X-Git-Url: http://lambda.jimpryor.net/git/gitweb.cgi?a=blobdiff_plain;f=advanced_topics%2Fmonads_in_category_theory.mdwn;h=4941e80b99875eb3f9179c704da83e63d4d3ae66;hb=4f87cdd50334c5d9dddbdad136fac4008d8ce6ff;hp=b148fe6e44c4a6f3ef2d8c31f0a2c5f979d94d53;hpb=24506613cb3b7c8e1cbe45a55f29de40d92bdbd3;p=lambda.git diff --git a/advanced_topics/monads_in_category_theory.mdwn b/advanced_topics/monads_in_category_theory.mdwn index b148fe6e..4941e80b 100644 --- a/advanced_topics/monads_in_category_theory.mdwn +++ b/advanced_topics/monads_in_category_theory.mdwn @@ -208,9 +208,13 @@ In other words, `<=<` is a binary operator that takes us from two members 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. -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: +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: 
 	    (i) γ <=< φ is also in T
@@ -257,60 +261,99 @@ 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)
+
+*	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:
+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:
-	(2) ((join MG') -v- (MM γ)) assigns to C1 the morphism join[MG'(C1)] ∘ MM(γ*).
+Next, consider the composite transformation ((join MG') -v- (MM γ)).
 
-Next, consider the composite transformation ((M γ) -v- (join G)).
+*	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:
+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 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:a→b in C:
+
+
 	(4) unit[b] ∘ f = M(f) ∘ unit[a]
+

+
+Next consider the composite transformation ((M γ) -v- (unit G)):
 
-Next consider the composite transformation ((M γ) -v- (unit G)). (5) This assigns to C1 the morphism M(γ*) ∘ unit[G(C1)].
+
+	(5) This assigns to C1 the morphism M(γ*) ∘ unit[G(C1)].
+

+
+Next consider the composite transformation ((unit MG') -v- γ).
 
-Next consider the composite transformation ((unit MG') -v- γ). (6) This assigns to C1 the morphism unit[MG'(C1)] ∘ 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) 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
@@ -379,10 +422,12 @@ 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:
 
+
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 @@ -392,6 +437,7 @@ 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 +