1 **Don't try to read this yet!!! Many substantial edits are still in process.
4 **Caveats**: I really don't know much category theory. Just enough to put this
5 together. Also, this really is "put together." I haven't yet found an
6 authoritative source (that's accessible to a category theory beginner like
7 myself) that discusses the correspondence between the category-theoretic and
8 functional programming uses of these notions in enough detail to be sure that
9 none of the pieces here is misguided. In particular, it wasn't completely
10 obvious how to map the polymorphism on the programming theory side into the
11 category theory. And I'm bothered by the fact that our `<=<` operation is only
12 partly defined on our domain of natural transformations. But these do seem to
13 me to be the reasonable way to put the pieces together. We very much welcome
14 feedback from anyone who understands these issues better, and will make
20 A **monoid** is a structure `(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:
22 for all s1,s2,s3 in S:
23 (i) s1*s2 etc are also in S
24 (ii) (s1*s2)*s3 = s1*(s2*s3)
25 (iii) z*s1 = s1 = s1*z
27 Some examples of monoids are:
29 * finite strings of an alphabet A, with * being concatenation and z being the empty string
30 * all functions X->X over a set X, with * being composition and z being the identity function over X
31 * 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**.)
32 * if we let * be multiplication and z be 1, we get different monoids over the same sets as in the previous item.
36 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."
38 When a morphism f in category C has source c1 and target c2, we'll write f:c1->c2.
40 To have a category, the elements and morphisms have to satisfy some constraints:
41 (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.
42 (ii) composition of morphisms has to be associative
43 (iii) every element e of the category has to have an identity morphism id[e], which is such that for every morphism f:c1->c2:
44 id[c2] o f = f = f o id[c1]
46 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.
48 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. (Every node must have at least one such, which is that node's identity morphism.)
51 Some examples of categories are:
53 * 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 is the identity function over that set.
55 * 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. 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.
57 * 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 `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).
61 * 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)
62 * sets ordered by size (this illustrates it too)
64 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.
70 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:
73 <LI>associate with every element c1 of C an element F(c1) of D
74 <LI>associate with every morphism f:c1->c2 of C a morphism F(f):F(c1)->F(c2) of D
75 <LI>"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:
76 F(id[c1]) = id[F(c1)].
77 <LI>"distribute over composition", that is for any morphisms f and g in C:
78 F(g o f) = F(g) o F(f)
81 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`.
83 How functors compose: If `G` is a functor from category `C` to category `D`, and `K` is a functor from category `D` to category `E`, then `KG` is a functor which maps every element `c1` of `C` to element `K(G(c1))` of `E`, and maps every morphism `f` of `C` to morphism `K(G(f))` of `E`.
85 I'll assert without proving that functor composition is associative.
89 4. Natural Transformation
90 -------------------------
91 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.
93 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:
94 for every morphism f:c1->c2 in C:
95 eta[c2] o G(f) = H(f) o eta[c1]
97 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).
100 How natural transformations compose:
102 Consider four categories B,C,D, and E.
103 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:
105 - B -+ +--- C --+ +---- D -----+ +-- E --
107 F: ------> G: ------> K: ------>
108 | | | | | eta | | | psi
110 | | H: ------> L: ------>
114 -----+ +--------+ +------------+ +-------
116 Then (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.
118 And (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.
121 (phi -v- eta) is a natural transformation from G to J; this is known as a "vertical composition". We will rely later on this, where f:c1->c2:
122 phi[c2] o H(f) o eta[c1] = phi[c2] o H(f) o eta[c1]
124 by naturalness of phi, is:
126 phi[c2] o H(f) o eta[c1] = J(f) o phi[c1] o eta[c1]
128 by naturalness of eta, is:
130 phi[c2] o eta[c2] o G(f) = J(f) o phi[c1] o eta[c1]
131 ----------------- -----------------
132 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:
133 ----------------- -----------------
134 (phi -v- eta)[c2] o G(f) = J(f) o (phi -v- eta)[c1]
136 I'll assert without proving that vertical composition is associative and has an identity, which we'll call "the identity transformation."
139 (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:
141 (phi -h- eta)[c1] = L(eta[c1]) o psi[G(c1)]
142 = psi[H(c1)] o K(eta[c1])
144 Horizontal composition is also associative, and has the same identity as vertical composition.
150 In earlier days, these were also called "triples."
152 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.
154 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.
156 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).
158 We also need to designate for M a "join" transformation, which is a natural transformation from the (composite) functor MM to M.
160 These two natural transformations have to satisfy some constraints ("the monad laws") which are most easily stated if we can introduce a defined notion.
162 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:
164 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
166 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.)
168 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.
170 Now we can specify the "monad laws" governing a monad as follows:
172 (T, <=<, unit) constitute a monoid
174 That's it. In other words:
177 (i) q <=< p etc are also in T
178 (ii) (r <=< q) <=< p = r <=< (q <=< p)
179 (iii.1) (unit P') <=< p = p
180 (iii.2) p = p <=< (unit P)
182 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.
186 6. The standard category-theory presentation of the monad laws
187 --------------------------------------------------------------
188 In category theory, the monad laws are usually stated in terms of unit and join instead of unit and <=<.
191 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].
192 P3. functors "preserve identity", that is for every element c1 in F's source category: F(id[c1]) = id[F(c1)].
195 Let's remind ourselves of some principles:
196 * composition of morphisms, functors, and natural compositions is associative
197 * 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)
198 * 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].
201 Let's use the definitions of naturalness, and of composition of natural transformations, to establish two lemmas.
204 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:
206 (1) join[b] o MM(f) = M(f) o join[a]
208 Next, consider the composite transformation ((join MQ') -v- (MM q)).
209 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*).
210 (join MQ') is a transformation from MMMQ' to MMQ' that assigns c1 the morphism join[MQ'(c1)].
212 (2) ((join MQ') -v- (MM q)) assigns to c1 the morphism join[MQ'(c1)] o MM(q*).
214 Next, consider the composite transformation ((M q) -v- (join Q)).
215 (3) This assigns to c1 the morphism M(q*) o join[Q(c1)].
217 So for every element c1 of C:
218 ((join MQ') -v- (MM q))[c1], by (2) is:
219 join[MQ'(c1)] o MM(q*), which by (1), with f=q*: Q(c1)->MQ'(c1) is:
220 M(q*) o join[Q(c1)], which by 3 is:
221 ((M q) -v- (join Q))[c1]
223 So our (lemma 1) is: ((join MQ') -v- (MM q)) = ((M q) -v- (join Q)), where q is a transformation from Q to MQ'.
226 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:
227 (4) unit[b] o f = M(f) o unit[a]
229 Next consider the composite transformation ((M q) -v- (unit Q)). (5) This assigns to c1 the morphism M(q*) o unit[Q(c1)].
231 Next consider the composite transformation ((unit MQ') -v- q). (6) This assigns to c1 the morphism unit[MQ'(c1)] o q*.
233 So for every element c1 of C:
234 ((M q) -v- (unit Q))[c1], by (5) =
235 M(q*) o unit[Q(c1)], which by (4), with f=q*: Q(c1)->MQ'(c1) is:
236 unit[MQ'(c1)] o q*, which by (6) =
237 ((unit MQ') -v- q)[c1]
239 So our lemma (2) is: (((M q) -v- (unit Q)) = ((unit MQ') -v- q)), where q is a transformation from Q to MQ'.
242 Finally, we substitute ((join Q') -v- (M q) -v- p) for q <=< p in the monad laws. For simplicity, I'll omit the "-v-".
244 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:
246 (i) q <=< p etc are also in T
248 (i') ((join Q') (M q) p) etc are also in T
251 (ii) (r <=< q) <=< p = r <=< (q <=< p)
253 (r <=< q) is a transformation from Q to MR', so:
254 (r <=< q) <=< p becomes: (join R') (M (r <=< q)) p
255 which is: (join R') (M ((join R') (M r) q)) p
256 substituting in (ii), and helping ourselves to associativity on the rhs, we get:
258 ((join R') (M ((join R') (M r) q)) p) = ((join R') (M r) (join Q') (M q) p)
259 ---------------------
260 which by the distributivity of functors over composition, and helping ourselves to associativity on the lhs, yields:
261 ------------------------
262 ((join R') (M join R') (MM r) (M q) p) = ((join R') (M r) (join Q') (M q) p)
264 which by lemma 1, with r a transformation from Q' to MR', yields:
266 ((join R') (M join R') (MM r) (M q) p) = ((join R') (join MR') (MM r) (M q) p)
268 which will be true for all r,q,p just in case:
270 ((join R') (M join R')) = ((join R') (join MR')), for any R'.
272 which will in turn be true just in case:
274 (ii') (join (M join)) = (join (join M))
277 (iii.1) (unit P') <=< p = p
279 (unit P') is a transformation from P' to MP', so:
280 (unit P') <=< p becomes: (join P') (M unit P') p
281 which is: (join P') (M unit P') p
282 substituting in (iii.1), we get:
283 ((join P') (M unit P') p) = p
285 which will be true for all p just in case:
287 ((join P') (M unit P')) = the identity transformation, for any P'
289 which will in turn be true just in case:
291 (iii.1') (join (M unit) = the identity transformation
294 (iii.2) p = p <=< (unit P)
296 p is a transformation from P to MP', so:
297 unit <=< p becomes: (join P') (M p) unit
298 substituting in (iii.2), we get:
299 p = ((join P') (M p) (unit P))
301 which by lemma (2), yields:
303 p = ((join P') ((unit MP') p)
305 which will be true for all p just in case:
307 ((join P') (unit MP')) = the identity transformation, for any P'
309 which will in turn be true just in case:
311 (iii.2') (join (unit M)) = the identity transformation
314 Collecting the results, our monad laws turn out in this format to be:
316 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:
318 (i') ((join Q') (M q) p) etc also in T
320 (ii') (join (M join)) = (join (join M))
322 (iii.1') (join (M unit)) = the identity transformation
324 (iii.2')(join (unit M)) = the identity transformation
328 7. The functional programming presentation of the monad laws
329 ------------------------------------------------------------
330 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.
332 Additionally, whereas in category-theory one works "monomorphically", in functional programming one usually works with "polymorphic" functions.
334 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.)
336 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,...].
341 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:
342 t[c2] o f = M(f) o t[c1]
344 The composite morphisms said here to be identical are morphisms from the type c1 to the type M(c2).
348 In functional programming, instead of working with natural transformations we work with "monadic values" and polymorphic functions "into the monad" in question.
350 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:
352 let p = fun c -> [(1,c), (2,c)]
354 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')].
356 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.
358 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.