M ((join R') (M ρ) γ)) φ substituting in (ii), and helping ourselves to associativity on the rhs, we get: ((join R') (M ((join R') (M ρ) γ)) φ) = ((join R') (M ρ) (join G') (M γ) φ) --------------------- which by the distributivity of functors over composition, and helping ourselves to associativity on the lhs, yields: ------------------------ ((join R') (M join R') (MM ρ) (M γ) φ) = ((join R') (M ρ) (join G') (M γ) φ) --------------- which by lemma 1, with ρ a transformation from G' to MR', yields: ----------------- ((join R') (M join R') (MM ρ) (M γ) φ) = ((join R') (join MR') (MM ρ) (M γ) φ) which will be true for all ρ,γ,φ just in case: ((join R') (M join R')) = ((join R') (join MR')), for any R'. which will in turn be true just in case: (ii') (join (M join)) = (join (join M)) (iii.1) (unit F') <=< φ = φ ==> (unit F') is a transformation from F' to MF', so: (unit F') <=< φ becomes: (join F') (M unit F') φ which is: (join F') (M unit F') φ substituting in (iii.1), we get: ((join F') (M unit F') φ) = φ which will be true for all φ just in case: ((join F') (M unit F')) = the identity transformation, for any F' which will in turn be true just in case: (iii.1') (join (M unit) = the identity transformation (iii.2) φ = φ <=< (unit F) ==> φ is a transformation from F to MF', so: unit <=< φ becomes: (join F') (M φ) unit substituting in (iii.2), we get: φ = ((join F') (M φ) (unit F)) -------------- which by lemma (2), yields: ------------ φ = ((join F') ((unit MF') φ) which will be true for all φ just in case: ((join F') (unit MF')) = the identity transformation, for any F' which will in turn be true just in case: (iii.2') (join (unit M)) = the identity transformationCollecting the results, our monad laws turn out in this format to be: when φ a transformation from F to MF', γ a transformation from F' to MG', ρ a transformation from G' to MR' all in T: (i') ((join G') (M γ) φ) etc also in T (ii') (join (M join)) = (join (join M)) (iii.1') (join (M unit)) = the identity transformation (iii.2')(join (unit M)) = the identity transformation 7. The functional programming presentation of the monad laws ------------------------------------------------------------ In functional programming, unit is usually called "return" and the monad laws are usually stated in terms of return and an operation called "bind" which is interdefinable with <=< or with join. Additionally, whereas in category-theory one works "monomorphically", in functional programming one usually works with "polymorphic" functions. The base category