-So for every element C1 of <b>C</b>:
- ((join MQ') -v- (MM q))[C1], by (2) is:
- join[MQ'(C1)] ∘ MM(q*), which by (1), with f=q*: Q(C1)→MQ'(C1) is:
- M(q*) ∘ join[Q(C1)], which by 3 is:
- ((M q) -v- (join Q))[C1]
+* `(join MG')` is a transformation from `MMMG'` to `MMG'` that assigns `C1` the morphism `join[MG'(C1)]`.
+
+Composing them:
+
+<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>.
+
+<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>