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).

  • $\begingroup$ I was mistaken took me time to see it. $\endgroup$ – JonesY Jul 19 '16 at 9:45
  • $\begingroup$ The functions $X \to Y$ satisfying $\operatorname{cl}(f^{-1}(B)) \supseteq f^{-1}(\operatorname{cl}(B))$ are the open functions, that is, $f(U)$ is open whenever $U$ is open. So functions satisfying $\operatorname{cl}(f^{-1}(B)) = f^{-1}(\operatorname{cl}(B))$ are open continuous maps. $\endgroup$ – Stefan Hamcke Jul 19 '16 at 13:22
  • $\begingroup$ @StefanHamcke JonesY initially said the same thing, but I still don't see how to prove this. However, I don't have a counter example either. $\endgroup$ – Thomas Klimpel Jul 19 '16 at 16:13
  • $\begingroup$ Take an open set $U$ in $X$ and a point $y\in f(U)$. Let $C=Y\setminus f(U)$. Then $f^{-1}(\text{cl}(C)) \subseteq \text{cl}(f^{-1}(C))$, and the later set is disjoint from $U$. Can you finish from here and show that $y\notin \text{cl}(C)$? $\endgroup$ – Stefan Hamcke Jul 19 '16 at 17:01
  • $\begingroup$ @StefanHamcke Yes, we get $y\not\in\operatorname{cl}(C)$. This proves that $\operatorname{cl}(f^{-1}(B))\supseteq f^{-1}(\operatorname{cl}(B))$ implies that $f$ is open. But what about the other direction? (i.e. you wrote "are the open functions", not "are open functions") Does every open continuous map satisfy the condition $\operatorname{cl}(f^{-1}(B)) = f^{-1}(\operatorname{cl}(B))$? $\endgroup$ – Thomas Klimpel Jul 19 '16 at 18:29

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.)


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