Continuous maps satisfying a natural stronger condition Denote the topological closure operator by $\operatorname{cl}(\ )$ (and the interior operator by $\operatorname{int}(\ )$). A map $f:X \to Y$ between topological spaces is continuous iff $\operatorname{cl}(f^{-1}(B)) \subset f^{-1}(\operatorname{cl}(B))$ for all $B \subset Y$.
Is there a name for a map which satisfies $\operatorname{cl}(f^{-1}(B)) = f^{-1}(\operatorname{cl}(B))$ for all $B \subset Y$? This condition is stronger than continuity. It is equivalent to the condition $\operatorname{int}(f^{-1}(B)) = f^{-1}(\operatorname{int}(B))$ for all $B \subset Y$. Hence this condition feels quite natural to me (and it seems like I will need it for what I try to do).
 A: Stefan Hamcke (and JonesY) is right, this condition gives exactly the open continuous maps. The trick is to notice $A \subseteq f^{-1}(f(A))$ and $f(f^{-1}(B)) \subseteq B$. So the following conditions are equivalent:


*

*$f$ is open

*$f(\operatorname{int}(A)) \subseteq \operatorname{int}(f(A))$ for all $A \subseteq X$

*$\operatorname{int}(f^{-1}(B)) \subseteq f^{-1}(\operatorname{int}(B))$ for all $B \subseteq Y$

*$\operatorname{cl}(f^{-1}(B)) \supseteq f^{-1}(\operatorname{cl}(B))$ for all $B \subseteq Y$


Similarly, the following conditions are equivalent:


*

*$f$ is continuous

*$f(\operatorname{cl}(A)) \subseteq \operatorname{cl}(f(A))$ for all $A \subseteq X$

*$\operatorname{cl}(f^{-1}(B)) \subseteq f^{-1}(\operatorname{cl}(B))$ for all $B \subseteq Y$

*$\operatorname{int}(f^{-1}(B)) \supseteq f^{-1}(\operatorname{int}(B))$ for all $B \subseteq Y$


Looks like the similarity between open and closed maps was misleading here.

Let's prove one of these equivalences:
$\operatorname{int}(f^{-1}(B)) \subseteq f^{-1}(f(\operatorname{int}(f^{-1}(B)))) \subseteq f^{-1}(\operatorname{int}(f(f^{-1}(B)))) 
\subseteq f^{-1}(\operatorname{int}(B))$
$f(\operatorname{int}(A)) \subseteq f(\operatorname{int}(f^{-1}(f(A)))) \subseteq f(f^{-1}(\operatorname{int}(f(A)))) \subseteq \operatorname{int}(f(A))$
Both lines of this proof used both $A \subseteq f^{-1}(f(A))$ and $f(f^{-1}(B)) \subseteq B$. For a partial function $p$, only $p(p^{-1}(B)) \subseteq B$ is true. Hence already the implications (probably) fail for partial functions. (I wanted to know/check this, hence I added that explicit proof.)
