*This question was [asked and bountied](https://math.stackexchange.com/questions/3767428/could-the-post-forcing-continuum-be-a-new-cardinality) at MSE, without success.*

My main question is whether, starting with a model of determinacy, a "generic $\mathbb{R}$" could be different in cardinality from any ground model set:

> Is there a well-founded $V\models\mathsf{ZF+AD}$, a forcing $\mathbb{P}\in V$, and a $G$ which is $\mathbb{P}$-generic over $V$ such that $V[G]\models$ "There is no bijection between $\mathbb{R}^{V[G]}$ and any set in $V$"?

In case the answer is yes, there is a natural follow-up question - whether the above can happen "canonically:"

> Are there $V\models\mathsf{ZF+AD}$, $\mathbb{P}\in V$, and $G,H$ mutually $\mathbb{P}$-generic over $V$ such that $V[G\times H]$ satisfies "$\mathbb{R}^{V[G]}$ and $\mathbb{R}^{V[H]}$ are in bijection with each other but are *not* in bijection with any set in $V$?"

If the answer to *this* question is yes, that would give a very surprising answer to [this old question of mine](https://mathoverflow.net/questions/192314/a-new-cardinality-living-in-every-forcing-extension). I suspect that the first question has an affirmative answer and strongly suspect that the second question has a negative answer, but I don't see how to prove either point.

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Here are a couple quick comments:

 - Since this is only interesting if we add reals, [$\mathsf{AD}$ will not be preserved](https://mathoverflow.net/a/325782/8133). So determinacy doesn't give us a lot of "leverage" in the forcing extension that I can see.

 - To preempt worries about triviality, we can have $V[G]\models$ "There is some $a$ which is not in bijection with any $b\in V$." For example, [Monro showed](https://www.cambridge.org/core/journals/journal-of-symbolic-logic/article/on-generic-extensions-without-the-axiom-of-choice1/D2CAA61A488CD4A3644F272CBF6DB969) that we can have an amorphous set in $V[G]$ even if there are no amorphous sets in $V$; since amorphousness is downwards-absolute, any amorphous set in such a $V[G]$ is not (in $V[G]$, anyways) in bijection with any set in $V$. *(This reference was pointed out to me by Asaf Karagila.)* 

 - That said, questions of this type are always trivial over $\mathsf{ZFC}$ since forcing preserves choice and adds no new ordinals.

 - But *that* said, the previous bulletpoint is quite fragile and I don't see that it gives any insight into my question - there's no obvious replacement for $Ord$ that I see here to serve the same role as something simultaneously "invariant" and "universal," especially since it would have to be "*generically* universal."