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.

  • $\begingroup$ jdh.hamkins.org/… sounds relevant. Perhaps it's possible to show that $M$ and $L^M$ have the same category, at least when $M$ is countable. $\endgroup$
    – Asaf Karagila
    Oct 31, 2014 at 17:12
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    $\begingroup$ 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? $\endgroup$ Nov 1, 2014 at 4:23
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    $\begingroup$ They are models, so they live in some meta-theory in which that makes sense. $\endgroup$
    – Zhen Lin
    Nov 1, 2014 at 9:08
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    $\begingroup$ 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. $\endgroup$
    – Asaf Karagila
    Nov 1, 2014 at 21:16
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    $\begingroup$ Nice question, but this is surely MathOverflow material. $\endgroup$ Apr 28, 2015 at 5:56

1 Answer 1


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).

  • $\begingroup$ 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. $\endgroup$ Nov 10, 2016 at 18:28
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    $\begingroup$ 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? $\endgroup$ Mar 20, 2022 at 17:07

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