I'm quite confident that there is the same number of models of $ZFC$ and of $ZF\neg C$, which means that there would exist a bijection between the former and the latter, by definition. However, I was wondering - can we somehow employ a forcing techniques (or some other tools) to find an explicit bijection between them? Far more relaxed version of this question is: is there always a definable bijection between these?


Thanks in advance.



Is there a "natural" bijection between models of $ZFC$ and $ZF\neg C$?
Well, the simple answer is that if there is a single model, there is a proper class of models, of each cardinality possible.


But let's instead restrict to transitive models. Then the answer is considerably different. It is consistent that all the transitive models of $\sf ZFC$ have the same height, say $\alpha$. Of course, in this case $\alpha<\omega\_1$ and we have that every transitive model is countable.


This means that there are at most $2^{\aleph\_0}$ models, and in fact there are at least $2^{\aleph\_0}$ models as well, since there is a perfect set of reals which are Cohen generic for the model.


But the work of Harvey Friedman showed that there are models of $\sf ZF$ of height $\alpha$, of cardinality $\beth\_\alpha$. This means that there are plenty of intermediate models in between, and that there are many many more transitive models of $\sf ZF$ than transitive models of $\sf ZFC$.



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> *Harvey Friedman*, [**Large models of countable height**](http://www.ams.org/mathscinet-getitem?mr=416903), *Trans. Amer. Math. Soc.* **201** (1975), 227--239.
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