Is this equivalent to continuity? I played around a little bit with the definition of continuity and I think I got the following relations that may be equivalent to continuity. Maybe there is somebody who can check this: 
$$ f(\overline{M}) \subset \overline{f(M)} \Leftrightarrow \overline{f^{-1}(A)} \subset f^{-1}(\overline{A}) \Leftrightarrow f^{-1}(A^°)\subset (f^{-1}(A))^°.$$ 
 A: It's not entirely clear how to interpret your
$$f(\overline{M}) \subset \overline{f(M)} \Leftrightarrow \overline{f^{-1}(A)} \subset f^{-1}(\overline{A}) \Leftrightarrow f^{-1}(A^°)\subset (f^{-1}(A))^°,$$
I think it is meant to say:
Let $X,Y$ be topological spaces, and $f\colon X \to Y$ a map. Then


*

*$f$ is continuous if and only if for all $M\subset X$ we have $f(\overline{M}) \subset \overline{f(M)}$.

*$f$ is continuous if and only if for all $A\subset Y$ we have $\overline{f^{-1}(A)} \subset f^{-1}(\overline{A})$.

*$f$ is continuous if and only if for all $A\subset Y$ we have $f^{-1}(A^°)\subset (f^{-1}(A))^°$.


In that case, these are indeed correct characterisations of continuous maps.
By definition, $f$ is continuous if and only if $f^{-1}(U)$ is open for all open $U\subset Y$. By taking complements, we have that $f$ is continuous if and only if $f^{-1}(F)$ is closed for every closed $F\subset Y$.
So if $f$ is continuous, we have $M \subset f^{-1}(\overline{f(M)})$, the latter being closed, hence $\overline{M} \subset f^{-1}(\overline{f(M)})$, or equivalently $f(\overline{M}) \subset \overline{f(M)}$ for every $M\subset X$. Conversely, if $f(\overline{M}) \subset \overline{f(M)}$ for all $M\subset X$, let $F\subset Y$ closed, and consider $M=f^{-1}(F)$. Then $f(\overline{f^{-1}(F)}) \subset \overline{f(f^{-1}(F))} \subset \overline{F} = F$, whence $\overline{f^{-1}(F)} \subset f^{-1}(F)$, and that means $f^{-1}(F)$ is closed. That holds for all closed $F\subset Y$, hence $f$ is continuous.
The arguments for the other cases are similar.
If $f$ is continuous then $f^{-1}(A)$ is a subset of the closed set $f^{-1}(\overline{A})$, so $\overline{f^{-1}(A)} \subset f^{-1}(\overline{A})$ for all $A\subset Y$. Conversely, if $\overline{f^{-1}(A)} \subset f^{-1}(\overline{A})$ for all $A\subset Y$, then for closed $F\subset Y$ we have $\overline{f^{-1}(F)} \subset f^{-1}(\overline{F}) = f^{-1}(F)$, hence $f^{-1}(F)$ is closed.
If $f$ is continuous, then $f^{-1}(A^°)$ is an open subset of $f^{-1}(A)$, hence $f^{-1}(A^°)\subset (f^{-1}(A))^°$ for all $A \subset Y$. Conversely, if $f^{-1}(A^°)\subset (f^{-1}(A))^°$ for all $A \subset Y$, then for open $U\subset Y$, we have $f^{-1}(U) = f^{-1}(U^°) \subset (f^{-1}(U))^°$, hence equality, and $f^{-1}(U)$ is open.
