21
$\begingroup$

What set-theoretic assumptions are necessary and sufficient to ensure the existence of a nontrivial (i.e. not isomorphic to the identity) endofunctor of the category Set which is logical (i.e. preserves finite limits and power objects—hence also finite colimits and exponentials)?

On the lower end, Andreas Blass proved ("Exact functors and measurable cardinals") that there exists a nontrivial exact endofunctor of Set (that is, preserving finite limits and colimits) iff there exists a measurable cardinal. Since logical functors are a fortiori exact, the existence of a measurable cardinal is a necessary condition. On the upper end, any nontrivial elementary embedding j:V→V surely induces a logical endofunctor of Set, so the existence of a Reinhardt cardinal is a sufficient condition. But can it be pinned down more precisely?

$\endgroup$
2
  • $\begingroup$ Here is a comment I learned from Nate Ackermann. Blass shows that any exact endofunctor of Set is a directed union of ultrapowers by countably-complete ultrafilters. Suppose our functor were just a single countably-complete ultrapower. Then it would also give an elementary embedding j: V → M for some transitive model M ⊆ V containing all the ordinals. But if j also preserves powersets, then M must also contain all the subsets of the ordinals, which implies it is all of V. Hence no single ultrapower can be a logical functor. $\endgroup$ Commented Nov 16, 2009 at 4:16
  • 1
    $\begingroup$ Strictly speaking, Joel is correct that I proved the result he cites, but it was proved earlier by Vera Trnkova (Comm. Math. Univ. Carolinae 12 (1971) 227-233). My paper on this (Pac. J. Math. 63 (1976) 335-346) was followed by a correction about this point (Pac. J. Math 73 (1977) 540). $\endgroup$ Commented Sep 29, 2010 at 13:26

1 Answer 1

9
$\begingroup$

There is a set theoretic axiom due to Paul Corazza called the Wholeness Axiom, which is stated in the language of ZFC augmented by a single unary function symbol $j$. The axiom expresses, as a scheme, that $j$ is a nontrivial elementary embedding from $V$ to $V$. That is, we have the elementary axiom scheme, expressing "for all $x$, $\phi(x)$ iff $\phi(j(x))$" and the nontriviality axiom, expressing "exists $x$, $j(x)\neq x$" and the critical point axiom, expressing "there is a least ordinal $\kappa$ such that $\kappa < j(\kappa)$".

Under this axiom, $j$ really is an elementary embedding from the universe $V$ to $V$, and so this presumably induces the kind of functor you want.

The point is that the large cardinal consistency strength of this axiom is weaker than a Reinhardt cardinal. In fact, it is strictly below an I3 cardinal.

But the situation with this axiom is not great, since the $j$ you get will not be a definable class in the usual sense of ZF. Also, you will not have the Replacement Axiom in the full language with $j$. So to make use of the axiom, you in effect give up a little of what you mean by the existence of such a $j$.

There is an ambiguity, isn't there, in the question when you ask about the existence of a proper class object. What kind of existence is desired? The question is not directly formalizable in ordinary set theory, since the question is itself a quantification over proper classes (although Kelly Morse set theory would accommodate this). Do you want a definable class? Do you want a class in the sense of Goedel-Bernays? Having a relaxed attitude about this allows the large cardinal consistency strength of the answer to come down.

$\endgroup$
2
  • $\begingroup$ Thanks! So certainly the consistency is below I3, since this Wholeness Axiom (where does that name come from?) will give a nontrivial logical endofunctor of Set. But it feels as though merely having a logical functor should be much weaker than any sort of elementary embedding. You're certainly right that there's an ambiguity in the question. But I'd be interested in answers to any of the possible ways of making it precise! $\endgroup$ Commented Nov 19, 2009 at 22:10
  • 2
    $\begingroup$ Was there any progress on the wholeness axiom since this rather old post? I ask so as to get the informal explanation, rather than digging into any paper. $\endgroup$
    – David Roberts
    Commented Sep 27, 2017 at 7:59

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .