Hi Peter - Am very sure I didn't see the fwd below - did you resend it to categories also? If so I haven't received that either which would be very strange... In any case, I can post the note below if you wish, and regret the delay... cheers, Bob ps I may not have written to say so and hope you assumed: I do plan to attend the CT Octoberfest (very much looking forward to it), but don't plan to give a talk. On Tue, 14 Sep 2010, Peter Selinger wrote:
Hi Bob,
I had submitted the below answer to Bob's question to the categories list yesterday. Seems that it was lost? It's not as "good" as Thorsten Palm's answer, but it's an alternative answer (and I sent it before Thorsten's answer arrived on the list). -- Peter
Forwarded message:
Subject: Re: categories: Canonical quotients To: pare@mathstat.dal.ca Date: Mon, 13 Sep 2010 11:09:19 -0300 (ADT) Cc: categories@mta.ca
Hi Bob,
how about this: using a sufficiently strong version of the axiom of choice, assume a well-ordering on the class of all sets (the "global well-ordering"). (Someone will be able to tell me whether this is strictly stronger than the ordinary axiom of choice, or equivalent to it. In any case, assuming set theory is consistent in the first place, and sufficiently large cardinals exist, there certainly exist models of set theory with such a property).
Given any equivalence relation ~ on a set X, say that an element x of X is a "canonical representative" of ~ if x is the (unique) least element in its equivalence class under the global well-ordering. Let X/~ be the set of canonical representatives, and define the map X -> X/~ to pick out the canonical representative of each class. I think this gives canonical quotients on (this version of) Set.
-- Peter
Robert Pare wrote:
Peter Freyd's and John Kennison's examples definitively settled Mike Barr's question about canonical subobjects that compose. But I had started thinking about it and had what I thought would be a nice example. The category of sets has canonical quotients (equivalence classes) but they don't compose. I think there is no choice that do, but so far I haven't been able to prove or disprove this. Anybody?
Bob