+<pre>
+ (3) 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:
+ 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:
+
+<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 <code>f:a→b</code> in <b>C</b>:
+
+<pre>