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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.

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    $\begingroup$ This question has been on MSE for less than a day - generally one should wait longer before moving to MO. $\endgroup$
    – Noah Schweber
    Commented Nov 22, 2017 at 16:08
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    $\begingroup$ 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 $\sigma_X$. $\endgroup$
    – Taras Banakh
    Commented Aug 16, 2018 at 7:39

1 Answer 1

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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.

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