# An equivalent definition of open functions

In any topological space:
How can I prove That $f$ is open if the pre-image of $\operatorname{cl}(A)$ is contained in the closure of pre-image of $A$,
Can some one give me a hint.
Actually, the relation is iff but I have proved one direction and I could not prove this direction.

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$\newcommand{\cl}{\operatorname{cl}}$Suppose that $f:X\to Y$ has the property that $f^{-1}[\cl A]\subseteq\cl f^{-1}[A]$ for all $A\subseteq Y$. Let $U$ be an open set in $X$, and suppose that $f[U]$ is not open in $Y$; then there is a $y\in f[U]$ such that every open nbhd of $y$ has non-empty intersection with $Y\setminus f[U]$. In other words, $y\in\cl(Y\setminus f[U]$). Choose $x\in U$ such that $y=f(x)$. Then $$x\in f^{-1}\big[\cl(Y\setminus f[U])\big]\subseteq\cl f^{-1}\big[Y\setminus f[U]\big]\;,$$

so $V\cap f^{-1}\big[Y\setminus f[U]\big]\ne\varnothing$ whenever $V$ is an open nbhd of $x$. This leads almost immediately to a contradiction, which I’ve spoiler-protected below.

In particular, $U\cap f^{-1}\big[Y\setminus f[U]\big]\ne\varnothing$, which is clearly impossible.

The relation $f^{-1}\left(\overline B\right)\subseteq\overline{f^{-1}(B)}$ can be dualized. Take the complements on both sides and turn around the inclusion, then make use of the facts that interior of complement is complement of closure and that complement and preimage commute:
\begin{eqnarray}
&f^{-1}\left(\overline B\right)\subseteq\overline{f^{-1}(B)}\\
\iff &X\setminus\overline{f^{-1}(B)}\subseteq X\setminus f^{-1}\left(\overline B\right)\\
\iff &\text{int }X\setminus f^{-1}(B)\subseteq f^{-1}\left(Y\setminus\overline B\right)\\
\iff &\text{int }f^{-1}(Y\setminus B)\subseteq f^{-1}(\text{int }Y\setminus B)&\ (*)
\end{eqnarray}

To show that $f$ is open, let $U$ be an open subset of $X$. Then
\begin{eqnarray}f(U)=f(\text{int }U)\subseteq f\left(\text{int }f^{-1}(f(U))\right)\subseteq\text{int }f(U)\subseteq f(U)\end{eqnarray} This implies that $f(U)$ is open. The first inclusion comes from replacing $U$ by $f^{-1}(f(U))$, for the second we replace $Y\setminus B$ by $f(U)$ in $(*)$.

I edited my answer based on Brian’s comment (We don’t really need continuity. I just assumed you were taking $f$ to be continuous):

Let $O$ be an open subset of $X$ where $f: X \rightarrow Y$. It suffices to show that $Y – f(O)$ is closed. Assume for contradiction that $Y – f(O)$ is not closed, i.e., that $Y – f(O)$ does not contain all its limit points. Let $y$ be a limit point of $Y – f(O)$ such that $y \in f(O)$. Let $a \in O$ such that $f(a) = y$. Since $y \in \overline{Y – f(O)}$, it follows from the inclusion $f^{-1}(\overline{Y – f(O)}) \subset \overline{f^{-1}(Y – f(O))}$, that $a \in \overline{f^{-1}(Y – f(O))}$. But, $f^{-1}(Y – f(O)) \subset X – O$ which is closed. What can you say about $\overline{f^{-1}(Y – f(O))}$ then? What does this say about $a$? Can you deduce a contradiction from our choice of $a$?

The contradiction shows that $Y – f(O)$ contains all its limit points, hence is closed.