The function $f: X \rightarrow Y$ is closed iff $\overline{f(A)} \subseteq f(\bar{A})$ The function $f: X \rightarrow Y$, is closed iff $\overline{f(A)} \subseteq f(\bar{A})$
My attempt:
(=>)
Suppose $f$ is a closed map.
Let $A$ be closed i.e $A = \overline{A}$, then $f(A) = f(\overline{A})$ and as $f$ is closed then $f(A) = \overline{f(A)} = f(\overline{A})$. As you see I ended in an equality not in a contention.
For (<=) Lets say $B$ is a set s.t $\overline{B} = f(\overline{A})$, then if $\overline{f(A)} \subseteq f(\bar{A})$ we have $f(A) \subseteq B$ and $A \subseteq f^{-1}(B)$. But I'm stuck here, don't know how to prove it is $A = \overline{A}$. Any tip you could give will be help full. Thanks!
 A: Suppose first that $f$ is closed. Let $A\subseteq X$ be arbitrary. Then as $A\subseteq\overline{A}$, $f(A)\subseteq f(\overline{A})$. As, $\overline{A}$ is closed and $f$ is closed, $f(\overline{A})$ is closed in $Y$. But $\overline{f(A)}$ is the intersection of all closed sets containing $f(A)$. In particular, $f(\overline{A})$ is in the intersection, so that $\overline{f(A)}\subseteq f(\overline{A})$.
Now suppose that $\overline{f(A)}\subseteq f(\overline{A})$ for all subsets $A$ of $X$. Let $B$ be a closed set in $X$. We want to show that $f(B)$ is closed in $Y$. However, $f(B) = f(\overline{B})$, and we have $\overline{f(B)}\subseteq f(\overline{B}) = f(B)$. But a set is always contained in its closure, so that $f(B) = \overline{f(B)}$ is closed, and we are finished.
A: If $f$ is closed then for all $A$ we have $f(\bar A)=Cl (f(\bar A))\supset Cl (f(A))$.... (We have $Cl (f(\bar A)) \supset Cl(f(A)$ because $f(\bar A)\supset f(A)$ because  $\bar A \supset A.)$
If $f$ is not closed then for some $B$ we have   $Cl(f(\bar B))\supsetneqq f(\bar B).$ So with $\bar B=A$ we have $\bar A =\bar B=A$, and $Cl (f(A)) \supsetneqq f(A)=f(\bar A)....$ So $ f(\bar A)\not \supset Cl(f(A)).$
