Continuous function (Topology) Theorem. Let $f:X\rightarrow{Y}$ be a continuous function. They are equivalent: $f(\overline{A})\subseteq\overline{f(A)}$ for all $A\subseteq X$ iff $f^{-1}(B^{\circ})\subseteq(f^{-1}(B))^{\circ}$ for all $B\subseteq Y$
Proof: Let $ B\subseteq{Y} $ be. Let's see that $f^{-1}(B^{\circ})\subseteq{(f^{-1} (B))^{\circ}} $. Note that $ B^{\circ} = Y\setminus{(\overline{Y\setminus {B}})} $. Then $$ f^{-1}(B^{\circ})= f^{- 1}(Y\setminus {(\overline {Y\setminus{B}})}) = X \setminus{f^{-1} (\overline {Y\setminus{B}})} $$ On the other hand, we have $$ (f^{- 1}(B))^{\circ} = X\setminus {\overline { (X \setminus {f^{- 1} (B)}})} = X \setminus {\overline {f ^ {- 1} (Y \setminus {B})}} $$ Now, let's see that $ \overline {f ^ {- 1} (Y \setminus {B})} \subseteq {f ^ {- 1} (\overline {Y \setminus {B}})} $. Let $ A = f ^ {- 1} (Y \setminus {B}) \subseteq {X} $ be. By hypothesis, we have $$ f (\overline {f ^ {- 1} (Y \setminus {B})}) \subseteq {\overline {f (f ^ {- 1} (Y \setminus {B})) }} \subseteq {\overline {Y \setminus {B}}} $$ For the above, we have $$ \overline {f ^ {- 1} (Y \setminus {B})} \subseteq {f ^ {- 1} (f (\overline {f ^ {- 1} (Y \setminus {B})}))} \subseteq {f ^ {- 1} (\overline {Y \setminus {B}})} $$ Therefore, $$ f ^ {- 1} (B ^ {\circ}) \subseteq {X \setminus {f ^ {- 1} (\overline {Y \setminus {B}})}} \subseteq { X \setminus {\overline {f ^ {- 1} (Y \setminus {B})}}} = {(f ^ {- 1} (B)) ^ {\circ}} $$ Reciprocally, Let $ A \subseteq {X} $ be. Let's see that $ f (\overline {A}) \subseteq {\overline {f (A)}} $. Note that $ \overline {A} = X \setminus {(X \setminus {A}) ^ {\circ}} $. Then $$ f (\overline {A}) = f (X \setminus {(X \setminus {A}) ^ {\circ}}) \subseteq {Y \setminus {f ((X \setminus {A}) ^ {\circ})}} $$ On the other hand, we have $$ \overline {f (A)} = Y \setminus {(Y \setminus {f (A)}) ^ {\circ}} $$ Now , let's see that $ (Y \setminus {f (A)}) ^ {\circ} \subseteq {f ((X \setminus {A}) ^ {\circ})} $. Let $ B = Y \setminus {f (A)} \subseteq {Y} $ be. By hypothesis, we have $$ f ^ {- 1} ((Y \setminus {f (A)}) ^ {\circ}) \subseteq {(f ^ {- 1} (Y \setminus {f (A)} )) ^ {\circ}} $$ Note that $ f ^ {- 1} (Y \setminus {f (A)}) = X \setminus {f ^ {- 1} (f (A))} \subseteq {X \setminus {A}} $. Then $$ f ^ {- 1} ((Y \setminus {f (A)}) ^ {\circ}) \subseteq {(X \setminus {A}) ^ {\circ}} $$ Thus, $ f (f ^ {- 1} ((Y \setminus {f (A)})^ {\circ})) \subseteq {f ((X \setminus {A})^ {\circ})} $
I don't know how to conclude...
 A: Referring to the link: Relation between interior and exterior.
Given that $f^{-1}$ exists, so $f$ is bijective and $f$ is given continuous so $f^{-1}$ is open.
Let, $A\subset X$ and $f(\overline A)\subset \overline{f(A)}$ holds.
Now, let $B\subset Y$ then $$B^\circ=Y-\overline{Y-B}\\ \implies f^{-1}(B^\circ)=f^{-1}(Y-\overline{Y-B})\\=X-f^{-1}(\overline{Y-B})\\ \subset X-\overline {f^{-1}(Y-B)}\\=X-\overline{X-f^{-1}(B)}\\=f^{-1}(B)^\circ$$
Now repeat the same process by setting $\overline A=X-(X-A)^\circ$.
A: The following two statements about a function $f:X \to Y$ between topological spaces are equivalent:


*

*$\forall A \subseteq X: f[\overline{A}] \subseteq \overline{f[A]}$.

*For all closed $C \subseteq Y$: $f^{-1}[C]$ is closed in $X$.


For 1 implies 2: let $C$ be closed in $Y$. Define $D = f^{-1}[C]$ and apply fact 1 to it: $$f[\overline{D}] \subseteq \overline{f[D]} = \overline{f[f^{-1}[C]]}\subseteq \overline{C} = C$$ as $C$ is closed. But this implies that $\overline{D} \subseteq f^{-1}[C] =D$ (all points of $\overline{D}$ map into $C$) and so $D = f^{-1}[C]$ is closed, as required.
For 2 implies 1: Let $A \subseteq X$ and note we have $$A \subseteq f^{-1}[f[A]] \subseteq f^{-1}[\overline{f[A]}]$$ and by 2. the latter set is closed. Hence 
$\overline{A} \subseteq f^{-1}[\overline{f[A]}]$ as well and this means that $f[\overline{A}]\subseteq \overline{f[A]}$ by the definitions of inverse images.

The following two statements about a function $f:X \to Y$ between topological spaces are equivalent:


*$\forall B \subseteq Y: f^{-1}[B^\circ] \subseteq f^{-1}[B]^\circ$.

*For all $O \subseteq Y$ open, $f^{-1}[O]$ is open in $X$.


For 3 implies 4: let $O$ be open in $Y$, then $O^\circ = O$ and by 3:
$$f^{-1}[O]= f^{-1}[O^\circ] \subseteq f^{-1}[O]^\circ \subseteq f^{-1}[O]$$
so $f^{-1}[O]^\circ = f^{-1}[O]$ and $f^{-1}[O]$ is open. 
For 4 implies 3: if $B \subseteq Y$ is any subset $B^\circ \subseteq B$ so
also $f^{-1}[B^\circ] \subseteq f^{-1}[B]$. The former is an open subset (by 4)  contained in $f^{-1}[B]$ and the interior of a set is its largest open subset so it follows that $f^{-1}[B^\circ] \subseteq f^{-1}[B]^\circ$, as required.

Now note that 2 and 4 are clearly equivalent as $f^{-1}[Y\setminus A]= X\setminus f^{-1}[A]$ for all subsets $B$ of $Y$, and open and closed sets are related via complements. 
So all conditions are equivalent to $f$ being continuous.
