1. Monoids
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A 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 ). 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 .)
(d) the natural numbers with * being multiplication and z being 1 constitute a different monoid over the same set as in (c).
2. Categories
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A is a generalization of a monoid. A category consists of a class of elements, and a class of 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 <= 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
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A 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 . 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. 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]
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by naturalness of phi, is:
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phi[c2] o H(f) o eta[c1] = J(f) o phi[c1] o eta[c1]
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by naturalness of eta, is:
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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
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In earlier days, these were also called "triples."
A 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.