Characterization of open maps

Question

I'm looking for different but equivalent definitions of the concept of open map. So Let $X,Y$ be topological spaces and $f:X\longrightarrow Y$ a function, not assumed to be continuous. I conjectured the following equivalence:

1) $f$ is open, i.e. sends open subsetes of $X$ in open subsets of $Y$;

2) $f(x)\in\overline{B}\Rightarrow x\in\overline{f^{-1}(B)}$, for every $x\in X,B\subseteq Y$.

For 1)$\Rightarrow$ 2) i did: let $f(x)\in\overline{B}$, suppose $x\notin\overline{f^{-1}(B)}$. Hence i can find an open $U$ around $x$ not intersecting $f^{-1}(B)$ and mapping $U$ via $f$, i can also find an open of $Y$ containing $f(x)$ and not intersecting $\overline{B}$, which is against assumptions.

Could someone help me proving (or disproving) the reverse implication? Thanks

Answer 1

As remarked by user59671, 2) is equivalent to $f^{-1}(\overline{B}) \subset \overline{f^{-1}(B)}$ for all $B \subset Y$. Assume this is true. Let $U \subset X$ be open. Then $f^{-1}(\overline{Y \backslash f(U)}) \subset \overline{f^{-1}(Y \backslash f(U))} = \overline{X \backslash f^{-1}(f(U))} \subset \overline{X \backslash U} = X \backslash U$. Therefore $f^{-1}(\overline{Y \backslash f(U)}) \cap U = \emptyset$ which is equivalent to $\overline{Y \backslash f(U)} \cap f(U) = \emptyset$. This shows $\overline{Y \backslash f(U)} \subset Y \backslash f(U)$. Therefore $Y \backslash f(U)$ must be closed, i.e. $f(U)$ open.