From: Jim Pryor Date: Tue, 2 Nov 2010 13:27:10 +0000 (-0400) Subject: cat theory tweaks X-Git-Url: http://lambda.jimpryor.net/git/gitweb.cgi?p=lambda.git;a=commitdiff_plain;h=2f14eb20bd2428f346f8d5c8caa35b67eb043096;hp=24506613cb3b7c8e1cbe45a55f29de40d92bdbd3 cat theory tweaks Signed-off-by: Jim Pryor --- diff --git a/advanced_topics/monads_in_category_theory.mdwn b/advanced_topics/monads_in_category_theory.mdwn index b148fe6e..fe12655b 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 +