Let $A, B \subset \omega^\omega$. The Wadge Game $(A, B)$ is played as followed: Player I plays $a_0 \in \omega$. Then Player II plays $b_0 \in \omega$. Then Player I choose $a_1 \in \omega$; afterward player II plays $b_1 \in \omega$. And so on. In the end player I produces a sequence $f = (a_0, a_1, a_2, ...)$ and player II produces a sequence $g = (b_0, b_1, b_2, ...)$.

We say that player II wins if $f \in A$ if and only if $g \in B$. Player I wins otherwise. The game $(A,B)$ is *determined* if one of the players has a winning strategy in this game.

Suppose that $\Lambda, \Gamma \subset \mathscr{P}(\omega^\omega)$. $(\Gamma, \Lambda)$ wadge determinacy is the statement that for all $A \in \Lambda$ and $B \in \Gamma$, the Wadge game $(A, B)$ is determined.

Define $\neg \Gamma = \{\omega^\omega - A : A \in \Gamma\}$. It is clear that $(\Gamma, \Lambda)$ determinacy implies $(\neg \Gamma, \neg\Lambda)$ determinacy. A winning strategy for player I (respectively player II) in $(A,B)$ is a winning strategy for player I (respectively player II) in the game $(\omega^\omega - A, \omega^\omega - B)$.

My question is does $(\Gamma, \Lambda)$-determinacy implies the determinacy of $(\neg \Gamma, \Lambda)$ or $(\Lambda, \neg\Gamma)$? This appears to be equivalent to whether the determinacy of $(\Gamma,\Lambda)$ implies the determinacy of $(\Lambda, \Gamma)$? Do these classes need to be closed under certain operations for this hold?

The cases I am most interested in is when $\Lambda, \Gamma$ are classes in the Borel Hierarchy, for example $(\Sigma_1^0, \Pi_1^0)$, i.e. (Open, Closed) and whether $(\Sigma_1^0, \Pi_1^0)$ Wadge determinacy is equivalent to $(\Sigma_1^0, \Sigma_1^0)$ Wadge determinacy.

Thanks for any insight anyone can provide.