What are the necessary and sufficient conditions on a function between two topological spaces such that it satisfies the following property? Let $(X,\tau_X)$ and $(Y,\tau_Y)$ be two topological spaces and $f:X\to Y$ is a function such that for all $A,B\subseteq X$, $$f(A)\subseteq \overline{f(B)} \implies A\subseteq \overline{B}$$where $\overline{f(B)}$ denotes the closure of $f(B)$ in $f(X)$ and $\overline{B}$ denotes the closure of $B$ in $X$. 
My Questions


*

*Does there exist any standard name for these type of functions?

*If $f$ is continuous, then what other conditions on $f$ are sufficient such that it satisfies the above property?

*Are the conditions also necessary?
My Attempt
I already tried asking the question here, however, as you can see I got no response.
 A: The condition
$\forall A,B\ \ (f(A)\subseteq \overline{f(B)}\Rightarrow A\subseteq \overline{B})$ is equivalent to
$$(*)\qquad\qquad \forall x, B\qquad f(x)\in \overline{f(B)}\Rightarrow x\in \overline{B}.$$
Under mild assumptions on $X,Y$ and continuity, this is equivalent to injectivity. 
Easy implication:
Suppose $X$ is $T_1$. If $(*)$ holds, then from $f(x)=f(y)\in \overline{f(y)}$ you get $x\in \overline{y}=y$ (because points are closed), hence $x=y$ and thus $f$ is injective.
Trickier one:
 Suppose $f$ is continuous and injective. Suppose $X$ is compact and $Y$ Hausdorff. Then if $f(x)\in \overline B$ there is a net $(b_i)$ such that $f(b_i)\to f(x)$. Since $X$ is compact, $b_i$ sub-converges to some $b\in B$, and by continuity $f(b_i)\to f(b)$. Since $Y$ is $T_2$ then you have uniqueness of the limit so $f(b)=f(x)$. By injectivity $b=x$, so $x\in \overline B$. 
If you dislike nets, you can (build a proof without them, or) work with first countable spaces and usual sequences.
I don't know if there is a general name for property $(*)$. Without compactness (but still with continuity and some separation axiom) it sounds to me as a kind of injectivity at infinity: you forbidd that $f(b_i)\to f(x)$ if $b_i$ does not converge to $x$.  
