Non smallness of the set of anafunctors without AC?
This is a supplementing answer to aws' answer. In this answer I sketch the a class-symmetric extension in which there is a proper class of Russell sets, and there is no set of Russell cardinals.
(Recall that a Russell set is a set which can be partitioned into countably many pairs, such that no infinite family of those pairs admits a choice function.)
We work in $\sf ZFC+GCH$, where $\sf GCH$ is assumed for simplicity. $\DeclareMathOperator{\dom}{dom} \DeclareMathOperator{\sym}{sym} \DeclareMathOperator{\fix}{fix} \newcommand{\PP}{\Bbb P} \newcommand{\cG}{\mathcal G} \newcommand{\HS}{\mathsf{HS}} \newcommand{\ZF}{\mathsf{ZF}} \newcommand{\cF}{\mathcal F} \newcommand{\id}{\operatorname{id}} \newcommand{\tup}[1]{\langle #1\rangle}$
If $\PP$ is a forcing, and $\{\dot x_i\mid i\in I\}$ is a set (or class) or $\PP$-names, we define $\{\dot x_i\mid i\in I\}^\bullet$ to be the $\PP$-name $\{\tup{1_\PP,\dot x_i}\mid i\in I\}$. This extends to ordered pairs, and naturally to sequences.
1 Localized failure
Fix a regular cardinal $\kappa$. Let $\PP_\kappa$ be the following presentation of adding a Cohen subset to $\kappa$. We mimic the construction of Cohen's second model.
A condition in $\PP_\kappa$ is a partial function $p\colon\omega\times 2\times\kappa\to2$, such that $|\dom p|<\kappa$.
We define the following names, for $(n,i,m)\in\omega\times2\times\omega$:
- $\dot x_{n,i,m}=\{\tup{p,\check\alpha}\mid p(n,i,m,\alpha)=1\}$,
- $\dot X_{n,i}=\{\dot x_{n,i,m}\mid m\in\omega\}^\bullet$, and
- $\dot S_n=\{\dot X_{n,i}\mid i<2\}^\bullet$.
Next we define the automorphism group $\cG_\kappa$, to be permutations $\pi$ of $\omega\times2\times\omega$ moving only finitely many points and satisfying that if $\pi(n,i,m)=(n',i',m')$, then $n=n'$; and for all $n$, either $i=i'$ or $i'=1-i$. It will be clearer to understand, once we see how these act on the names defined above:
If $p\in\PP_\kappa$, then $\pi p(\pi(n,i,m),\alpha)=p(n,i,m,\alpha)$. This is the standard way this action is defined. It extends to $\PP_\kappa$-names recursively: $\pi\dot x=\{\tup{\pi p,\pi\dot y}\mid\tup{p,\dot y}\in\dot x\}$.
$\pi\dot x_{n,i,m}=\dot x_{\pi(n,i,m)}$,
- $\pi\dot X_{n,i}=\dot X_{n,i'}$ if there are some $m$ and $m'$ such that $\pi(n,i,m)=\pi(n,i',m')$, and
- $\pi\dot S_n=\dot S_n$.
In other words, for every pair $A_n$, we decide whether or not we switch the elements in that pair, and separately, we may permute the elements of each set in the pair.
Finally, we take $\cF_\kappa$ to be the normal filter of subgroups generated by fixing finitely many points. Namely, if $E\subseteq\omega\times2\times\omega$ is finite, $\fix(E)$ is the group of all automorphisms which fix all the points in $E$ pointwise; and $\cF_\kappa$ is generated by $\fix(E)$ for $E$ finite.
We say that a name is symmetric if $\{\pi\mid \pi\dot x=\dot x\}\in\cF_\kappa$; and hereditarily symmetric if also every name which appears inside $\dot x$ is hereditarily symmetric.
We denote by $\HS$ the class of hereditarily symmetric names. If $G$ is a generic filter, then $\HS^G=\{\dot x^G\mid\dot x\in\HS\}$ is a model of $\ZF$. Let $M$ denote $\HS^G$, and we omit the dots to indicate the interpretations of the names, e.g. $\dot A_n^G$ is $A_n$ and so on.
Standard arguments show that:
- Each $\dot x_{n,i,m}$ is symmetric (and thus hereditarily symmetric),
- Each $\dot X_{n,i}$ is symmetric (and thus ...),
- The sequence $\tup{\dot A_n\mid n<\omega}^\bullet$ is hereditarily symmetric.
- The name $\dot A=\{\dot X_{n,i}\mid(n,i)\in\omega\times 2\}^\bullet$ is hereditarily symmetric.
So all these will then be in $M$. Moreover,
- For all $n$ and $i$, $X_{n,i}$, in $M$ is a Dedekind-finite set.
- $A$ is a Russell set in $M$. Specifically, $\{A_n\mid n<\omega\}$ is a partition witnessing that.
2 Global failure
Let $D$ be a class of regular cardinals. For example, all regular cardinals. For every $\kappa$ in $D$, note that $\PP_\kappa$ is $\kappa$-closed. Let $\PP$ be the Easton product of the $\PP_\kappa$'s. We then define an Easton support product of the groups and filters.
The class of hereditarily symmetric names of this system will include all the Russell sets we added by each $\PP_\kappa$. So it remains to show that there is no set of Russell sets which catch all of them up to equi-cardinality.
But then we get this quite easily. Any set of Russell sets will be added by some condition, and then some large enough $\kappa$ will add a new Russell set.
You can find details of a similar construction here:
Karagila, Asaf, Embedding orders into the cardinals with $\mathsf {DC}_{\kappa} $, Fundam. Math. 226, No. 2, 143-156 (2014). ZBL1341.03068.
I don't know much about class forcing, so I'll just sketch a Frankel Mostowski style model that I think gives a model of $\mathbf{ZFA}$ where local smallness fails.
We take the domain to be the discrete category on $\mathbb{N}$ and the codomain to be the group $\mathbb{Z}/2$, i.e. the category with one object, and a single nontrivial morphism, which we'll write as $g$, and the identity, $1$. We'll show that local smallness for this domain and codomain implies a certain set theoretic statement and then show that the statement is independent of $\mathbf{ZFA}$.
Say that a family of pairs is a countable family of sets $X = (X_n)_{n \in \mathbb{N}}$ such that for every $n$, $X_n$ has cardinality $2$. We say two families of pairs $X$ and $Y$ are isomorphic if there is a choice of bijections for each $n$, $\phi_n \colon X_n \rightarrow Y_n$. Note that since there exists a bijection $X_n \cong 2$ for every $n$, if we assume countable choice then we can choose such a bijection for each $n$ to show $X$ is isomorphic to the family of pairs constantly equal to 2. We'll show that local smallness implies there is a set of families of pairs $P$ that contains a representative of every isomorphism class.
Given a family of pairs $X$, we view it as an anafunctor from $\mathbb{N}$ to $\mathbb{Z}/2$ as follows. Let $Z$ be the category with objects $\coprod_{n \in \mathbb{N}} X_n$ and morphisms consisting of isomorphisms $f_n \colon x \rightarrow y$ for every $n$ and for $x \neq y \in X_n$ and otherwise just identity morphisms. We are forced to take $F(f_n) = 1_n$ and we take $G(f_n) := g$ (the nontrivial morphism in $\mathbb{Z}/2$). Then we have
Lemma Two families of pairs are isomorphic if and only if the corresponding anafunctors are equivalent.
Proof Suppose the ananfunctors corresponding to $X$ and $Y$ are equivalent. Then, by the characterisation of equivalence in the question, there is a natural transformation $\theta$ between the two functors $X \times_{\mathbb{N}} Y \rightarrow \mathbb{Z}/2$. Given $x \in X_n$ and $y \neq y' \in Y_n$, note that the naturality of $\theta$ implies that one of $\theta_{x, y}$ and $\theta_{x, y'}$ must be $g$ and the other 1. We define $\phi_n(x)$ to be the unique $y \in Y_n$ such that $\theta_{x, y} = 1$. This gives a choice of bijections. The converse is similar.
Lemma Any anafunctor from $\mathbb{N}$ to $\mathbb{Z}/2$ is equivalent to one given by a family of pairs.
Sketch proof Suppose we are given a category $Z$ and functors $F \colon Z \rightarrow \mathbb{N}$ and $G \colon Z \rightarrow \mathbb{Z}/2$ making an anafunctor. We define an equivalence relation $\sim$ on each fibre $F^{-1}(n)$. Note that since $F$ is full and faithful, for any $x, x' \in F^{-1}(n)$ there is a unique isomorphism $f \colon x \rightarrow x'$. We set $x \sim x'$ if $G(f) = 1$. This gives an equivalence relation and there are at most two equivalence classes. We define a family of pairs $X$ by taking $X_n$ to be the quotient if there are two equivalence classes and to be $2$ if there is only one equivalence class. These turn out to be equivalent anafunctors.
The above two lemmas together imply that local smallness implies the existence of a set of families of pairs containing a representative of every isomorphism class.
We now sketch a proof that this is independent of $\mathbf{ZFA}$. In the metatheory assume choice, and the existence of an inaccessible, $\kappa$. Let $G$ be the group $(\operatorname{Sym}(2))^\kappa$. Then $G$ acts on $\kappa \times 2$ by $g.(\alpha, i) := g(\alpha).i$, and this restricts to an action on $\alpha \times 2$ for every $\alpha < \kappa$.
We define $V_{\alpha, \beta}$ for $\alpha, \beta < \kappa$ by recursion on $\alpha$. Take $V_{\alpha + 1, \beta} := \mathcal{P}((\beta \times 2) \amalg V_{\alpha, \beta})$ and take unions at limit stages. We then set $V := \bigcup_{\alpha, \beta < \kappa} V_{\alpha, \beta}$. The action of $G$ on $\kappa \times 2$ lifts to an action on $V$ in the usual way, and we take $V^{\mathrm{fs}}$ to be the subset of elements with hereditary finite support relative to $\kappa \times 2$. I think $\mathbf{ZFA}$ should hold in this model.
Now to sketch a proof that there is no set of families of pairs containing a representative of every isomorphism class in the model. Suppose for a contradiction that $P$ is such a set. Let $\lambda < \kappa$ be such that $P \in V_{\lambda, \lambda}$. Define a family of pairs $X$ by taking $X_n$ to be the pair of atoms $\{(\lambda + n, 0), (\lambda + n, 1) \}$. By assumption there is a family of pairs $Y$ in $P$ and a function $\phi \colon \coprod_\mathbb{N} Y_n \rightarrow X_n$ giving a bijection for each $n$. Since $\phi$ has a finite support, there must be some $N$ such that $\lambda + N$ contains a support for $\phi$. Let $g \in G$ be the element that swaps $(\lambda + N, 0)$ and $(\lambda + N, 1)$ and fixes everything else. Then $g$ must fix both $\phi$, and everything in the transitive closure of $P$. So for $y \in Y_N$ we have $g.(\phi(N, y)) = \phi(N, y)$ giving a contradiction.