X-Git-Url: http://lambda.jimpryor.net/git/gitweb.cgi?p=lambda.git;a=blobdiff_plain;f=advanced_topics%2Fmonads_in_category_theory.mdwn;h=f05747f675b416fa5d5af01dee2d0f0278258874;hp=4941e80b99875eb3f9179c704da83e63d4d3ae66;hb=704abdbe4afe78e553738662a3cb0bb6944b54ff;hpb=4f87cdd50334c5d9dddbdad136fac4008d8ce6ff diff --git a/advanced_topics/monads_in_category_theory.mdwn b/advanced_topics/monads_in_category_theory.mdwn index 4941e80b..f05747f6 100644 --- a/advanced_topics/monads_in_category_theory.mdwn +++ b/advanced_topics/monads_in_category_theory.mdwn @@ -33,8 +33,8 @@ 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. +* 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 ---------- @@ -58,7 +58,7 @@ To have a category, the elements and morphisms have to satisfy some constraints: 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.) +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: @@ -67,7 +67,7 @@ Some examples of categories are: * 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: +* 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) @@ -136,7 +136,7 @@ Then (η F) is a natural transformation from the (composite) fun 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 -v- η) is a natural transformation from `G` to `J`; this is known as a "vertical composition". For any morphism f:C1→C2 in C: 
 	φ[C2] ∘ H(f) ∘ η[C1] = φ[C2] ∘ H(f) ∘ η[C1]
@@ -186,22 +186,20 @@ 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.
+Let `T` be a set of natural transformations φ, each being between some arbitrary endofunctor `F` on C and another functor which is the composite `MF'` of `M` and another arbitrary endofunctor `F'` on C. 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)`.
+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`.
+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:
+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. 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)
 

 
-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 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`.
@@ -212,48 +210,66 @@ 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 φ is a transformation to some `MF'` and γ is a transformation from `F'`. But wherever `<=<` is defined, the monoid laws are satisfied: +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 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.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)
+	(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: +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
-	  ??
+	  = (((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') <=< 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: +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 = (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 = (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 `MF'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 ρ, γ, φ in T for which ρ <=< γ and γ <=< φ are defined:
+
+	    (i) γ <=< φ etc are 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'G
+
@@ -272,40 +288,47 @@ Let's remind ourselves of some principles: * 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]. +* if η is a natural transformation from `G` to `H`, then for every f:C1→C2 in `G` and `H`'s source category C: η[C2] ∘ G(f) = H(f) ∘ η[C1]. + +* (η F)[E] = η[F(E)] + +* (K η)[E} = K(η[E]) + +* ((φ -v- η) F) = ((φ F) -v- (η F)) 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:C1→C2 in C: +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:C1→C2 in C: 
 	(1) join[C2] ∘ MM(f)  =  M(f) ∘ join[C1]
-Next, consider the composite transformation ((join MG') -v- (MM γ)). +Next, let γ be a transformation from `G` to `MG'`, and + 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\*). +* γ 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)]`. +* `(join MG')` is a transformation from `MM(MG')` to `M(MG')` 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*).
+	(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)). +Next: 
-	(3) This assigns to C1 the morphism M(γ*) ∘ join[G(C1)].
+	(3) Consider the composite transformation ((M γ) -v- (join G)). 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:
+	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]
@@ -313,33 +336,34 @@ So for every element `C1` of C: So our **(lemma 1)** is: 
-	((join MG') -v- (MM γ))  =  ((M γ) -v- (join G)), where γ is a transformation from G to MG'.
+	((join MG') -v- (MM γ))  =  ((M γ) -v- (join G)),
+	where as we said γ is a natural 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: +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:C1→C2 in C: 
-	(4) unit[b] ∘ f = M(f) ∘ unit[a]
+	(4) unit[C2] ∘ f = M(f) ∘ unit[C1]
-Next consider the composite transformation ((M γ) -v- (unit G)): +Next: 
-	(5) This assigns to C1 the morphism M(γ*) ∘ unit[G(C1)].
+	(5) Consider the composite transformation ((M γ) -v- (unit G)). This assigns to C1 the morphism M(γ*) ∘ unit[G(C1)].
-Next consider the composite transformation ((unit MG') -v- γ). +Next: 
-	(6) This assigns to C1 the morphism unit[MG'(C1)] ∘ γ*.
+	(6) Consider the composite transformation ((unit MG') -v- γ). 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:
+	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]
@@ -347,11 +371,12 @@ So for every element C1 of C: So our **(lemma 2)** is: 
-	(((M γ) -v- (unit G))  =  ((unit MG') -v- γ)), where γ is a transformation from G to MG'.
+	(((M γ) -v- (unit G))  =  ((unit MG') -v- γ)),
+	where as we said γ is a natural 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-". +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: