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Suppose we want to prove that (classical!) $\mathsf{ZF}$ does not prove, say, “for every infinite set $A \subseteq \{0,1\}^{\mathbb{N}}$ there exists an injection $\mathbb{N} \to A$” (I take this example as about the simplest possible example of a refutation of Choice). The approach originating in Cohen's work proceeds roughly as follows (see Jech, Set Theory: Third Millennium Edition, example 15.52 for details):

Step 1. Choose a notion of forcing (here we will take partial functions $\mathbb{N}^2 \dashrightarrow \{0,1\}$ with finite domain, ordered by reverse extension), which in turn defines a complete Boolean algebra $\mathbf{B}$ (here the Boolean algebra of regular open sets of the product topology on $\{0,1\}^{\mathbb{N}^2}$), which in turn defines a Boolean-valued model $V^{\mathbf{B}}$ of $\mathsf{ZFC}$ (here, essentially, adding $\omega$ Cohen-generic elements of $\{0,1\}^{\mathbb{N}}$, precisely the $(a_n)$ where $a_n := a(n,—)$ is the slice of the generic function $a: \mathbb{N}^2 \to \{0,1\}$ inside $V^{\mathbf{B}}$ such that the truth value of $a(n,p)=i$ is given by the partial function $\{(n,p,i)\}$).

Step 2. Choose a group of permutations acting on the notion of forcing (and hence the Boolean algebra, and hence on the Boolean-valued model) of step 1, and a normal filter of subgroups of this group, and call “symmetric” the elements of the Boolean-valued model which are symmetric under a subgroup in the filter, and “hereditarily symmetric” those symmetric elements whose elements are themselves hereditarily symmetric: then the hereditarily symmetric elements will form a submodel $N$ of $V^{\mathbf{B}}$ which models $\mathsf{ZF}$ and, in general, not $\mathsf{ZFC}$. (Here we will take the group of permutations of $\mathbb{N}$ acting on finite partial functions $\mathbb{N}^2 \dashrightarrow \{0,1\}$ by permuting the first component of elements of $\mathbb{N}^2$, so effectively it permutes the $a_n$, and the normal filter of subgroups generated by fixators of finitely many elements.)

Now I would like to describe these two steps as operations on topoi. Let us restrict my goal to explaining why there is a Boolean topos in which the quoted sentence in the first paragraph of this question fails (i.e., forget about producing a model of $\mathsf{ZF}$, though a topos should probably be enough to get one of $\mathsf{Z}$).

I know how to describe step 1 in topos-theoretic terms: what we are doing is, essentially, taking the topos of $(\neg\neg)$-sheaves on the topos of sheaves on $\{0,1\}^{\mathbb{N}^2}$; more precisely, starting from a notion of forcing, we are taking its frame completion (here this will be the frame of open sets of $\{0,1\}^{\mathbb{N}^2}$), then taking the frame $\mathbf{B}$ of regular open subsets of that (i.e., the minimal dense sublocale, or “booleanization”) in order to keep classical logic, and we are dealing with the topos of sheaves on this locale. (The words “essentially” and “dealing with” are pushing some dust under the rug here: this question has more to say with the relation between $V^{\mathbf{B}}$ and the topos of sheaves on $\mathbf{B}$. But this is not what concerns me here, since as I just said I am willing to get less than a model of full $\mathsf{ZFC}$.)

What I want to understand, now, is step 2, i.e., how from the topos of sheaves on $\mathbf{B}$ (and a group of symmetries of it, and a normal filter of subgroups of that) we get a topos which “corresponds” to the symmetric submodel $N$ (putting the word “correspond” in quotes, because I'm not sure exactly what I should be asking for) and which refutes the quoted statement. So:

Question. How should we think of step 2 in the above approach, topos-theoretically? To what topos-theoretic construction does it correspond and what topos corresponding to $N$ does it construct? What functor (or geometric morphism) does this step 2 give between the topos of sheaves on $\mathbf{B}$ (corresponding to $V^{\mathbf{B}}$) and the topos corresponding to $N$?

I should note that there is an attempt to explain how to falsify Choice using a topos-theoretic construction in section VI.§4 of MacLane & Moerdijk's book Sheaves in Geometry and Logic, but it doesn't mention any connection with symmetric models, and I don't understand how it relates to the set-theoretic approach outlined above.

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  • $\begingroup$ Something like G-equivariant sheaves on the boolean algebra, for G a certain topological subgroup of the automorphism group of the ba? The construction of the rigid accessible pointed trees in that topos to get the model of ZF means that you get all that hereditary symmetry, I think. $\endgroup$
    – David Roberts
    Commented Sep 28 at 10:39

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Working topos-theoretically — or in a material set theory with atoms, e.g. ZFA (aka ZFU) — your step 1 is essentially unnecessary. Indeed, this is historically the background of Cohen’s construction: Frankel and Mostowski developed permutation models showing choice was independent of ZFA, using a collection of atoms with a group action; Cohen later showed how to adapt it to pure ZF by adding what you call “step 1”. Seen this way, the point of step 1 is necessary in ZF-like theories (with foundation, and without atoms) to obtain a suitable collection of “atom-like” objects inside the well-founded hierarchy, with a suitable group action.

So looking just at step 2: categorically, this corresponds to taking the topos of continuous $G$-sets for a given topological group $G$ — in the case of the original Fraenkel–Mostowski model, the group $\mathrm{Aut}(\mathbb{N})$. This is always a 2-valued topos, and fairly easy to calculate in. The Fraenkel–Mostowski failure of choice can be read (without much difficulty) as presenting an object in this topos for which choice fails — this is worked out in the Fourman 1980 paper Sheaf models for set theory (doi:10.1016/0022-4049(80)90096-1).

The combination with step 2 is analysed very nicely in the monograph Blass, Ščedrov, Freyd’s models for the independence of the axiom of choice, showing how a topos presentation of Cohen’s model arises as a pullback of the Fourman/Fränkel-Mostowski topos described above, and another sheaf topos previously presented by Freyd.

The “step 2” side, of symmetric models and their analogues, has been followed up in a lot of further topos-theoretic work. As far as I know, though, the “step 1” side — in particular, general techniques for converting ZFA-based independence results into ZF-compatible models — has not been pursued so much in the topos-theoretic setting, since it’s not so essential for obtaining independence results there.

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    $\begingroup$ 'Unnecessary' is relative to what you want to prove. I think you can phrase one of the consequence of Cohen's argument in more purely 'structural' or topos-theoretic terms like this: Cohen showed that there is a Boolean quotient of the free topos with NNO in which choice fails. In particular, choice can fail already in the power set of the reals. I don't think the Frankel-Mostowski method can establish this. $\endgroup$ Commented Sep 27 at 16:37
  • $\begingroup$ @JamesEHanson: I completely agree! $\endgroup$ Commented Sep 27 at 17:03
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    $\begingroup$ Andrej Scedrov and I wrote up most of this material in the first part of "Freyd's models for the independence of the axiom of choice" [Memoirs A.M.S. 79 (1989)], trying to make the main part of the paper accessible to both topos theorists and set theorists. Another source, from a more thoroughly topos point of view, is Freyd's "All topoi are localic"; the localic groupoids there correspond to the topological groups in this answer. $\endgroup$ Commented Sep 27 at 21:02
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    $\begingroup$ Too late to edit my previous comment so here's a new one. (1) The volume number 79 in my citation is from the A.M.S. website, but my records show 404 instead. I think it's the 404th Memoir but bundled with others in the 79th volume. (2) I used Andre Scedrov's old first name and his current last name. He used to be Andrej Scedov, with hacek accents on the S and c. People messed that up often enough to motivate him to drop the j and the accents. $\endgroup$ Commented Sep 27 at 21:12
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    $\begingroup$ Oops; accidentally hit "Return", so continuation of anecdote: he'd like to find a way to keep the j that he'd deleted from his name, perhaps as a "fake" middle initial, not standing for anything. A few minutes later, Michael Makkai mentioned that, when he westernized his name, he dropped a long, Hungarian middle name. It so happened that his middle name begins with J, so of course I suggested that he donate his unused middle name to Andre. $\endgroup$ Commented Sep 28 at 16:45

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