Thomas Streicher <streicher@mathematik.tu-darmstadt.de> Tuesday, March 18, 2008 6:09 pm Wrote:

In your Replacement axiom (p.48 of your "Philosophia" article) you psotultathe existence of a map f : S -> A such that S_x \cong x^*f for all x : 1->X. Can you prove that this f is unique up to isomorphism, i.e. that wellpointedness for maps entails wellpointedness for families?

Sure. It takes the axiom of choice of course, since without choice the result may be false (even two countably infinite families of countably infinite sets need not be isomorphic). It is the obvious argument by Zorn's lemma, which follows from choice: Given two families S-->A and S'-->A with corresponding fibers isomorphic, consider the set of all pairs <U,i> with U a subset of A, and i an isomorphism over U from the restriction of S to the restriction of S'. By Zorn at least one of these is maximal (for the obvious ordering by inclusion) so call it <U,i>. Since well-pointedness implies Boolean, U has a complement in A--which must be empty or else we could extend the isomorphism. best, Colin