# 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 stationary st.) in $BM(X)$. Then can we say the player has the same strategy in $BM(Y)$ ?

My attempt: 1) $\sigma_Y(U)=f(\sigma_X(f^{-1}(U)))$ while $\sigma_i$ is stationery st. in $BM(i)$

2) Trying to find a continuous, open, surjection function from a scattered space to rational numbers.

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

Thanks for any help.

• 2) There may be no such functions. For instance, there is no continuous surjection from a convergent sequence to rational sumbers, because the first space is compact whereas the second is not. – Alex Ravsky Jan 4 '18 at 9:07
• 1) Your attempt looks OK, but needs an accurate check. – Alex Ravsky Jan 4 '18 at 10:57
• @ Alex Ravsky 2) Most probably you are right, 1) For stattionary st. that attempt did not work or I couldn't. But eventually for winning st. I did. The question also right here mathoverflow.net/questions/286719/… but some people don't love it since the time of its appearing between MO and MSE is short. – S.S.Danyal Jan 8 '18 at 17:46