# If non-empty player has a winning strategy in Banach-Mazur game BM(X), then it also has in BM(Y)?

Let $f:X\rightarrow Y$ be a continuous, open, surjection function and second player (non-empty) has a winning strategy (not important which one, say for simplicity stationery st.) in $BM(X)$. Then can we say the player has the same strategy in $BM(Y)$ ?

My attempts: 1) To say yes, $\sigma_Y(U)=f(\sigma_X(f^{-1}(U)))$ while $\sigma_i$ is stationery st. which depends on only last move of the opponent in $BM(i)$. But it didn't work.

2)To say no, Trying to find a continuous, open, surjection function from a scattered space to rational numbers.

Some definitions right here https://dantopology.wordpress.com/2012/06/08/the-banach-mazur-game/

At first I asked the question right here but no one answered yet https://math.stackexchange.com/questions/2531030/if-non-empty-player-has-a-winnig-strategy-in-banach-mazur-game-bmx-then-it-al

Thanks for any help.

• This question has been on MSE for less than a day - generally one should wait longer before moving to MO. – Noah Schweber Nov 22 '17 at 16:08
• It seems that for non-stationary strategies your first attempt should work after a suitable modification taking intersections of preimages with the open sets suggested by the strategy $\sigma_X$. – Taras Banakh Aug 16 '18 at 7:39

Topological spaces $$X$$ for which the second player (Non-empty) has a winning strategy in the Banach-Mazur game $$BM(X)$$ are called weakly $$\alpha$$-favorable by White and Choquet by Kechris.
According to White, a open continuous image of a weakly $$\alpha$$-favorable space is weakly $$\alpha$$-favorable.