+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)]`.
+
+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>