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=0db16cb2d5f141b5201d6bafc8dbcd27d3e97141;hp=966fc95c7e6994e66ee5e014cb6a93a91a810c04;hb=c0b681e8273859ba1a6d526265bb451853ddeb7d;hpb=f11472452da56243652ffeed1f56b418f6b6c1dd
diff --git a/advanced_topics/monads_in_category_theory.mdwn b/advanced_topics/monads_in_category_theory.mdwn
index 966fc95c..0db16cb2 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] @@ -186,66 +186,90 @@ 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 transformationswhen 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 @@ -388,6 +453,7 @@ Collecting the results, our monad laws turn out in this format to be: (iii.1') (join (M unit)) = the identity transformation (iii.2')(join (unit M)) = the identity transformation +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 transformationsφ
, 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. -Letφ
andγ
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 transformationγ
. 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)
, andφ
, and abbreviate it as follows: +Letφ
andγ
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 transformationγ
. 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)
, andφ
, 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 membersφ
andγ
of `T` to a composite natural transformation. (In functional programming, at least, this is called the "Kleisli composition operator". Sometimes it's writtenφ >=> γ
where that's the same asγ <=< 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 compositeγ <=< 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 whenφ
is a transformation to some `MF'` andγ
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 whenφ
is a transformation to some `MF'` andγ
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)-Ifφ
is a natural transformation from `F` to `M(1C)` andγ
is(φ G')
, that is, a natural transformation from `FG` to `MG`, then we can extend (iii.1) as follows: +Ifφ
is a natural transformation from `F` to `M(1C)` andγ
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 saidγ
is a natural transformation from some `FG'` to `MG'`. -Similarly, ifφ
is a natural transformation from `1C` to `MF'`, andγ
is(φ G)
, that is, a natural transformation from `G` to `MF'G`, then we can extend (iii.2) as follows: +Similarly, ifρ
is a natural transformation from `1C` to `MR'`, andγ
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 saidγ
is a natural transformation from `G` to some `MR'G`. -where as we saidγ
is a natural transformation from `G` to some `MF'G`. +Summarizing then, the monad laws can be expressed as: ++ For all γ, φ in T for which ρ <=< γ and γ <=< φ are defined: + + (i) γ <=< φ is 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 +@@ -253,60 +277,99 @@ 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 `<=<`. -(* + 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]. +* 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)
+ +* ifη
is a natural transformation from `F` to `G`, then for everyf: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: +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 morphismf:C1→C2
in C: + ++ (1) join[C2] ∘ MM(f) = M(f) ∘ join[C1] ++ +Next, consider the composite transformation((join MG') -v- (MM γ))
. - (1) join[b] ∘ MM(f) = M(f) ∘ join[a] +*γ
is a transformation from `G` to `MG'`, and assigns elements `C1` in C a morphismγ\*: G(C1) → MG'(C1)
.(MM γ)
is a transformation that instead assigns `C1` the morphismMM(γ\*)
. -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: +* `(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))
. -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: +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'. +-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 morphismf: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: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((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*. +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'. +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-". +Finally, we substitute((join G') -v- (M γ) -v- φ)
forγ <=< 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 @@ -375,10 +438,12 @@ Finally, we substitute ((join G') -v- (M γ) -v- φ) for γ <=< &ph 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: +