Signed-off-by: Jim Pryor <profjim@jimpryor.net>
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In category theory, the monad laws are usually stated in terms of `unit` and `join` instead of `unit` and `<=<`.
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In category theory, the monad laws are usually stated in terms of `unit` and `join` instead of `unit` and `<=<`.
P2. every element C1 of a category <b>C</b> has an identity morphism 1<sub>C1</sub> such that for every morphism f:C1→C2 in <b>C</b>: 1<sub>C2</sub> ∘ f = f = f ∘ 1<sub>C1</sub>.
P3. functors "preserve identity", that is for every element C1 in F's source category: F(1<sub>C1</sub>) = 1<sub>F(C1)</sub>.
P2. every element C1 of a category <b>C</b> has an identity morphism 1<sub>C1</sub> such that for every morphism f:C1→C2 in <b>C</b>: 1<sub>C2</sub> ∘ f = f = f ∘ 1<sub>C1</sub>.
P3. functors "preserve identity", that is for every element C1 in F's source category: F(1<sub>C1</sub>) = 1<sub>F(C1)</sub>.
Let's remind ourselves of some principles:
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 <b>C</b>: 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: <code>F(g ∘ f) = F(g) ∘ F(f)</code>
+
+* if <code>η</code> is a natural transformation from `F` to `G`, then for every <code>f:C1→C2</code> in `F` and `G`'s source category <b>C</b>: <code>η[C2] ∘ F(f) = G(f) ∘ η[C1]</code>.
Let's use the definitions of naturalness, and of composition of natural transformations, to establish two lemmas.
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 <b>C</b>, join[C1] will be a morphism from MM(C1) to M(C1). And for any morphism f:a→b in <b>C</b>:
+Recall that join is a natural transformation from the (composite) functor `MM` to `M`. So for elements `C1` in <b>C</b>, `join[C1]` will be a morphism from `MM(C1)` to `M(C1)`. And for any morphism <code>f:C1→C2</code> in <b>C</b>:
+
+<pre>
+ (1) join[C2] ∘ MM(f) = M(f) ∘ join[C1]
+</pre>
+
+Next, consider the composite transformation <code>((join MG') -v- (MM γ))</code>.
+
+* <code>γ</code> is a transformation from `G` to `MG'`, and assigns elements `C1` in <b>C</b> a morphism <code>γ\*: G(C1) → MG'(C1)</code>. <code>(MM γ)</code> is a transformation that instead assigns `C1` the morphism <code>MM(γ\*)</code>.
+
+* `(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:
+
+<pre>
+ (2) <code>((join MG') -v- (MM γ))</code> assigns to `C1` the morphism <code>join[MG'(C1)] ∘ MM(γ*)</code>.
+</pre>
-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 MG') -v- (MM γ)) assigns to C1 the morphism join[MG'(C1)] ∘ MM(γ*).
+Next, consider the composite transformation <code>((M γ) -v- (join G))</code>.
-Next, consider the composite transformation ((M γ) -v- (join G)).
(3) This assigns to C1 the morphism M(γ*) ∘ join[G(C1)].
(3) This assigns to C1 the morphism M(γ*) ∘ join[G(C1)].
-So for every element C1 of <b>C</b>:
+So for every element `C1` of <b>C</b>:
+
+<pre>
((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]
((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]
+</pre>
+
+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'.
+<pre>
+ ((join MG') -v- (MM γ)) = ((M γ) -v- (join G)), where γ is a transformation from G to MG'.
+</pre>
-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>:
+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 <code>f:a→b</code> in <b>C</b>:
+
+<pre>
(4) unit[b] ∘ f = M(f) ∘ unit[a]
(4) unit[b] ∘ f = M(f) ∘ unit[a]
+</pre>
+
+Next consider the composite transformation <code>((M γ) -v- (unit G))</code>:
+
+<pre>
+ (5) This assigns to C1 the morphism M(γ*) ∘ unit[G(C1)].
+</pre>
-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 <code>((unit MG') -v- γ)</code>.
-Next consider the composite transformation ((unit MG') -v- γ). (6) This assigns to C1 the morphism unit[MG'(C1)] ∘ γ*.
+<pre>
+ (6) This assigns to C1 the morphism unit[MG'(C1)] ∘ γ*.
+</pre>
So for every element C1 of <b>C</b>:
So for every element C1 of <b>C</b>:
((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]
((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]
+</pre>
+
+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'.
+<pre>
+ (((M γ) -v- (unit G)) = ((unit MG') -v- γ)), where γ is a transformation from G to MG'.
+</pre>
-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 <code>((join G') -v- (M γ) -v- φ)</code> for <code>γ <=< φ</code> 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
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
which will in turn be true just in case:
(iii.2') (join (unit M)) = the identity transformation
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:
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
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
(iii.1') (join (M unit)) = the identity transformation
(iii.2')(join (unit M)) = the identity transformation
(iii.1') (join (M unit)) = the identity transformation
(iii.2')(join (unit M)) = the identity transformation