Signed-off-by: Jim Pryor <profjim@jimpryor.net>
* 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>
* 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>.
+* if <code>η</code> is a natural transformation from `G` to `H`, then for every <code>f:C1→C2</code> in `G` and `H`'s source category <b>C</b>: <code>η[C2] ∘ G(f) = H(f) ∘ η[C1]</code>.
+
+* <code>(η F)[E] = η[F(E)]</code>
+
+* <code>(K η)[E} = K(η[E])</code>
+
+* <code>((φ -v- η) F) = ((φ F) -v- (η F))</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 <code>f:C1→C2</code> 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>
<pre>
(1) join[C2] ∘ MM(f) = M(f) ∘ join[C1]
</pre>
-Next, consider the composite transformation <code>((join MG') -v- (MM γ))</code>.
+Next, let <code>γ</code> be a transformation from `G` to `MG'`, and
+ 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>.
+* <code>γ</code> 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)]`.
+* `(join MG')` is a transformation from `MM(MG')` to `M(MG')` that assigns `C1` the morphism `join[MG'(C1)]`.
(2) ((join MG') -v- (MM γ)) assigns to C1 the morphism join[MG'(C1)] ∘ MM(γ*).
</pre>
(2) ((join MG') -v- (MM γ)) assigns to C1 the morphism join[MG'(C1)] ∘ MM(γ*).
</pre>
-Next, consider the composite transformation <code>((M γ) -v- (join G))</code>.
- (3) This assigns to C1 the morphism M(γ*) ∘ join[G(C1)].
+ (3) Consider the composite transformation <code>((M γ) -v- (join G))</code>. This assigns to C1 the morphism M(γ*) ∘ join[G(C1)].
</pre>
So for every element `C1` of <b>C</b>:
<pre>
((join MG') -v- (MM γ))[C1], by (2) is:
</pre>
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:
+ 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>
M(γ*) ∘ join[G(C1)], which by 3 is:
((M γ) -v- (join G))[C1]
</pre>
So our **(lemma 1)** is:
<pre>
So our **(lemma 1)** is:
<pre>
- ((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 as we said γ is a natural 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 <code>f:a→b</code> 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:C1→C2</code> in <b>C</b>:
- (4) unit[b] ∘ f = M(f) ∘ unit[a]
+ (4) unit[C2] ∘ f = M(f) ∘ unit[C1]
-Next consider the composite transformation <code>((M γ) -v- (unit G))</code>:
- (5) This assigns to C1 the morphism M(γ*) ∘ unit[G(C1)].
+ (5) Consider the composite transformation ((M γ) -v- (unit G)). This assigns to C1 the morphism M(γ*) ∘ unit[G(C1)].
-Next consider the composite transformation <code>((unit MG') -v- γ)</code>.
- (6) This assigns to C1 the morphism unit[MG'(C1)] ∘ γ*.
+ (6) Consider the composite transformation ((unit MG') -v- γ). This assigns to C1 the morphism unit[MG'(C1)] ∘ γ*.
</pre>
So for every element C1 of <b>C</b>:
<pre>
((M γ) -v- (unit G))[C1], by (5) =
</pre>
So for every element C1 of <b>C</b>:
<pre>
((M γ) -v- (unit G))[C1], by (5) =
- M(γ*) ∘ unit[G(C1)], which by (4), with f=γ*: G(C1)→MG'(C1) is:
+ 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>
unit[MG'(C1)] ∘ γ*, which by (6) =
((unit MG') -v- γ)[C1]
</pre>
So our **(lemma 2)** is:
<pre>
So our **(lemma 2)** is:
<pre>
- (((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 as we said γ is a natural transformation from G to MG'.