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From $\sf L-S$ theorems we can have an external bijection $j$ from a set to its powerset. Obviously, $j$ cannot be fully internalized (used in all instances of Replacement). However, partial internalization is possible. The question here is about if $j$ can be partially internalized in the particular manner described below:

$\exists \alpha: \text{ limit} (\alpha) \land j: V_\alpha \to V_{\alpha +1} \land j \text{ is a bijection }$

$ \forall A \ \exists B: B=\{\{j(m),j(n)\} \mid \{n,m\} \in A\}$

$ \forall A \ \exists B: B=\{\{n,m\} \mid \{j(m),j(n)\} \in A\}$

In the other words, is it consistent to add to the language of $\sf ZF$ a partial function symbol $j$, and add [to axioms of $\sf ZF$] the above three sentences as axioms? Provided, of course, that Replacement doesn't use $j$.

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    $\begingroup$ Interesting question! For what it's worth, even the following version of this question (which I'll phrase model-theoretically for clarity) is unclear to me: suppose $M$ is a countable transitive model of $\mathsf{ZF}$ and $X,Y\in M$ are infinite. Must there be an external bijection $j:X\rightarrow Y$ such that for each $S\in M$ both $$_2\mathsf{Forth}_j(S):=\{\{j(m),j(n)\}:\{m,n\}\in S\}$$ and $$_2\mathsf{Back}_j(S):=\{\{m,n\}:\{j(m),j(n)\}\in S\}$$ are in $M$ as well? $\endgroup$
    – Noah Schweber
    Commented Oct 6, 2021 at 5:37
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    $\begingroup$ Is the answer known for the version where you just demand that for each set $S$, we have both $j[S]$ and $j^{-1}[S]$ are also sets? $\endgroup$
    – Farmer S
    Commented Oct 6, 2021 at 21:26
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    $\begingroup$ Actually I noticed now that the version I asked about is consistent; in fact if $M\models\mathrm{ZFC}$ then one can force it over $M$: Fix infinite sets $X,Y\in M$. Consider the following forcing $\mathbb{P}$, which attempts to build such bijection with this property... $\endgroup$
    – Farmer S
    Commented Oct 7, 2021 at 9:09
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    $\begingroup$ The conditions are pairs $(f,P,Q,F)$ where $f$ is a finite partial injective function $A\to Y$ where $A\subseteq X$, $P$ is a partition of $X$ into finitely many subsets, each of which is infinite, $Q$ is a partition of $Y$ into finitely many subsets, each of which is infinite, $F:P\to Q$ is a bijection, and $f$ respects $F$, i.e. $f(b)\in F(B)$ for each $b\in A$ and $B\in P$ with $b\in B$. And $(f',P',Q',F')\leq(f,P,Q,F)$ iff $f'$ extends $f$, $P'$ refines $P$, $Q'$ refines $Q$, and $F'$ refines $F$, in that if $B\in P$ and $C\in Q$ and $C\subseteq B$ then $F'(C)\subseteq F(B)$... $\endgroup$
    – Farmer S
    Commented Oct 7, 2021 at 9:10
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    $\begingroup$ If $G$ is $M$-generic then $j=\bigcup_{p\in G}f^p$ is a bijection $j:X\to Y$ as desired, i.e. $j[S]\in M$ for each $S\in\mathcal{P}(X)\cap M$, and $j^{-1}[S]\in M$ for each $S\in\mathcal{P}(Y)\cap M$. This is a straightforward density calculation. The inifiniteness of the sets $B\in P$ and $C\in Q$, and finiteness of the functions $f$, ensure that $j$ will end up being a bijection... $\endgroup$
    – Farmer S
    Commented Oct 7, 2021 at 9:11

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