I know I’ve written a whole ream of entries on this subject, but as I said before, it’s proved pretty darned fruitful.
Remember that the ultrafunction is defined thus:
u(f(a),b) = f(f(f(…f(a)…) (applied b times)
Now, consider the transformation t(g(a)) from the set of all functions into the set of all functions. Then, consider the ultrafunction of the transformation of the function:
u(t(f(a)),b) = t(f(t(f(t(f(…t(f(a))…) (as before, b repeats)
Would it be possible to “factor out” the transformation? Well, if that were possible, then we’d have:
t(f(f(f(…f(a)…) = t(f(t(f(t(f(…t(f(a))…)
But, since the transformations of two functions are equal only if they are taken of the same function, then:
f(f(f(…f(a)…) = f(t(f(t(f(t(…t(f(a))…)
Which will obviously only be true for a relatively small subset of the set of all functions.