*This question was previously [asked and bountied](https://math.stackexchange.com/questions/3769001/comparing-the-cardinalities-of-generic-mathbbrs) at MSE, unsuccessfully.*




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I'm currently interested in the behavior of cardinalities in generic extensions of models of $\mathsf{ZF+\neg AC}$, and especially of $\mathsf{ZF+AD}$. The following seems both natural and much simpler than various other questions I've asked on the topic:



> 
> Suppose $V$ is a transitive model of $\mathsf{ZF+AD}$. Let $c,d$ be mutually Cohen generic over $V$; is there in $V[c,d]$ a bijection between $\mathbb{R}^{V[c]}$ and $\mathbb{R}^{V[d]}$?
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> 
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"Obviously" the answer should be **yes**, but I don't see how to prove that. We have a set $\mathcal{X}\in V$ of names for reals such that every real in a Cohen extension is named by some element of $\mathcal{X}$, and each Cohen real $a$ induces an equivalence relation $\sim\_a$ on $\mathcal{X}$ as $\nu\sim\_a\mu\leftrightarrow\nu[a]=\mu[a]$. So really this question is asking for a bijection in $V[c,d]$ between $\mathcal{X}/\sim\_c$ and $\mathcal{X}/\sim\_d$. Intuitively this should exist since Cohen forcing is as homogeneous as one could hope; however, in the absence of choice in $V[c,d]$ I don't actually see how to build one.


I would also be very interested in a partial negative answer for $\mathsf{ZF}$-models, but the determinacy case is really my main point of focus.




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To preempt one natural attempt, note that [Cohen forcing kills determinacy](https://mathoverflow.net/a/325782/8133) so we can't use determinacy in $V[c,d]$ even though we have it in $V$. While at first glance this might appear to contradict (say) the generic absoluteness of the theory of $L(\mathbb{R})$ given large cardinals, there is no discrepancy since $(L(\mathbb{R}))^V[G]\not=(L(\mathbb{R}))^{V[G]}$ in general.



Comparing generic versions of $\mathbb{R}$
The answer seems to be no. Moreover: Suppose that every set of reals has the property of Baire. Let $\mathbb{C}$ be Cohen forcing and let $P$ be any wellorderable partial order. If $(c,d)$ is generic for $\mathbb{C} \times P$, then there is no injection in $V[c,d]$ from the Cantor space of $V[c]$ to any set in $V[d]$.


Here's a proof, which is a variation on the standard argument showing that if all sets of reals have the property of Baire, then there is no function choosing between complementary $\mathbb{E}\_{0}$ degrees. We will call members of the Cantor space reals. Suppose that $\tau$ is a $\mathbb{C} \times P$ name for a function mapping the reals of $V[g\_{left}]$ into $V[g\_{right}]$, where $g\_{left}$ and $g\_{right}$ are the $\mathbb{C} \times P$-names for the left and right coordinates of the generic. We will find a condition forcing the realization of $\tau$ not to be injective.


Let $N$ be the set of nice Cohen names $\sigma$ for elements of the Cantor space which are also functions with domain $\mathbb{C}$, where each value $\sigma(p)$ has the form $\check{t}$, for $t$ an element of $\mathbb{C}$ of the same length as $p$. That is, $N$ is the set of nice $\mathbb{C}$-names for reals such that each condition of $\mathbb{C}$ decides the initial segment of the realization of the name up to the length of the condition (and maybe more).


Let $\langle q\_n : n \in \omega \rangle$ be the natural enumeration of $\mathbb{C}$, where shorter sequences are listed before longer ones, and sequences of the same length are listed in lexicographic order (any enumeration in ordertype $\omega$ would work). Given $p \in \mathbb{C}$, let $n\_p$ be such that $p = q\_{n\_{p}}$. There is a natural bijection $b$ between the Cantor space and $N$, where we let $x$ in the Cantor space correspond to the set of pairs $(p, \check{t})$, where $t = \langle x(n\_{p \upharpoonright i}) : i \leq |p| \rangle$.


Applying the assumption that every subset of the Cantor space has the property of Baire, and the Kuratowski-Ulam theorem to deal with the wellorderable poset $P$, we get $s$ in $\mathbb{C}$ and $(p\_0, p\_1) \in \mathbb{C} \times P$ such that, for comeagerly many $x$ extending $s$, there is some $\mathbb{C}$-name $\rho$ with $(p\_0,p\_1)$ forcing $(b(x)\_{g\_{left}}, \rho\_{g\_{right}}) \in \tau$. That is, $(p\_0,p\_1)$ forces some $P$-name $\rho$ to represent via $g\_{right}$ the $\tau$-value for the $g\_{left}$-realization of $b(x)$. However, we can find $x$ and $x'$ extending s and in this comeager set such that $x$ and $x'$ disagree at exactly one value $n^\*$ with $q\_{n^\*}$ compatible with $p\_0$, and with this $q\_{n^\*}$ a proper extension of $p\_0$. So $p\_0$ doesn't decide whether or not the realizations of $b(x)$ and $b(x')$ will be the same.


Moreover, there exist $P$-names $\rho$ and $\rho'$ such that $(p\_0,p\_1)$ forces both


$(b(x)\_{g\_{left}}, \rho\_{g\_{right}}) \in \tau$


and


$(b(x')\_{g\_{left}}, \rho'\_{g\_{right}}) \in \tau$.


Let $p'\_1$ be a strengthening of $p\_1$ deciding whether or not $\rho$ and $\rho'$ have the same realization.


Since there are extensions of $p\_0$ forcing that the realizations of $b(x)$ and $b(x')$ will be the same, and $\tau$ is a name for a function, it must be that $p'\_1$ forces the realizations of $\rho$ and $\rho'$ to be the same. The condition $q\_{n^\*}$ however extends $p\_0$ and forces that the realizations of $b(x)$ and $b(x')$ will be different, and therefore $(q\_{n^{\*}}, p'\_1)$ forces that the realization of $\tau$ will not be an injection.


