Proof of a soft version of Moschovakis's lemma

The following fact, which I've heard being called "soft version of Moschovakis's lemma" (see top answer here) 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.

• This is Theorem 28.15 in Kanamori's book. – Andrés E. Caicedo Jun 14 '15 at 21:28
• @AndresCaicedo Can you confirm that Kanamori's book has the proof I am talking about? Note that I am looking for a specific proof of the fact. – Wojowu Jun 15 '15 at 5:58
• The proof is basically like you say, except that it is also combined with an induction of surjections onto $P(\beta)$ for $\beta<\alpha$. You play a game where player I tries to play a real coding an initial segment of the image of $X$ (using the earlier surjections), and player II tries to play a longer initial segment. It is then not difficult to show that if $X\neq Y$ then no winning strategy in $G(X)$ is winning in $G(Y)$. – Joel David Hamkins Jun 17 '15 at 20:40
• @AndresCaicedo I feel dumb for not doing this myself before, but I have just now checked that Kanamori's book indeed has the proof I was thinking about. Feel free to post this as the answer, and I'll award you the bounty. – Wojowu Jun 18 '15 at 13:00
• @AndresCaicedo If you wish to get a bounty for this question, now is your last chance, because bounty is ending. – Wojowu Jun 24 '15 at 20:14