Name for a continuous surjection such that $\operatorname{cl}(f^{-1}(A)) = f^{-1}(A') \implies \operatorname{cl}(A) = A'$ Consider a continuous surjective map $f \colon (X, \tau) \to (X', \tau')$ satisfying $$\operatorname{cl}(f^{-1}(A)) = f^{-1}(A') \implies \operatorname{cl}(A) = A'$$ for all $A, A' \in \wp(X')$
I wish to know if such maps has been studied before and if so do they have a name. 

These are quotient maps.
Proof: We want to show the equivalence of these two conditions:


*

*For all $A,A'\subseteq X'$, $\operatorname{cl}(f^{-1}(A)) = f^{-1}(A') \implies \operatorname{cl} A = A'$.

*For all $A\subseteq X'$, $A$ is closed iff $f^{-1}(A)$ is closed.


(1 $\Rightarrow$ 2)
Suppose (1).  Let $A\subseteq X'$.  The implication
$$ \text{$A$ is closed} \implies \text{$f^{-1}(A)$ is closed} $$
holds because $f$ is continuous.  The reverse implication is a special case of (1):
$$ \operatorname{cl}(f^{-1}(A)) = f^{-1}(A)
\implies \operatorname{cl} A = A
$$
(2 $\Rightarrow$ 1)
Suppose (2).  Let $A,A'\subseteq X'$, and suppose $\operatorname{cl}(f^{-1}(A)) = f^{-1}(A')$.  Since $f$ is continuous,
$$ f(f^{-1}(A))
\subseteq f(\operatorname{cl}(f^{-1}(A)))
\subseteq \operatorname{cl}(f(f^{-1}(A)))
$$
Applying the hypothesis yields
$$ f(f^{-1}(A)) \subseteq f(f^{-1}(A')) \subseteq \operatorname{cl}(f(f^{-1}(A))) $$
Since $f$ is surjective, this means
$$ A \subseteq A' \subseteq \operatorname{cl}(A) $$
But by hypothesis $f^{-1}(A')$ is the closure of some set, so it is closed; by (2), $A'$ is closed; so this yields $A' = \operatorname{cl}(A)$.
