Let $a,b$ be sets, we write $a\leq^\ast b$ if either $a=\varnothing$ or there exists a surjection $f\colon b\to a$. With the axiom of choice this is a linear ordering equivalent to the usual ordering of cardinals $a\leq b$ if there is an injection from $a$ into $b$).


Without the axiom of choice this order need not be linear, or even a partial order. Indeed there are models where $a < b$ (so trivially $a\leq^\ast b$) but also $b\leq^\ast a$.


Let $\Theta(a)$ denote the least ordinal $\alpha$ such that $\alpha\nleq^\ast a$. This is a dual notion to $\aleph(a)$ which is the least ordinal $\alpha$ such that $\alpha\nleq a$. It is not very hard to see that both these ordinals are in fact cardinals, and that they exist for every set in $\mathrm{ZF}$. In the case where $a=\mathbb R$ we simply write $\Theta$ for $\Theta(\mathbb R)$.


It is consistent that for some set $a$ it holds $\aleph(a)<\Theta(a)$. For example if $V$ is a Solovay model then $\aleph(\mathbb R)=\aleph\_1\leq^\ast\mathbb R$.


We also know that if $a\leq^\ast b$ then $a\leq 2^b$, where the injection is simply the sending a point in $a$ to its fiber under a fixed surjection from $b$. It follows, if so, that $\Theta(a)\leq\aleph(2^a)$. Even in $\mathrm{ZFC}$ one can see that it is consistent that there is an equality, and it is consistent that there is no equality (e.g. $a=\omega$ and take a model in which $\frak c=\aleph\_1$ and another where it is is $\aleph\_2$).



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> **Question:** Suppose that $V$ is a model of $\mathrm{ZF+AD}$, we know that if $\alpha\leq^\ast\mathbb R$ then $2^\alpha\leq^\ast\mathbb R$, and therefore $\alpha^+\leq^\ast\mathbb R$. Is it consistent that $\Theta<\aleph(2^\mathbb R)$?
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> Equivalently, suppose $\alpha<2^\mathbb R$, can we find a surjection from $\mathbb R$ onto $\alpha$?
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$\Theta$ and the Hartogs of $2^\mathbb R$
Yes, it is consistent with AD. In $L(\mathbb{R})$, if AD holds then $\Theta$ injects into $\mathcal{P}(\mathbb{R})$, or equivalently, into $2^{\mathbb{R}}$. The OD sets of reals are Wadge-cofinal and $\Theta$ is regular, so we can define an injection $F : \Theta \to \mathcal{P}(\mathbb{R})$ by induction. Let $F(\alpha)$ be the $<\_{OD}$-least OD set of reals with Wadge rank greater than the Wadge ranks of all the sets of reals $F(\gamma)$, $\gamma < \alpha$.


Going further, we can determine the Hartogs number $\aleph(2^{\mathbb{R}})$ in all models of $AD + V=L(\mathcal{P}(\mathbb{R}))$. The proof splits into two cases corresponding to the length of the Solovay sequence being a successor ordinal or a limit ordinal.


If the length of the Solovay sequence is a successor ordinal, this means that there is a set of reals $A$ such that every set of reals is OD from $A$ and a real. In this case $\aleph(2^{\mathbb{R}})$ has the largest possible value, namely $\Theta(2^{\mathbb{R}})$.
Given a surjection $F : \mathcal{P}(\mathbb{R}) \to Z$ for any set $Z$ we can construct an injection $Z \to \mathcal{P}(\mathbb{R})$. Let $G(z)$ be the set of pairs $(x,y)$ of reals such that $y$ is in the least $OD\_{A,x}$ set of reals $B$ with $F(B) = z$ (if it exists.)


On the other hand, if the length of the Solovay sequence is a limit ordinal then $\aleph(2^{\mathbb{R}})$ is equal to $\Theta$; that is, to $\Theta(\mathbb{R})$.
We have $\aleph(2^{\mathbb{R}}) \ge \Theta$ on general grounds like you said in the question. But any function $F : \Theta \to \mathcal{P}(\mathbb{R})$ is OD from a set $A$ of reals because $V=L(\mathcal{P}(\mathbb{R}))$, so the range of $F$ consists of $OD\_A$ sets of reals. The Solovay sequence has limit length, so the range of $F$ cannot be all of $\mathcal{P}(\mathbb{R})$.


