<pre>
(φ -h- η)[C1] = L(η[C1]) ∘ ψ[G(C1)]
- = ψ[H(C1)] ∘ K(η[C1])
+ = ψ[H(C1)] ∘ K(η[C1])
</pre>
Horizontal composition is also associative, and has the same identity as vertical composition.
A **monad** is a structure consisting of an (endo)functor `M` from some category <b>C</b> to itself, along with some natural transformations, which we'll specify in a moment.
-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 <b>C</b>, `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.
+Let `T` be a set of natural transformations <code>φ</code>, 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 <b>C</b>, <code>φ</code> 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, <code>φ</code> is a transformation from functor `F` to `MF'`, <code>γ</code> 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 <b>C</b> to `M(1C)`. So it will assign to `C1` a morphism from `C1` to `M(C1)`.
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 `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:
+Let <code>φ</code> and <code>γ</code> 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 <code>γ</code>. 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')`, <code>(M γ)</code>, and <code>φ</code>, and abbreviate it as follows:
- q <=< p =def. ((join Q') -v- (M q) -v- p)
+<pre>
+ γ <=< φ =def. ((join G') -v- (M γ) -v- φ)
+</pre>
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 `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`.)
+In other words, `<=<` is a binary operator that takes us from two members <code>φ</code> and <code>γ</code> of `T` to a composite natural transformation. (In functional programming, at least, this is called the "Kleisli composition operator". Sometimes its written <code>φ >=> γ</code> where that's the same as <code>γ <=< φ</code>.)
-`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`.
+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, perhaps we're cheating a bit, because `q <=< p` isn't fully defined on `T`, but only when `P` is a functor to `MP'` and `Q` is a functor from `P'`. But wherever `<=<` is defined, the monoid laws are satisfied:
+That's it. (Well, perhaps we're cheating a bit, because 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) q <=< p is also in T
- (ii) (r <=< q) <=< p = r <=< (q <=< p)
- (iii.1) unit <=< p = p (here p has to be a natural transformation to M(1C))
- (iii.2) p = p <=< unit (here p has to be a natural transformation from 1C)
+ (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 `p` is a natural transformation from `P` to `M(1C)` and `q` is `(p Q')`, that is, a natural transformation from `PQ` to `MQ`, 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 `PG` to `MG`, then we can extend (iii.1) as follows:
- q = (p Q')
- = ((unit <=< p) Q')
- = ((join -v- (M unit) -v- p) Q')
- = (join Q') -v- ((M unit) Q') -v- (p Q')
- = (join Q') -v- (M (unit Q')) -v- q
+ 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 Q') <=< q
+ = (unit G') <=< γ
-where as we said `q` is a natural transformation from some `PQ'` to `MQ'`.
+where as we said γ is a natural transformation from some `PG'` to `MG'`.
-Similarly, if `p` is a natural transformation from `1C` to `MP'`, and `q` is `(p Q)`, that is, a natural transformation from `Q` to `MP'Q`, then we can extend (iii.2) as follows:
+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:
- q = (p Q)
- = ((p <=< unit) Q)
- = (((join P') -v- (M p) -v- unit) Q)
- = ((join P'Q) -v- ((M p) Q) -v- (unit Q))
- = ((join P'Q) -v- (M (p Q)) -v- (unit Q))
+ 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))
??
- = q <=< (unit Q)
+ = γ <=< (unit G)
-where as we said `q` is a natural transformation from `Q` to some `MP'Q`.
+where as we said γ is a natural transformation from `G` to some `MF'G`.
(1) join[b] ∘ MM(f) = M(f) ∘ join[a]
-Next, consider the composite transformation ((join MQ') -v- (MM q)).
- q is a transformation from Q to MQ', and assigns elements C1 in <b>C</b> a morphism q*: Q(C1) → MQ'(C1). (MM q) is a transformation that instead assigns C1 the morphism MM(q*).
- (join MQ') is a transformation from MMMQ' to MMQ' that assigns C1 the morphism join[MQ'(C1)].
+Next, consider the composite transformation ((join MG') -v- (MM γ)).
+ γ is a transformation from G to MG', and assigns elements C1 in <b>C</b> 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 MQ') -v- (MM q)) assigns to C1 the morphism join[MQ'(C1)] ∘ MM(q*).
+ (2) ((join MG') -v- (MM γ)) assigns to C1 the morphism join[MG'(C1)] ∘ MM(γ*).
-Next, consider the composite transformation ((M q) -v- (join Q)).
- (3) This assigns to C1 the morphism M(q*) ∘ join[Q(C1)].
+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 <b>C</b>:
- ((join MQ') -v- (MM q))[C1], by (2) is:
- join[MQ'(C1)] ∘ MM(q*), which by (1), with f=q*: Q(C1)→MQ'(C1) is:
- M(q*) ∘ join[Q(C1)], which by 3 is:
- ((M q) -v- (join Q))[C1]
+ ((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 MQ') -v- (MM q)) = ((M q) -v- (join Q)), where q is a transformation from Q to MQ'.
+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 <b>C</b>, unit[C1] will be a morphism from C1 to M(C1). And for any morphism f:a→b in <b>C</b>:
(4) unit[b] ∘ f = M(f) ∘ unit[a]
-Next consider the composite transformation ((M q) -v- (unit Q)). (5) This assigns to C1 the morphism M(q*) ∘ unit[Q(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 MQ') -v- q). (6) This assigns to C1 the morphism unit[MQ'(C1)] ∘ q*.
+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 <b>C</b>:
- ((M q) -v- (unit Q))[C1], by (5) =
- M(q*) ∘ unit[Q(C1)], which by (4), with f=q*: Q(C1)→MQ'(C1) is:
- unit[MQ'(C1)] ∘ q*, which by (6) =
- ((unit MQ') -v- q)[C1]
+ ((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 q) -v- (unit Q)) = ((unit MQ') -v- q)), where q is a transformation from Q to MQ'.
+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 Q') -v- (M q) -v- p) for q <=< p 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 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:
+ 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) q <=< p etc are also in T
+ (i) γ <=< φ etc are also in T
==>
- (i') ((join Q') (M q) p) etc are also in T
+ (i') ((join G') (M γ) φ) etc are also in T
- (ii) (r <=< q) <=< p = r <=< (q <=< p)
+ (ii) (ρ <=< γ) <=< φ = ρ <=< (γ <=< φ)
==>
- (r <=< q) is a transformation from Q to MR', so:
- (r <=< q) <=< p becomes: (join R') (M (r <=< q)) p
- which is: (join R') (M ((join R') (M r) q)) p
+ (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 r) q)) p) = ((join R') (M r) (join Q') (M q) p)
+ ((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 r) (M q) p) = ((join R') (M r) (join Q') (M q) p)
+ ((join R') (M join R') (MM ρ) (M γ) φ) = ((join R') (M ρ) (join G') (M γ) φ)
---------------
- which by lemma 1, with r a transformation from Q' to MR', yields:
+ which by lemma 1, with ρ a transformation from G' to MR', yields:
-----------------
- ((join R') (M join R') (MM r) (M q) p) = ((join R') (join MR') (MM r) (M q) p)
+ ((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 r,q,p just in case:
+ which will be true for all ρ,γ,φ just in case:
((join R') (M join R')) = ((join R') (join MR')), for any R'.
(ii') (join (M join)) = (join (join M))
- (iii.1) (unit P') <=< p = p
+ (iii.1) (unit F') <=< φ = φ
==>
- (unit P') is a transformation from P' to MP', so:
- (unit P') <=< p becomes: (join P') (M unit P') p
- which is: (join P') (M unit P') p
+ (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 P') (M unit P') p) = p
+ ((join F') (M unit F') φ) = φ
- which will be true for all p just in case:
+ which will be true for all φ just in case:
- ((join P') (M unit P')) = the identity transformation, for any P'
+ ((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) p = p <=< (unit P)
+ (iii.2) φ = φ <=< (unit F)
==>
- p is a transformation from P to MP', so:
- unit <=< p becomes: (join P') (M p) unit
+ φ is a transformation from F to MF', so:
+ unit <=< φ becomes: (join F') (M φ) unit
substituting in (iii.2), we get:
- p = ((join P') (M p) (unit P))
+ φ = ((join F') (M φ) (unit F))
--------------
which by lemma (2), yields:
------------
- p = ((join P') ((unit MP') p)
+ φ = ((join F') ((unit MF') φ)
- which will be true for all p just in case:
+ which will be true for all φ just in case:
- ((join P') (unit MP')) = the identity transformation, for any P'
+ ((join F') (unit MF')) = the identity transformation, for any F'
which will in turn be true just in case:
Collecting the results, our monad laws turn out in this format to be:
- 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:
+ 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 Q') (M q) p) etc also in T
+ (i') ((join G') (M γ) φ) etc also in T
(ii') (join (M join)) = (join (join M))
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 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:
+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 p = fun c → [(1,c), (2,c)]
+ let φ = fun c → [(1,c), (2,c)]
-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')].
+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 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.
+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 (p : C1 → M(C1')) to an argument of type 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.