7
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ seems 2 is $f^{-1}(\overline{B})\subseteq \overline{f^{-1}(B)}$ for each $B\subseteq Y$. for 1-1 function seems correct. $\endgroup$ – user59671 May 16 '13 at 9:54
  • 4
    $\begingroup$ 2) and $\ \text{int }f^{-1}(B)\subseteq f^{-1}(\text{int }B)\ $ are equivalent to $f$ being open. These are just the reversed inclusion characterizing continuity. See math.stackexchange.com/questions/364767/… for proofs of this fact. $\endgroup$ – Stefan Hamcke May 16 '13 at 10:13
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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