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=24670796b21e1fc3cd230352e0049772bf17a2b3;hp=a6165e9deae3a47b9401f946e8c3087d85a33cdb;hb=ac47fea0dc57a0f496d14dba80383206397a35f9;hpb=7b00a5a3dc3f7208f67ed5c87faf22b351e14b0c
diff --git a/advanced_topics/monads_in_category_theory.mdwn b/advanced_topics/monads_in_category_theory.mdwn
index a6165e9d..24670796 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]
@@ -186,208 +186,296 @@ 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.)
-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`.
+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`.
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:
+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.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:
+ (iii.1) unit <=< φ = φ
+ (here φ has to be a natural transformation to M(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:
+
+
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 `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)
-where as we said γ is a natural transformation from `G` to some `MF'G`.
+ (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
+
-The standard category-theory presentation of the monad laws
------------------------------------------------------------
+Getting to 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)
+
+* 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: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 morphism f:C1→C2 in C:
- (1) join[b] ∘ MM(f) = M(f) ∘ join[a]
+
+ (1) join[C2] ∘ MM(f) = M(f) ∘ join[C1]
+
-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:
+Next, let γ be a transformation from `G` to `MG'`, and
+ consider the composite transformation ((join MG') -v- (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 `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*).
+
-Next, consider the composite transformation ((M γ) -v- (join G)).
- (3) This assigns to C1 the morphism M(γ*) ∘ join[G(C1)].
+Next, consider the composite transformation ((M γ) -v- (join G)):
-So for every element C1 of C:
+
+ (3) ((M γ) -v- (join G)) 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]
+
-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 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:
- (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 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:
-Next consider the composite transformation ((unit MG') -v- γ). (6) This assigns to C1 the morphism unit[MG'(C1)] ∘ γ*.
+
+ (4) unit[C2] ∘ f = M(f) ∘ unit[C1]
+
+
+Next, consider the composite transformation ((M γ) -v- (unit G)):
+
+
+ (5) ((M γ) -v- (unit G)) assigns to C1 the morphism M(γ*) ∘ unit[G(C1)].
+
+
+Next, consider the composite transformation ((unit MG') -v- γ):
+
+
+ (6) ((unit MG') -v- γ) 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]
+
-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 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-".
- 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:
+Finally, we substitute ((join G') -v- (M γ) -v- φ) for γ <=< φ in the monad laws. For simplicity, I'll omit the "-v-".
- (i) γ <=< φ etc are also in T
+
+ For all ρ, γ, φ in T,
+ where φ is a transformation from F to MF',
+ γ is a transformation from G to MG',
+ ρ 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
+ (i') ((join G') (M γ) φ) etc are also in T
- (ii) (ρ <=< γ) <=< φ = ρ <=< (γ <=< φ)
+
+ (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:
+ (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)
- ((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)
+ similarly, ρ <=< (γ <=< φ) is:
+ ((join R') (M ρ) ((join G') (M γ) φ))
- which will be true for all ρ,γ,φ just in case:
+ substituting these into (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)
- ((join R') (M join R')) = ((join R') (join MR')), for any R'.
+ which will be true for all ρ,γ,φ only when:
+ ((join R') (M join R')) = ((join R') (join MR')), for any R'.
- which will in turn be true just in case:
+ which will in turn be true when:
+ (ii') (join (M join)) = (join (join M))
- (ii') (join (M join)) = (join (join M))
- (iii.1) (unit F') <=< φ = φ
+ (iii.1) (unit G') <=< γ = γ
+ when γ is a natural transformation from some FG' to MG'
==>
- (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:
+ (unit G') is a transformation from G' to MG', so:
+ (unit G') <=< γ becomes: ((join G') (M unit G') γ)
- ((join F') (M unit F')) = the identity transformation, for any F'
+ substituting in (iii.1), we get:
+ ((join G') (M unit G') γ) = γ
- which will in turn be true just in case:
+ which will be true for all γ just in case:
+ ((join G') (M unit G')) = the identity transformation, for any G'
+ 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) γ = γ <=< (unit G)
+ when γ is a natural transformation from G to some MR'G
+ ==>
+ unit <=< γ becomes: ((join R'G) (M γ) unit)
+
+ substituting in (iii.2), we get:
+ γ = ((join R'G) (M γ) (unit G))
+
+ which by lemma 2, yields:
+ γ = ((join R'G) ((unit MR'G) γ)
+
+ which will be true for all γ just in case:
+ ((join R'G) (unit MR'G)) = the identity transformation, for any R'G
+
+ 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:
+
+ For all ρ, γ, φ in T,
+ where φ is a transformation from F to MF',
+ γ is a transformation from G to MG',
+ ρ is a transformation from R to MR',
+ and F'=G and G'=R:
- (i') ((join G') (M γ) φ) etc also in T
+ (i') ((join G') (M γ) φ) etc also in T
- (ii') (join (M join)) = (join (join M))
+ (ii') (join (M join)) = (join (join M))
(iii.1') (join (M unit)) = the identity transformation
- (iii.2')(join (unit M)) = the identity transformation
+ (iii.2') (join (unit M)) = the identity transformation
+
-7. The functional programming presentation of the monad laws
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+Getting to 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.