John wrote:

> Maybe someone can expound a bit more on the notion of 'consistency' that I'm vaguely alluding to. All I mean is that a bunch of squares commute... but it's good to try to tease out the meaning of this, and express it in something resembling plain English.

Well, I'm not sure if this is exactly what you're referring to, but maybe it's close?

Generally speaking, homomorphisms identify a substructure of a target object that resembles (i.e. preserves the behavior of) the source object. This is why group homomorphisms identify subgroups of the codomain -- though you could also take the perspective that this is _why_ we define subgroups the way we do.

Functors are "just" homomorphisms between categories, so a functor identifies a subcategory of the codomain. Thus, given two functors from a common source category, we obtain two subcategories of the target category. It seems like a natural transformation ought to be a homomorphism between these two subcategories. Perhaps the strange part is that this homomorphism is "internal" to the containing category: it decomposes into morphisms between pairs of objects across the subcategories. I'm tempted to call this a "representation" of a functor between the two subcategories, but I know that word is heavily loaded in category theory.

On the other hand, there seems to be no guarantee that an object \$$F(X)\$$ must correspond with exactly one object \$$G(X)\$$. For instance, we can let \$$F\$$ project \$$C\$$ all the way down to the one-object one-morphism category, and let \$$G\$$ leave \$$C\$$ unchanged. If a natural transformation \$$\alpha\$$ exists between \$$F\$$ and \$$G\$$, it must necessarily associate a morphism \$$\alpha_X : \mathbf{1} \to X\$$ for every object in \$$C\$$. This is an extreme example, but if a situation even remotely like this is possible, then natural transformations _can't_ induce a functor, since it would need to associate multiple objects in \$$G(C)\$$ with a single object in \$$F(C)\$$.