+
+
+1. Monoids
+----------
+A <monoid> is a structure 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*z
+
+Some examples of monoids are:
+
+ (a) finite strings of an alphabet A, with * being concatenation and z being the empty string
+
+ (b) all functions X->X over a set X, with * being composition and z being the identity function over X
+
+ (c) 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>.)
+
+ (d) the natural numbers with * being multiplication and z being 1 constitute a different monoid over the same set as in (c).
+
+
+
+2. 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 given 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 C has source c1 and target c2, we'll write 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 o 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 id[e], which is such that for every morphism f:a->b:
+ id[b] o f = f = f o id[a]
+
+These 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.
+
+
+Some examples of categories are:
+
+ (a) any category 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 is the identity function over that set.
+
+ (b) any monoid (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. The result of composing the morphism consisting of s1 with the morphism s2 is the morphism s3, where s3=s1+s2. The identity morphism on the (single) category element x is the monoid's identity z.
+
+ (c) a <preorder> <= is a binary relation on a set S which is reflexive and transitive. It need not be connected (that is, there may be members x,y of S such that neither x<=y nor y<=x). 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).
+ * sets ordered by size
+ 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.
+
+
+
+3. Functors
+-----------
+A <functor> is a "homomorphism", that is, a structure-preserving mapping, between categories. In particular, a functor F from category C to category D must:
+ (i) associate with every element c1 of C an element F(c1) of D
+ (ii) associate with every morphism f:c1->c2 of C a morphism F(f):F(c1)->F(c2) of D
+ (iii) "preserve identity", that is, for every element c1 of C: F of c1's identity morphism in C must be the identity morphism of F(c1) in D:
+ F(id[c1]) = id[F(c1)].
+ (iv) "distribute over composition", that is for any morphisms f and g in C:
+ F(g o f) = F(g) o F(f)
+
+A functor that maps a category to itself is called an <endofunctor>. The (endo)functor that maps every element and morphism of C to itself is denoted 1C.
+
+How functors compose:
+If F is a functor from category C to category D, and H is a functor from category D to category E, then HF is a functor which maps every element c1 of C to element H(F(c1)) of E, and maps every morphism f of C to morphism H(F(f)) of E.
+
+I'll assert without proving that functor composition is associative.
+
+
+
+4. Natural Transformation
+-------------------------
+So categories include elements and morphisms. Functors consist of mappings from the elements and morphisms of one category to those of another (or the same) category. <Natural transformations> are a third level of mappings, from one functor to another.
+
+Where G and H are functors from category C to category D, a natural transformation eta between G and H is a family of morphisms eta[c1]:G(c1)->H(c1) in D for each element c1 of C. That is, eta[c1] has as source c1's image under G in D, and as target c1's image under H in D. The morphisms in this family must also satisfy the constraint:
+ for every morphism f:c1->c2 in C:
+ eta[c2] o G(f) = H(f) o eta[c1]
+
+That is, the morphism via G(f) from G(c1) to G(c2), and then via eta[c2] to H(c2), is identical to the morphism from G(c1) via eta[c1] to H(c1), and then via H(f) from H(c1) to H(c2).
+
+
+How natural transformations compose:
+
+Consider four categories B,C,D, and E.
+Let F be a functor from B to C; G,H, and J be functors from C to D; and K and L be functors from D to E. Let eta be a natural transformation from G to H; phi be a natural transformation from H to J; and psi be a natural transformation from K to L. Pictorally:
+
+- B -+ +--- C --+ +---- D -----+ +-- E --
+ | | | | | |
+ F: ------> G: ------> K: ------>
+ | | | | | eta | | | psi
+ | | | | v | | v
+ | | H: ------> L: ------>
+ | | | | | phi | |
+ | | | | v | |
+ | | J: ------> | |
+-----+ +--------+ +------------+ +-------
+
+(eta F) is a natural transformation from the (composite) functor GF to the composite functor HF, such that where b1 is an element of category B, (eta F)[b1] = eta[F(b1)]---that is, the morphism in D that eta assigns to the element F(b1) of C.
+
+(K eta) is a natural transformation from the (composite) functor KG to the (composite) functor KH, such that where c1 is an element of category C, (K eta)[c1] = K(eta[c1])---that is, the morphism in E that K assigns to the morphism eta[c1] of D.
+
+
+(phi -v- eta) is a natural transformation from G to J; this is known as a "vertical composition". We will rely later on this:
+ phi[c2] o H(f) o eta[c1] = phi[c2] o H(f) o eta[c1]
+ -------------
+ by naturalness of phi, is:
+ --------------
+ phi[c2] o H(f) o eta[c1] = J(f) o phi[c1] o eta[c1]
+ --------------
+ by naturalness of eta, is:
+ --------------
+ phi[c2] o eta[c2] o G(f) = J(f) o phi[c1] o eta[c1]
+ ----------------- -----------------
+Hence, we can define (phi -v- eta)[c1] as: phi[c1] o eta[c1] and rely on it to satisfy the constraints for a natural transformation from G to J:
+ ----------------- -----------------
+ (phi -v- eta)[c2] o G(f) = J(f) o (phi -v- eta)[c1]
+
+I'll assert without proving that vertical composition is associative and has an identity, which we'll call "the identity transformation."
+
+
+(psi -h- eta) 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:
+
+ (phi -h- eta)[c1] = L(eta[c1]) o psi[G(c1)]
+ = psi[H(c1)] o K(eta[c1])
+
+Horizontal composition is also associative, and has the same identity as vertical composition.
+
+
+
+5. Monads
+---------
+In earlier days, these were also called "triples."
+
+A <monad> is a structure consisting of an (endo)functor M from some category C to itself, along with some natural transformations, which we'll specify in a moment.
+
+Let T be a set of natural transformations p, each being between some (variable) functor P and another functor which is the composite MP' of M and a (variable) functor P'. That is, for each element c1 in C, p assigns c1 a morphism from element P(c1) to element MP'(c1), satisfying the constraints detailed in the previous section. For different members of T, the relevant functors may differ; that is, p is a transformation from functor P to MP', q is a transformation from functor Q to MQ', and none of P,P',Q,Q' 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 on C to M(1C). So it will assign to c1 a morphism from c1 to M(c1).
+
+We also need to designate for M a "join" transformation, which is a natural transformation from the (composite) functor MM to M.
+
+These two natural transformations have to satisfy some constraints ("the monad laws") which are most easily stated if we can introduce a defined notion.
+
+Let p and q be members of T, that is they are natural transformations from P to MP' and from Q to MQ', respectively. Let them be such that P' = Q. Now (M q) will also be a natural transformation, formed by composing the functor M with the natural transformation q. Similarly, (join Q') will be a natural transformation, formed by composing the natural transformation join with the functor Q'; it will transform the functor MMQ' to the functor MQ'. Now take the vertical composition of the three natural transformations (join Q'), (M q), and p, and abbreviate it as follows:
+
+ q <=< p =def. ((join Q') -v- (M q) -v- p) --- since composition is associative I don't specify the order of composition on the rhs
+
+In other words, <=< is a binary operator that takes us from two members p and q of T to a composite natural transformation. (In functional programming, at least, this is called the "Kleisli composition operator". Sometimes its written p >=> q where that's the same as q <=< p.)
+
+p is a transformation from P to MP' which = MQ; (M q) is a transformation from MQ to MMQ'; and (join Q') is a transformation from MMQ' to MQ'. So the composite q <=< p will be a transformation from P to MQ', 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 monoid
+
+That's it. In other words:
+
+ for all p,q,r in T:
+ (i) q <=< p etc are also in T
+ (ii) (r <=< q) <=< p = r <=< (q <=< p)
+ (iii.1) (unit P') <=< p = p
+ (iii.2) p = p <=< (unit P)
+
+A word about the P' and P in (iii.1) and (iii.2): since unit on its own is a transformation from 1C to M(1C), it doesn't have the appropriate "type" for unit <=< p or p <=< unit to be defined, for arbitrary p. However, if p is a transformation from P to MP', then (unit P') <=< p and p <=< (unit P) will both be defined.
+
+
+
+6. The 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 <=<.
+
+(*
+ P2. every element c1 of a category C has an identity morphism id[c1] such that for every morphism f:c1->c2 in C: id[c2] o f = f = f o id[c1].
+ P3. functors "preserve identity", that is for every element c1 in F's source category: F(id[c1]) = id[F(c1)].
+*)
+
+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 o f) = F(g) o F(f)
+ * if eta is a natural transformation from F to G, then for every f:c1->c2 in F and G's source category C: eta[c2] o F(f) = G(f) o eta[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 C, join[c1] will be a morphism from MM(c1) to M(c1). And for any morphism f:a->b in C:
+
+ (1) join[b] o MM(f) = M(f) o join[a]
+
+Next, consider the composite transformation ((join MQ') -v- (MM q)).
+ q is a transformation from Q to MQ', and assigns elements c1 in C a morphism q*: Q(c1) -> MQ'(c1). (MM q) is a transformation that instead assigns c1 the morphism MM(q*).
+ (join MQ') is a transformation from MMMQ' to MMQ' that assigns c1 the morphism join[MQ'(c1)].
+ Composing them:
+ (2) ((join MQ') -v- (MM q)) assigns to c1 the morphism join[MQ'(c1)] o MM(q*).
+
+Next, consider the composite transformation ((M q) -v- (join Q)).
+ (3) This assigns to c1 the morphism M(q*) o join[Q(c1)].
+
+So for every element c1 of C:
+ ((join MQ') -v- (MM q))[c1], by (2) is:
+ join[MQ'(c1)] o MM(q*), which by (1), with f=q*: Q(c1)->MQ'(c1) is:
+ M(q*) o join[Q(c1)], which by 3 is:
+ ((M q) -v- (join Q))[c1]
+
+So our (lemma 1) is: ((join MQ') -v- (MM q)) = ((M q) -v- (join Q)), where q is a transformation from Q to MQ'.
+
+
+Next recall that unit is a natural transformation from 1C to M. So for elements c1 in C, unit[c1] will be a morphism from c1 to M(c1). And for any morphism f:a->b in C:
+ (4) unit[b] o f = M(f) o unit[a]
+
+Next consider the composite transformation ((M q) -v- (unit Q)). (5) This assigns to c1 the morphism M(q*) o unit[Q(c1)].
+
+Next consider the composite transformation ((unit MQ') -v- q). (6) This assigns to c1 the morphism unit[MQ'(c1)] o q*.
+
+So for every element c1 of C:
+ ((M q) -v- (unit Q))[c1], by (5) =
+ M(q*) o unit[Q(c1)], which by (4), with f=q*: Q(c1)->MQ'(c1) is:
+ unit[MQ'(c1)] o q*, which by (6) =
+ ((unit MQ') -v- q)[c1]
+
+So our lemma (2) is: (((M q) -v- (unit Q)) = ((unit MQ') -v- q)), where q is a transformation from Q to MQ'.
+
+
+Finally, we substitute ((join Q') -v- (M q) -v- p) for q <=< p in the monad laws. For simplicity, I'll omit the "-v-".
+
+ for all p,q,r in T, where p is a transformation from P to MP', q is a transformation from Q to MQ', R is a transformation from R to MR', and P'=Q and Q'=R:
+
+ (i) q <=< p etc are also in T
+ ==>
+ (i') ((join Q') (M q) p) etc are also in T
+
+
+ (ii) (r <=< q) <=< p = r <=< (q <=< p)
+ ==>
+ (r <=< q) is a transformation from Q to MR', so:
+ (r <=< q) <=< p becomes: (join R') (M (r <=< q)) p
+ which is: (join R') (M ((join R') (M r) q)) p
+ substituting in (ii), and helping ourselves to associativity on the rhs, we get:
+
+ ((join R') (M ((join R') (M r) q)) p) = ((join R') (M r) (join Q') (M q) p)
+ ---------------------
+ which by the distributivity of functors over composition, and helping ourselves to associativity on the lhs, yields:
+ ------------------------
+ ((join R') (M join R') (MM r) (M q) p) = ((join R') (M r) (join Q') (M q) p)
+ ---------------
+ which by lemma 1, with r a transformation from Q' to MR', yields:
+ -----------------
+ ((join R') (M join R') (MM r) (M q) p) = ((join R') (join MR') (MM r) (M q) p)
+
+ which will be true for all r,q,p 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 P') <=< p = p
+ ==>
+ (unit P') is a transformation from P' to MP', so:
+ (unit P') <=< p becomes: (join P') (M unit P') p
+ which is: (join P') (M unit P') p
+ substituting in (iii.1), we get:
+ ((join P') (M unit P') p) = p
+
+ which will be true for all p just in case:
+
+ ((join P') (M unit P')) = the identity transformation, for any P'
+
+ which will in turn be true just in case:
+
+ (iii.1') (join (M unit) = the identity transformation
+
+
+ (iii.2) p = p <=< (unit P)
+ ==>
+ p is a transformation from P to MP', so:
+ unit <=< p becomes: (join P') (M p) unit
+ substituting in (iii.2), we get:
+ p = ((join P') (M p) (unit P))
+ --------------
+ which by lemma (2), yields:
+ ------------
+ p = ((join P') ((unit MP') p)
+
+ which will be true for all p just in case:
+
+ ((join P') (unit MP')) = the identity transformation, for any P'
+
+ which will in turn be true just in case:
+
+ (iii.2') (join (unit M)) = the identity transformation
+
+
+Collecting the results, our monad laws turn out in this format to be:
+
+ when p a transformation from P to MP', q a transformation from P' to MQ', r a transformation from Q' to MR' all in T:
+
+ (i') ((join Q') (M q) p) 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 C will have types as elements, and monadic functions as its morphisms. The source and target of a morphism will be the types of its argument and its result. (As always, there can be multiple distinct morphisms from the same source to the same target.)
+
+A monad M will consist of a mapping from types c1 to types M(c1), and a mapping from functions f:c1->c2 to functions M(f):M(c1)->M(c2). This is also known as "fmap f" or "liftM f" for M, and is called "function f lifted into the monad M." For example, where M is the list monad, M maps every type X into the type "list of Xs", and maps every function f:x->y into the function that maps [x1,x2...] to [y1,y2,...].
+
+
+
+
+A natural transformation t assigns to each type c1 in C a morphism t[c1]: c1->M(c1) such that, for every f:c1->c2:
+ t[c2] o f = M(f) o t[c1]
+
+The composite morphisms said here to be identical are morphisms from the type c1 to the type M(c2).
+
+
+
+In functional programming, instead of working with natural transformations we work with "monadic values" and polymorphic functions "into the monad" in question.
+
+For an example of the latter, let p be a function that takes arguments of some (schematic, polymorphic) type c1 and yields results of some (schematic, polymorphic) type M(c2). An example with M being the list monad, and c2 being the tuple type schema int * c1:
+
+ let p = fun c -> [(1,c), (2,c)]
+
+p is polymorphic: when you apply it to the int 0 you get a result of type "list of int * int": [(1,0), (2,0)]. When you apply it to the char 'e' you get a result of type "list of int * char": [(1,'e'), (2,'e')].
+
+However, to keep things simple, we'll work instead with functions whose type is settled. So instead of the polymorphic p, we'll work with (p : c1 -> M(int * c1)). This only accepts arguments of type c1. For generality, I'll talk of functions with the type (p : c1 -> M(c1')), where we assume that c1' is a function of c1.
+
+A "monadic value" is any member of a type M(c1), for any type c1. For example, a list is a monadic value for the list monad. We can think of these monadic values as the result of applying some function (p : c1 -> M(c1')) to an argument of type c1.
+
+