This question is vaguely related to this MO question of mine.
Suppose $V$ is a (for simplicity, well-founded) model of $\mathsf{ZF+AD}$. Let $\mathbb{P}$ be a forcing notion in $V$, and let $G$ be $\mathbb{P}$-generic over $V$. My question is whether a certain "new cardinality" phenomenon can occur:
Could it be the case that $V[G]\models$ "There is no bijection between $\mathbb{R}^{V[G]}$ and any $x\in V$"?
One wrinkle here is that $V[G]$ need not satisfy $\mathsf{AD}$ anymore; indeed, preserving $\mathsf{AD}$ is quite hard. So $\mathsf{AD}$ doesn't give us a lot of "leverage" in the forcing extension that I can see.
I would be happy to replace $\mathsf{ZF+AD}$ with any stronger (consistent :P) theory; in particular, it might be easier to get a negative answer for something more "canonical" like $\mathsf{ZF+AD}+V=L(\mathbb{R})$.
Here are a couple quick comments:
On the one hand, to preempt worries about triviality note that we can have $V[G]\models$ "There is some $a$ which is not in bijection with any $b\in V$." For example, Monro showed 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.)
On the other hand, questions of this type are always trivial over $\mathsf{ZFC}$ for the following reason:
Forcing preserves choice, so $V[G]\models\mathsf{ZFC}$.
That means that for every name $\nu$ we have in $V[G]$ that $\nu[G]$ is in bijection with some ordinal $\alpha\in Ord^{V[G]}$.
But $Ord^V=Ord^{V[G]}$, so $V[G]$ sees a bijection between $\nu[G]$ and some set in $V$.
On the other other hand, the previous hand's argument is quite fragile and I don't see that it gives any insight into my question. Specifically, the third bulletpoint is crucial but has no relevant analogue for the $\mathsf{ZF+AD}$-setting that I can see.