First, this question is strictly in the context of Reverse Mathematics where various set comprehension and various axiom of choice may not be available.
Question: Over $\text{RCA}_0$, if one has $\Sigma_1^0$-DET in Baire Space, does one have $\Pi_1^0$-DET in Baire Space? (Games in Baire Space are the only kind considered in Simpson's book).
For example, it is clear that $\Sigma_1^0$-DET in Cantor space implies $\Pi_1^0$-DET in Cantor Space. Given a $\Pi_1^0$ formula $\varphi(f)$. Consider the new $\Sigma_1^0$ game $\neg\varphi(0*f)$ and $\neg\varphi(1*f)$.
Case 1: If there exists an $i$ ($i = 0$ or $i = 1$) such that Player II has a winning strategy T' in $\neg\varphi(i*f)$, then Player I has a winning strategy $S$ for $\varphi(f)$ where $S$ plays $i$ on the first move and then follows T'.
Case 2 : If player I has a winning strategy S' for $\neg\varphi(0*f)$ and a winning strategy S'' for $\neg\varphi(1*f)$, then player II has a winning $T$ for $\varphi(f)$ which is described as follows, if Player I plays $0$, then player II follows S' and if Player I plays $1$, then Player II follows S''.
Note that the above does not require the axiom of choice because in the second case, I constructed $T$ by merely combining two strategies S' and S''.
This argument would not hold for games in Baire Space since the analog for case II would be that for all $i \in \mathbb{N}$, player I has a winning strategy S_i' for $\neg\varphi(i*f)$. However, now I can not combine all these strategy together to get a winning strategy for $\varphi(f)$.
It seems that I may have heard that this fact is true even for games in Baire Space. If this claim known to be true and how is it proved. Thanks for any help you can provide.