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-**Don't try to read this yet!!! Many substantial edits are still in process.
-Will be ready soon.**
-
-Caveats
--------
-I really don't know much category theory. Just enough to put this
-together. Also, this really is "put together." I haven't yet found an
-authoritative source (that's accessible to a category theory beginner like
-myself) that discusses the correspondence between the category-theoretic and
-functional programming uses of these notions in enough detail to be sure that
-none of the pieces here is misguided. In particular, it wasn't completely
-obvious how to map the polymorphism on the programming theory side into the
-category theory. And I'm bothered by the fact that our `<=<` operation is only
-partly defined on our domain of natural transformations. But this does seem to
-me to be the reasonable way to put the pieces together. We very much welcome
-feedback from anyone who understands these issues better, and will make
-corrections.
-
-
-Monoids
--------
-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:
-
-
-
- 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: - -* 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 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 ∘ 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 1E, which is such that for every morphism f:C1→C2: - 1C2 ∘ f = f = f ∘ 1C1 -- -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. - -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.) - - -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
(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 `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; 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 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(1C1) = 1F(C1). - - (iv) "distribute over composition", that is for any morphisms f and g in C: - F(g ∘ f) = F(g) ∘ 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 `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. - -I'll assert without proving that functor composition is associative. - - - -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 η between `G` and `H` is a family of morphisms
η[C1]:G(C1)→H(C1)
in D for each element `C1` of C. That is, η[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: - η[C2] ∘ G(f) = H(f) ∘ η[C1] -- -That is, the morphism via `G(f)` from `G(C1)` to `G(C2)`, and then via
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 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 η be a natural transformation from `G` to `H`; φ be a natural transformation from `H` to `J`; and ψ be a natural transformation from `K` to `L`. Pictorally:
-
-- - B -+ +--- C --+ +---- D -----+ +-- E -- - | | | | | | - F: ------> G: ------> K: ------> - | | | | | η | | | ψ - | | | | v | | v - | | H: ------> L: ------> - | | | | | φ | | - | | | | v | | - | | J: ------> | | - -----+ +--------+ +------------+ +------- -- -Then
(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 B, (η F)[B1] = η[F(B1)]
---that is, the morphism in D that η
assigns to the element `F(B1)` of C.
-
-And (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 C, (K η)[C1] = K(η[C1])
---that is, the morphism in E that `K` assigns to the morphism η[C1]
of D.
-
-
-(φ -v- η)
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
:
-
-- 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 C to itself, along with some natural transformations, which we'll specify in a moment. - -Let `T` be a set of natural transformations
φ
, each being between some (variable) functor `F` and another functor which is the composite `MF'` of `M` and a (variable) functor `F'`. That is, for each element `C1` in C, 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 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 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:
-
-- γ <=< φ =def. ((join G') -v- (M γ) -v- φ) -- -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
φ
and γ
of `T` to a composite natural transformation. (In functional programming, at least, this is called the "Kleisli composition operator". Sometimes its written φ >=> γ
where that's the same as γ <=< φ
.)
-
-φ
is a transformation from `F` to `MF'` which = `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 monoid
-
-That'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 `F` is a functor to `MF'` and `G` is a functor from `F'`. But wherever `<=<` is defined, the monoid laws are satisfied:
-
- (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) φ = φ <=< unit (here φ 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 -v- (M unit) -v- φ) G')
- = (join G') -v- ((M unit) G') -v- (φ G')
- = (join G') -v- (M (unit G')) -v- γ
- ??
- = (unit G') <=< γ
-
-where as we said γ
is a natural transformation from some `FG'` to `MG'`.
-
-Similarly, if φ is a natural transformation from `1C` to `MF'`, and γ is `(φ G)`, that is, a natural transformation from `G` to `MF'G`, then we can extend (iii.2) as follows:
-
- γ = (φ G)
- = ((φ <=< unit) G)
- = (((join F') -v- (M φ) -v- unit) G)
- = ((join F'G) -v- ((M φ) G) -v- (unit G))
- = ((join F'G) -v- (M (φ G)) -v- (unit G))
- ??
- = γ <=< (unit G)
-
-where as we said γ is a natural transformation from `G` to some `MF'G`.
-
-
-
-
-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 1C1 such that for every morphism f:C1→C2 in C: 1C2 ∘ f = f = f ∘ 1C1.
- P3. functors "preserve identity", that is for every element C1 in F's source category: F(1C1) = 1F(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 ∘ 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 C: 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 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] ∘ MM(f) = M(f) ∘ join[a]
-
-Next, consider the composite transformation ((join MG') -v- (MM γ)).
- γ is a transformation from G to MG', and assigns elements C1 in C a morphism 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 C:
- ((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 C, unit[C1] will be a morphism from C1 to M(C1). And for any morphism f:a→b in C:
- (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 C:
- ((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 transformation
-
-
-Collecting 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 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] ∘ f = M(f) ∘ 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 φ 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 φ = fun c → [(1,c), (2,c)]
-
-φ 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 φ, we'll work with (φ : C1 → M(int * C1)). This only accepts arguments of type C1. For generality, I'll talk of functions with the type (φ : 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 (φ : C1 → M(C1')) to an argument of type C1.
-
-