Is it possible to put restrictions on the set $A$ such that $\overline{f^{-1}(A)} = f^{-1}(\bar{A})$ for a continuous $f$?

Question

Suppose $f: X \to Y$ is continuous where $X, Y$ are metric spaces. For any open set $A \subset Y$. It is clear by continuity \begin{align*} \overline{f^{-1}(A)} \subset f^{-1}(\bar{A}). \end{align*} Is it possible to just choose very 'nice' $A$ to make above relation be an equality? We exclude the case $A=Y$. By 'nice' $A$, I mean choose an $A$ with properties, for example, bounded, regular open, connected, etc, such that we have \begin{align*} \overline{f^{-1}(A)} = f^{-1}(\bar{A}). \end{align*} Or this depends on $f$?

Answer 1

Let $A$ be an open set in $Y$ such that $A$ is not closed. This implies that $A\not=\emptyset$ and $A\not=Y$. Then, fix $y\in \overline{A}\setminus A$ (which exists since $A$ is not closed). Consider the constant function $f:X\rightarrow Y$ given by $f(x)=y$ for all $x\in X$. Then $f^{-1}(A)=\emptyset$ and $f^{-1}(\overline{A})=X$.

Therefore the requested condition depends on the function $f$ (and is mostly independent of $X$ and $Y$ and choice of $A$).

Answer 2

Not really.

Theorem: Let $(Y,d_Y)$ be a metric space and $A\subset Y$ such that $A\neq\overline{A}$. Then there exists a metric space $X$ and a continuous function $f:X\to Y$ such that $\overline{f^{-1}(A)}\neq f^{-1}(\overline{A})$.

Proof: Let $x\in\overline{A}\setminus A$. Define $X:=A\cup\{\alpha\}$, where $\alpha\not\in Y$, with metric $d_X(a,b):=d_Y(a,b)$ and $d_X(a,\alpha):=d_X(\alpha,a):=1+d_Y(x,a)$ for $a,b\in A$ and $d(\alpha,\alpha):=0$. Then define $f:X\to Y$ by $f(a)=a$ for $a\in A$ and $f(\alpha)=x$. Then $\alpha\in f^{-1}(\overline{A})\setminus\overline{f^{-1}(A)}$.

Of course, do not forget to verify that $d_X$ is a metric and that $f$ is continuous.