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=44ab78665ab4af653c1e91c0aabde0997ebb6274;hp=b148fe6e44c4a6f3ef2d8c31f0a2c5f979d94d53;hb=b27052e5da57c36b2cf9dc2f18d7dcf102a841f6;hpb=24506613cb3b7c8e1cbe45a55f29de40d92bdbd3
diff --git a/advanced_topics/monads_in_category_theory.mdwn b/advanced_topics/monads_in_category_theory.mdwn
index b148fe6e..44ab7866 100644
--- a/advanced_topics/monads_in_category_theory.mdwn
+++ b/advanced_topics/monads_in_category_theory.mdwn
@@ -257,60 +257,99 @@ 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[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)]`.
- (1) join[b] ∘ MM(f) = M(f) ∘ join[a]
+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)). - 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))
.
-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: -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 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))
:
+
++ (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)
.
-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: -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 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 +418,12 @@ 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 +433,7 @@ 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 +