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.

  • 8
    $\begingroup$ This question has been on MSE for less than a day - generally one should wait longer before moving to MO. $\endgroup$ – Noah Schweber Nov 22 '17 at 16:08
  • 1
    $\begingroup$ 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 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.


Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.