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Question. Let $M$ be a model of enough set theory. Then we can form a category $\mathbf{Set}_M$ whose objects are the elements of $M$ and whose morphisms are the functions in $M$. To what extent is $M$ determined by $\mathbf{Set}_M$ as a category up to equivalence?

For example, suppose $M$ and $N$ are models of ZF. Then are $\mathbf{Set}_M$ and $\mathbf{Set}_N$ equivalent as categories if and only if $M$ and $N$ are isomorphic?

I expect the answer will depend on exactly what we assume about $M$.

For instance, let $M$ be a model of ZFA and let $M'$ be the universe of pure sets in $M$. Then $M \cong M'$ if and only if $M$ has no atoms; but the inclusion $\mathbf{Set}_{M'} \hookrightarrow \mathbf{Set}_M$ is an equivalence as soon as $M$ satisfies the axiom "each set is in bijection with some pure set", which happens if e.g. $M$ satisfies the axiom of choice.

On the other hand, suppose $M$ is a transitive model of ZF. By transitive closure / Mostowski collapse, every set in $M$ is obtained from a "ZF-tree" in $M$, i.e. a set $T$ (in $M$) equipped with a well-founded extensional binary relation $E$ and a unique $E$-maximal element. The notion of ZF-tree is one that can be formulated in the internal language of a topos, so the collection of ZF-trees is recoverable from $\mathbf{Set}_M$ up to equivalence, and hence, $M$ is (exactly!) recoverable from $\mathbf{Set}_M$ up to equivalence.

Following Benedikt Löwe, a somewhat more sophisticated version of the above should work to recover well-founded models $M$ of ZFA with ($M$-)countably many atoms from $\mathbf{Set}_M$.

But what about, say:

  • Non-well-founded models of ZF(A)?
  • Weaker fragments of ZF, e.g. Mac Lane set theory?
  • Set theories where the category of sets is not a topos, e.g. NBG or NF(U)?

To keep the question from being too open-ended, let me say that I would be happy to know the answer just for (possibly non-well-founded) models of ZF.

Zhen Lin
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  • http://jdh.hamkins.org/every-model-embeds-into-own-constructible-universe/ sounds relevant. Perhaps it's possible to show that $M$ and $L^M$ have the same category, at least when $M$ is countable. – Asaf Karagila Oct 31 '14 at 17:12
  • Well, another way to phrase the question to ask which properties of $M$ can be detected in $\mathbf{Set}_M$. One of them is the axiom of choice, so we can't have $M$ and $L^M$ related in that way in general. The axiom of global choice is more subtle though, I think, and it would be interesting to know if global choice for $M$ can be detected in $\mathbf{Set}_M$. – Zhen Lin Oct 31 '14 at 17:23
  • I was pretty sure that Joel's work is about models of $\sf ZFC$. Of course that the axiom of choice is a way to differentiate between models. But perhaps it's not possible to go much further than that between models of choice. – Asaf Karagila Oct 31 '14 at 17:33
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    It's not clear to me that it makes sense a priori to directly compare versions of the category of sets coming from two models of set theory like that. The problem is that the two models don't agree on what a set is, so in what sense can they agree on what a functor between categories (which is in particular a map of sets on objects and a map of sets on morphisms) is so you can define what an equivalence between the two categories is? – Qiaochu Yuan Nov 01 '14 at 04:23
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    They are models, so they live in some meta-theory in which that makes sense. – Zhen Lin Nov 01 '14 at 09:08
  • Zhen Lin, it seems to me, a little bit, that two models with the same cardinality and same [internal] cardinals structure will necessarily have the same category. – Asaf Karagila Nov 01 '14 at 14:23
  • It sounds plausible. Certainly if they have equivalent categories then they must have isomorphic cardinal arithmetic, so it's a necessary condition. – Zhen Lin Nov 01 '14 at 21:10
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    Well, I'm also guessing that it's equivalent. Since you can really just notice that two objects are isomorphic in this category if and only if they have the same cardinality in $M$; and every cardinal has a proper class (of the size of the model) of sets with its cardinality, except $0$ which always has a singleton. So models of the same size and the same cardinal arithmetic will necessarily have an isomorphism between the categories (it might not be "natural" though). More to the point, [countable] models of $\sf ZFC$ will be determined by their height, which is the type of their cardinals. – Asaf Karagila Nov 01 '14 at 21:16
  • Therefore non-isomorphic models can generate the same category. For example $M$ and $L^M$ will have isomorphic categories, when $M$ satisfies choice. But if $M\neq L^M$ then they are not isomorphic (at least if they are well-founded). – Asaf Karagila Nov 01 '14 at 21:17
  • Again, it's certainly plausible, but there are details to be checked. After all, we also need the maps between objects to match up in a way that respects composition, and it's not obvious to me that having isomorphic cardinal arithmetic is a sufficient condition. – Zhen Lin Nov 01 '14 at 23:50
  • Oh yeah, you have to take care of the morphisms too... I think that at least in the countable case it might be doable by induction. – Asaf Karagila Nov 02 '14 at 04:12
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    Nice question, but this is surely MathOverflow material. – goblin GONE Apr 28 '15 at 05:56

2 Answers2

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This is not an answer, but it's just more convenient to write a big chunk of text here instead of in a comment. I mostly just have a few small comments (that might be flawed in one way or the other). First, $\textbf{Set}_{\mathcal M}\simeq\textbf{Set}_{\mathcal N}$ holds for any two countable models of ZFC $\mathcal M$ and $\mathcal N$. The reason for this is that for every $\alpha$ we can fix bijections $F_\alpha:\aleph_\alpha^{\mathcal M}\to\omega$ and $G_\alpha:\aleph_\alpha^{\mathcal N}\to\omega$, and then define a functor $e:\textbf{Set}_{\mathcal M}\to\textbf{Set}_{\mathcal N}$ as

$e(f:\aleph_\alpha^{\mathcal M}\to\aleph_\beta^{\mathcal M}):=F_\beta^{-1}\circ G_\beta\circ f\circ F_\alpha^{-1}\circ G_\alpha:\aleph_\alpha^{\mathcal N}\to\aleph_\beta^{\mathcal N}$,

which is both essentially surjective and fully faithful, so an equivalence.

But if either model is uncountable, then an equivalence $e$ would at least require that

  1. $\text{Card}^V(\kappa)=\text{Card}^V(e(\kappa))$ (since $\hom_{\mathcal M}(1,\kappa)\approx\hom_{\mathcal N}(1,e(\kappa))$)
  2. $\text{Card}^V((\kappa^\lambda)^{\mathcal M})=\text{Card}^V((e(\kappa)^{e(\lambda)})^{\mathcal N})$ (since $\hom_{\mathcal M}(\lambda,\kappa)\approx\hom_{\mathcal N}(e(\lambda),e(\kappa))$)

So, (1) at least requires that $|\mathcal M|=|\mathcal N|$. Also, $\textsf{Set}_V\simeq\textsf{Set}_L$ would at least imply $\textsf{GCH}$ by (2).

Dan Saattrup Nielsen
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  • I know this is about $\textsf{ZFC}$-models rather than $\textsf{ZF}$-models, as it seems the original question was intended to be about. My apologies. – Dan Saattrup Nielsen Nov 10 '16 at 18:28
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    Is this not assuming that $\mathcal{M}$ and $\mathcal{N}$ have the same ordinals? How does this work if $\mathcal{M}$ is well-founded and $\mathcal{N}$ is not? – James E Hanson Mar 20 '22 at 17:07
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(Note: This doesn't fully answer the question, which is very deep.)

I didn't see a reference in the comments to Michael Shulman's "Comparing material and structural set theories", which is both comprehensive and very well written. It seems very pertinent to the topic at hand.

https://arxiv.org/abs/1808.05204

I'm not sure whether it addresses the question as written, which is about whether isomorphism of models of a given material set theory implies equivalence of the categories of their sets. The paper is more about characterizing categories of sets based on the specific theory (axioms) of a material set theory, rather than a specific model. That having been said, it seems like the results of the paper would provide a framework for how to address the question in case the answer is negative using topos theory and/or categorical logic.

E.g. consider theorem 3.7 or proposition 6.4. One could seemingly use those results, combined with a characterization of when Heyting pretoposes (with certain additional properties) are equivalent.

So for theorem 3.7 (v), if we know the answer to "under what conditions are Heyting pretoposes with natural numbers objects equivalent?", then we could possibly answer this question for models of "core axioms of material set theory together with the axiom of infinity and exponentials".

Of course, this might just be "reducing" a difficult question to a more difficult question, but in case it's useful to future readers I wanted to throw this reference out here, and this is a little too long for a comment.

hasManyStupidQuestions
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