Proving an equality in continuity and closure on topological spaces So i have a continuous function $f : X \rightarrow Y$ where $X,Y$ are topological spaces. 
So i have to prove this equality:
$f$ is closed  if and only if for every $A$ subset in $X$ this is true: $f(Cl(A))=Cl(f(A))$
So one implication(from left to right) is that since f is closed it means that since for any $A$ subset in $X$, ($Cl(A)$ is the smallest closed set in X, that has A as a subset) , f(Cl(A)) is also a closed set.
I don't know how i can formulate this any further.
Also the second implication starts with premise: for every $A$ subset in $X$ this is true: $f(Cl(A))=Cl(f(A))$
How can i prove that f is closed with that.
Any help would be great. 
Thank you in advance.
 A: $\implies$ To show that $\overline{f(A)}=f(\overline A)$, you need to show that $f(\overline A)$ is closed (which you've done), and that for that for every closed set $K$ for which $f(A)\subset K$, we have $f(\overline A)\subset K$. 
Letting $K$ be a closed set for which $f(A)\subset K$, we then then have $A\subset f^{-1}(K)$. Since $f^{-1}(K)$ is closed, this implies $\overline A\subset f^{-1}(K)$, which then implies $f(\overline A)\subset K$.
$\impliedby$ For any closed set $K$, you must prove $f(K)$ is closed. This is equivalent to showing  $ \overline{f(K)}=f(K)$. Can you see how to use the assumption to prove that last equation?
A: Let $f: X \to Y$.
Then $f$ is continuous iff $$\forall A \subseteq X: f[\overline{A}] \subseteq \overline{f[A]}$$
And $f$ is closed iff
$$\forall A \subseteq X: \overline{f[A}] \subseteq f[\overline{A}]$$.
So $\forall A \subseteq X: \overline{f[A}] = f[\overline{A}]$ happens iff $f$ is both closed and continuous.
The first equivalence is classical, but I'll reprove it: 
If  $f$ is continuous and $A \subseteq X$ we have that $\overline{f[A]}$ is closed and so $A \subseteq f^{-1}[f[A]] \subseteq f^{-1}[\overline{f[A]}]$ and the right hand side is closed by continuity. Hence $\overline{A} \subseteq f^{-1}[\overline{f[A]}]$ which implies $f[\overline{A}]\ \subseteq \overline{f[A]}$, as required.
If we have the inclusion as a given we want to see continuity of $f$, so let $C$ be closed in $Y$ and define $A = f^{-1}[C]$, then by the assumption we know that $f[\overline{A}] \subseteq \overline{f[A]}$. But $f[A] \subseteq C$ so $\overline{f[A]} \subseteq C$ as well, as $C$ is closed. So $f[\overline{A}] \subseteq C$ and from this we conclude $\overline{A} \subseteq f^{-1}[C] = A$ which means that $A$ is closed. So the inverse image of a closed set is closed and $f$ is indeed continuous.
Now for the closed part: if $f$ is closed then for any $A \subseteq X$ we have $f[A] \subseteq f[\overline{A}]$ and the right hand side is closed, so $\overline{f[A]} \subseteq f[\overline{A}]$ as required.
If that second inclusion always holds let $C \subseteq X$ be closed and then we have, by the inclusion assumption for $A=C$, that $\overline{f[C]} \subseteq f[\overline{C}] = f[C]$ and so $f[C]$ is closed. Hence $f$ is a closed function.  
