# Characterization of open and closed maps in terms of filters

Understanding continuity and compactness in terms of filters has been very clarifying for me. Is there a convenient characterization of open and closed maps in terms of filters?

For instance, it seems to me that a map $f:X \to Y$ is open iff for each $x \in X$ and each neighborhood base $\mathcal{N}_x$ of $x$, $f(\mathcal{N_x})$ generates a coarser filter than the neighborhood filter at $f(x)$. Does this seem right?

• I've played with a half dozen ideas but don't see a good filter way to characterize closed maps. Annoying! It might just be too late at night. The two filters I played with are that of sets contained a closed containing $x$, which is dumb, and that of sets whose closure contains $x$, which is even dumber, since it's almost always the discrete filter. I'm also messing with taking the $\forall$ image of some filters instead of the $\exists$, with the only real result being that the $\forall$ of the neighborhood filter is coarse means (over $T_0$ spaces) that $f$ is open and injective. – Kevin Arlin Jun 9 '15 at 7:30
• @KevinCarlson I figured out by just dualizing that $f(\overline{S}) \supset \overline{f(S)} \iff f_\forall ((S^c)^\circ) \subset f_\forall (S^c)^\circ$...therefore relabeling $S^c$ by $S$, a characterization of closed maps is that $f_\forall(S^\circ) \subset f_\forall (S)^\circ$ for all $S \subset X$. That might get somewhere. – Eric Auld Jun 9 '15 at 20:25
• These are pretty ugly properties. Consider that a function $f:\mathbb{R}\to\mathbb{R}$ that maps every open interval onto $\mathbb{R}$ is open. Such functions are constructed by inserting a Cantor set in every element of a countable basis, and using an arbitrary bijection between that set and $\mathbb{R}$. Not much respect for topology there. – user147263 Jun 10 '15 at 22:54

One possibility for closed maps is $$f \text{ closed}\iff \bigg( f(\mathcal{F}) \to y \implies \mathcal{F} \to f^{-1}(\{y\}) \bigg),$$ or equivalently $$f \text{ closed}\iff \bigg( f(\mathcal{F}) \text{ clusters at } y \implies \mathcal{F} \text{ clusters at } f^{-1}(\{y\}) \bigg), \tag{1}$$ where the notion of a filter $\mathcal{F}$ clustering at a set $S$ means that every $F$ in $\mathcal{F}$ meets every neighborhood of $S$, and similarly for convergence. Note that we can allow $f^{-1}(\{y\})$ to be empty, if we consider "clustering at the empty set" to be a vacuous condition.
Note if $f$ happens to be proper, then $f^{-1}(\{y\})$ is compact, and "$\mathcal{F}$ clusters at $f^{-1}(\{y\})$" is equivalent to "$\mathcal{F}$ clusters at some point in $f^{-1}(\{y\})$". Interestingly, the condition "$f(\mathcal{F})$ clusters at $y \implies \mathcal{F}$ clusters at some $x \in f^{-1}(\{y\})$" is equivalent to $f$ being universally closed (see Bourbaki Topology section 10.2).
Another, more esoteric possibility for closed maps is the following: consider the functor $f_\forall :\mathcal{P}(X) \to \mathcal{P}(Y)$ given by $$f_\forall (S) := \{ y \in Y: f(x)=y \implies x \in S\},$$ and denote by $f_\exists$ the normal direct image. Then $f_\exists$ takes closed sets to closed sets iff $f_\forall$ takes open sets to open sets, since $f_\exists (S^c) = f_\forall(S)^c$. So assuming the point in the original post is correct, a map is closed iff $\langle f_\forall (\mathcal{N}_x) \rangle \subseteq \langle \mathcal{N}_{f(x)} \rangle$ for all $x$, where $\langle \mathcal{B} \rangle$ represents the filter generated by filter base $\mathcal{B}$.