In this question, Andres Caicedo's answer describe a mapping $\pi^{-1}$that shows why $\bf AD_{2} \Rightarrow \bf AD_{\omega} $ .
We define a function $\pi$ by setting $\pi(x)=(n_0,n_1,\dots)\in\omega^\omega$. Say that the first $k$ moves of I are $1$s and the $k+1$-st is $0$. Then $n_0=k$. It is irrelevant what II has been moving so far. Say that, once I has played this first $0$, the next $j$ moves of II are $1$, and II's move after is $0$. Then $n_1=j$. It is irrelevant what I has been playing meanwhile. Say that, once II has played this $0$, the next $l$ moves of I are $1$s, and their $l+1$-st is $0$. Then $n_2=l$. Etc.
What can we say about this set-valued mapping $\pi^{-1}$? If $A$ is a Bore set, it seems to me $\pi^{-1}[A]$ should have a close ranking with $A$ in the Borel hierachy.
I tried the simplest case that $A$ is an uncountable closed set in $\omega^{\omega}$.Despite that $\pi^{-1}[A]$ is an uncountable union of closed sets that're neither open nor compact, I can't even tell whether it must be closed or not.