Generalization of $f(\overline{S}) \subset \overline{f(S)} \iff f$ continuous

Question

A common characterization of a continuous function $f: X \to Y$ is that for any $S \subset X$, $f(\overline{S}) \subset \overline{f(S)}$. Similarly, closed maps are such that $f(\overline{S}) \supset \overline{f(S)}$, and there are some in terms of the interior as well. Trying to extrapolate, I came up with the following implications: $$f(S^\circ) \subset f(S)^\circ \iff f\text{ open} \qquad f^{-1}(T^\circ) \subset f^{-1}(T)^\circ \iff f\text{ continuous}$$ $$f(S^\circ) \supset f(S)^\circ \quad (\text{relation?}) \qquad f^{-1}(T^\circ) \supset f^{-1}(T)^\circ \iff f\text{ open}$$ $$f(\overline{S}) \subset \overline{f(S)} \iff f\text{ continuous}\qquad f^{-1}(\overline{T}) \subset \overline{f^{-1}(T)} \iff f\text{ open}$$ $$f(\overline{S}) \supset \overline{f(S)} \iff f\text{ closed} \qquad f^{-1}(\overline{T}) \supset \overline{f^{-1}(T)} \iff f\text{ continuous}$$

First of all, I am interested to know whether these are correct. Secondly, is there a way that I can see these as less a collection of facts, and more as a unified whole? Why should these be the correct implications? Perhaps, as I ask in a related question, a characterization in terms of filters would help?

Edit: I fixed some of the implications after Stefan's and Alan's helpful input.

Answer 1

The links between these properties become a lot simpler once we examine their duals. Consider the functor $f_\forall :\mathcal{P}(X) \to \mathcal{P}(Y)$ given by $$f_\forall (S) := \{ y \in Y: f(x)=y \implies x \in S\}.$$ One could say it is dual to the normal direct image functor which we might call $f_\exists$. $f_\forall$ is right adjoint to the functor $f^{-1}$, and $f_\exists$ is left adjoint to $f^{-1}$. In other words, $$f(S) \subseteq T \iff S \subseteq f^{-1}(T), \qquad \text{and}$$ $$f^{-1}(T) \subseteq S \iff T \subseteq f_\forall (S).$$ Note that $f(S^c)=f_\forall (S)^c$. This implies that any statement about particular sets, containment, and images remains true if you replace the set with its complement, replace $f_\exists$ with $f_\forall$, and switch the order of containment. For example, $$f^{-1}(f(A)) \subseteq B \iff f^{-1}(f(A))^c \supseteq B^c \iff f^{-1}(f_\forall (A^c)) \supseteq B^c.$$ Also, we can dualize statements involving the interior or closure operators if we simply switch them. So for instance $$f(\overline{S}) \subseteq f_\forall(f^{-1}(T^c)^\circ) \iff f_\forall((S^c)^\circ) \supseteq f(\overline{f^{-1}(T)}).$$

Since the relations we are looking at are supposed to hold for all sets, we don't have to worry about whether we're considering the complement or not, and we can say that $$f(S^\circ) \subseteq f(S)^\circ \quad \text{for all }S \subseteq X \iff f_{\forall}(\overline{S}) \supseteq \overline{f_\forall(S)}\quad \text{for all }S \subseteq X. \tag{*}$$ Note finally that $f$ takes open sets to open sets iff $f_{\forall}$ takes closed sets to closed sets and vice versa.

Now it becomes clear that these conditions are linked up. Concerning the relations in the second column, $$f^{-1}(T^\circ) \supseteq f^{-1}(T)^\circ \quad \forall T \iff f^{-1}(\overline{T}) \subseteq \overline{f^{-1}(T)} \quad \forall T,\,\, \text{and} \tag{**}$$ $$f^{-1}(T^\circ) \subseteq f^{-1}(T)^\circ \quad \forall T \iff f^{-1}(\overline{T}) \supseteq \overline{f^{-1}(T)} \quad \forall T.$$ It so happens that $\bigg( f^{-1}(T^\circ) \supseteq f^{-1}(T)^\circ \,\, \forall T \iff f$ continuous$\bigg)$ is fairly obvious, because both conditions are in terms of the behavior of $f^{-1}$. On the other hand, $f$ being open is not handily related to the behavior of $f^{-1}$, and I would argue that $\bigg( f^{-1}(T^\circ) \supset f^{-1}(T)^\circ \iff f\text{ open} \bigg)$ is less obvious (my strategy would be to start by using the left adjointness of $f_\exists)$.

Now for the relations on the left hand side, note that $$ \bigg( f(S^\circ) \subset f(S)^\circ \,\, \forall S \iff f \text{ open} \bigg) \iff \bigg( f(\overline{S}) \supset \overline{f(S)}\,\, \forall S \iff f \text{ closed} \bigg), \tag{#}$$ by (*) and since $f$ is open iff $f_\forall$ is closed and vice versa. The reason we can obtain this equivalence is that the dual properties openness and closedness are formulated terms of the behavior of $f_\exists$, as are the conditions. Again it so happens that $f(S^\circ) \subset f(S)^\circ \,\, \forall S \iff f \text{ open}$ is fairly obvious because both are in terms of the behavior of $f_\exists$.

This leaves only conditions $f(S^\circ) \supset f(S)^\circ$ and $f(\overline{S}) \subset \overline{f(S)}$. Unfortunately, we we have both inconveniences at once: the conditions are not equivalent like in (**), and the property that adheres to the latter (continuity) is not in terms of the behavior of $f_\exists$ so it is not easy to dualize it and find a convenient formulation for the former, as we did in (#).

Answer 2

Concerning the mystery relation $f(S^\circ)\supseteq f(S)^\circ$, I've spent a while trying to relate this to continuity in some way, and all of the inequalities go the wrong way. So instead of looking for positive theorems, I'm looking at counterexamples. Consider the functions: $$f(x)=\left\{\begin{array}{11}x&x\in\mathbb{Q}\\-x&x\in\mathbb{R}\setminus\mathbb{Q}\end{array}\right.$$ and $$g(x)=\left\{\begin{array}{11}\vert x\vert&x\in\mathbb{Q}\\-\vert x\vert&x\in\mathbb{R}\setminus\mathbb{Q}.\end{array}\right.$$ The function $f$ fails the test, but $g$ passes the test. This is because the range of $g$ has no interior. But the range of $f$ equals $\mathbb{R}$. Then if you look at the set $f(S)^\circ$ for $S=(\mathbb{R}^-\setminus\mathbb{Q})\cup\mathbb{Q}^+$, you find $f(S^\circ)=f(\emptyset)=\emptyset$ is not a superset of $f(S)^\circ=\mathbb{R}^+$. But in the case of $g$, you find $g(S)^\circ=\emptyset$ for all $S\subseteq\mathbb{R}$. So the test passes trivially.

But the functions are really the same if you restrict them to one half of the real line. Thus the condition $f(S^\circ)\supseteq f(S)^\circ$ has a highly non-local character. It doesn't seem to have much relation to continuity. But maybe I'll change my mind if I look at it further. This relation should mean something.

PS. According to my investigation of this, if $f$ is injective and the range of $f$ is an open subset of $Y$, then $f$ is continuous if and only if $f(S^\circ)\supseteq f(S)^\circ$ for all $S\subseteq X$. I can't see how to remove the requirement for the range to be an open set. Maybe it can't be done. I'm fairly sure the injectivity is going to be very difficult to remove as a condition. The need for constraints on $f$ seems to make this condition different to the others.

I should also mention that the open range condition is only required in order to show that the relation implies continuity. You only need injectivity and continuity to get the relation $f(S^\circ)\supseteq f(S)^\circ$ for all $S\subseteq X$.

And one more comment. Define $h:[0,1]\to\mathbb{R}$ like this. $$h(x)=\left\{\begin{array}{11}x&x<1\\2&x=1\end{array}\right.$$ This function is injective, and it fairly clearly satisfies $h(S^\circ)\supseteq h(S)^\circ$ for all $S\subseteq X=[0,1]$. But it is very clearly not continuous. The range is also not an open subset of $\mathbb{R}$. So I think your 8 conditions are unfortunately lacking some symmetry for just this one condition.

And as just one more "last comment", I found the formulas in the question to resemble in some ways the sequent calculus. I wonder if there is some kind of connection there.

Answer 3

The comments section for this question is too big. So I'm putting my comment down here in the answers section.

It seems to me that $f$ is continuous if and only if $f^{-1}(\bar T)\supseteq\overline{f^{-1}(T)}$ for all subsets of $Y$. I checked this a couple of times and could find no error. Then I looked it up in Kelley, "General topology", 1955, page 86. He agrees with me. So you can replace that arrow with a double arrow.