9
$\begingroup$

The Mostowski collapse lemma (see here for a quick ref) is one of the key basic tools in the set-theory arsenal. I wonder if the collapse is natural, in the functorial sense.

More precisely, is this a reflection from the large category of well-founded models of ZF to the subcategory of transitive models?

My taste would say yes, but I have not thought it through ((apologies if the answer turns out to be trivial).

MOTIVATION: still thinking a bit about the MULTIVERSE Category. If the answer is affirmative, then it makes good sense to simply work with the subcategory of transitive models of ZF, which is certainly more manageable, and simpler to ponder.

ADDENDUM TO THE MOTIVATION: on a quick after-thought, I partially retract what I just said: there could still be some interest in considering the larger category of not necessarily well-founded models. In this case, perhaps someone could provide some speculations as to this larger cat and what can be found there (exotic models)

$\endgroup$
2
  • $\begingroup$ I've asked almost the same question here: mathoverflow.net/questions/37639/… $\endgroup$ Jul 23, 2012 at 9:27
  • $\begingroup$ yes, almost. Thanks for the link Martin. Take a look at Andreas' answer and subsequent comments, I am sure you will find it to your taste. That the mostowksy collpase is a reflection (actually an equivalence) is no surprise, but I confess I never thought of the other answer Andreas gave me, namely that taking the wekk-founded part is (likely) a coreflection into the well-founded universe (and therefore into the mostowsky cat). That, I think, needs further investigation, something cool lurks there. $\endgroup$ Jul 23, 2012 at 10:00

2 Answers 2

13
$\begingroup$

You didn't say what the morphisms in your categories are supposed to be; Trevor assumed you meant elementary embeddings, but you could also have meant mere embeddings, or something else. Nevertheless, unless you make a very strange choice of morphisms, the answer to your question is yes. Not only is the Mostowski collapse a reflection, it's an equivalence of categories. The transitive models constitute a skeleton of the category of all well-founded models; that is, every well-founded model $M$ is isomorphic to exactly one transitive model. Better yet, the isomorphism is uniquely determined by $M$. (All this information is part of the full statement of Mostowski's collapsing theorem.) So, from a category-theoretic point of view, it makes no difference whether you work with arbitrary well-founded models or with only the transitive models. Note, though, that in some situations, non-transitive well-founded models arise naturally, for example as elementary substructures of transitive ones, and in such cases your desire to work only with transitive models would require you to immediately apply the Mostowski collapse as soon as such a model enters your considerations.

$\endgroup$
4
  • $\begingroup$ Yes, Andreas, I was admittedly sloppy (see my comment to Trevor). Accepted without reservation! Now, can you tell me what is the "categorical status" of the not-mostowkian models? How does the category of well founded models (and its equivalent version, the transitive universe) sits in the category of models at large? I suspect there is no simple characterization, and if this is the case, looks like that in a theory of the multiverse(s) one cannot reduce oneself to well founded ones. Thoughts? $\endgroup$ Jul 21, 2012 at 18:58
  • $\begingroup$ In your comment to Trevor, you described the intended morphisms as "all maps between models, not necessarily elementary." I assume you intend "all maps" preserving at least the membership relation. (Otherwise, we might as well forget that the objects are models of ZF and regard them as unstructured sets.) Is that right? If so, do you also intend that the maps preserve non-membership? that they be one-to-one? $\endgroup$ Jul 21, 2012 at 19:14
  • $\begingroup$ Note that non-well-founded models of ZF can look very different from well-founded ones. For example, they can fail to satisfy Con(ZF). On the other hand, every model M has a well-founded part, its largest well-founded submodel containing all members (in the sense of M) of its members. With a suitable notion of morphism, this might provide a coreflection from arbitrary models to well-founded ones. $\endgroup$ Jul 21, 2012 at 19:17
  • $\begingroup$ 1) Yes, I meant all maps AS models of ZF, so they do respect membership. In other words, my cat is simply the cat of models of the theory ZF, as, say, the category of groups is MOD(Theory of groups). No, I do not demand anything else but being a map of models. 2) As for the second part, again great!!! I see, the "take my well founded part" provides a (candidate, but likely) co-reflection from the large multiverse to its well founded part. Very very interesting... then there is probably more to dig here (coalgebras). But for now I have my answers. Thanks again: your answers are gold $\endgroup$ Jul 21, 2012 at 19:33
5
$\begingroup$

If I understand the definition correctly at https://en.wikipedia.org/wiki/Reflective_subcategory, the question boils down to showing that every elementary embedding $f: B \to A$ between wellfounded models uniquely factors through the transitive collapse of $B$. This is true: It factors through the transitive collapse of $B$ because the transitive collapse map is an isomorphism. Uniqueness follows from the fact that isomorphic wellfounded models (in particular, $B$, its transitive collapse, and the range of $f$) are uniquely isomorphic.

$\endgroup$
1
  • $\begingroup$ Trevor, as usual my question was a bit sloppy. I mentioned the category of models of ZF without saying what the maps are. This category is simply the category of $L_ZF$ structures which happens to be models of ZF, so the maps are all maps between models, not necessarily elementary. However, I think your point remains true, so you have my vote (which will become an "accepted" if there are no objections). Now, I am curious to know about the not well founded models (ie the remainder of the large cat of all models). The candidate reflection is not there, but is there something weaker in place? $\endgroup$ Jul 21, 2012 at 18:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.