-That's it. (Well, perhaps we're cheating a bit, because γ <=< φ isn't fully defined on `T`, but only when `F` is a functor to `MF'` and `G` is a functor from `F'`. But wherever `<=<` is defined, the monoid laws are satisfied:
+That's it. Well, there may be a wrinkle here. I don't know whether the definition of a monoid requires the operation to be defined for every pair in its set. In the present case, <code>γ <=< φ</code> isn't fully defined on `T`, but only when <code>φ</code> is a transformation to some `MF'` and <code>γ</code> is a transformation from `F'`. But wherever `<=<` is defined, the monoid laws are satisfied:
+
+<pre>
+ (i) γ <=< φ is also in T
+
+ (ii) (ρ <=< γ) <=< φ = ρ <=< (γ <=< φ)