The following fact, which I've heard being called "soft version of Moschovakis's lemma" (see top answer [here](https://mathoverflow.net/questions/129036/counterintuitive-consequences-of-the-axiom-of-determinacy)) is the following:



> 
> Under AD, if there is a surjection $\Bbb R\rightarrow\alpha$, then there is a surjection $\Bbb R\rightarrow\mathcal{P}(\alpha)$.
> 
> 
> 


I remember few days ago I have seen a really nice proof of this result, which directly used AD (and possibly DC, can't remember) and not some complex workaround using scales and things like that. However, when yesterday I wanted to show the proof to a friend of mine, I couldn't have find it anywhere, and I would be very thankful if anyone pointed out where I could have seen this proof.


The rough idea of the proof is the following:


1. Fix a surjection $f:\Bbb R\rightarrow\alpha$.
2. For each $X\subseteq\alpha$, using $f$, define a game $G(X)$ on $\omega$.
3. Show that, for any $X\neq Y$, no winning strategy for $G(X)$ is a winning strategy for $G(Y)$.
4. Thus, the function $F:\Bbb R\rightarrow\mathcal{P}(\alpha)$, defined as $F(r)=X$ if $r$ codes a winning strategy for $G(X)$, and $F(r)=\varnothing$ if $r$ codes no winning strategy, is well-defined.
5. By AD, every $G(X)$ has a winning strategy for some player, so $F$ is surjective.


Hope this helps to find the proof I meant. Thanks in advance.



Proof of a soft version of Moschovakis's lemma
The argument you are looking for is given in Kanamori's book, see Theorem 28.15. 


For the more nuanced version of the lemma, see section 7D in Moschovakis's descriptive set theory book (particularly 7.D.5-8), or section 3.1 in the Koellner-Woodin chapter of the Handbook.


