Intuition for an open mapping

Open mapping: were it invertible, its inverse would be continuous! :-)

Take an open mapping $$ f: A \to B. $$ If $b \in B$ is in the image of $f$, then, if $b_{\lambda}$ approaches $b$, $a_{\lambda}$ such that $f(a_{\lambda}) = b_{\lambda}$ "approaches" $f^{-1}(b)$.

Technically, it works as follows...

If $f(A) \subsetneq B$, then lets just trim $B$ down and assume $f$ is surjective. That is, $B = f(A)$.

Now, consider the quotient space $A/\sim$, where $x \sim y$ iff $f(x) = f(y)$. Embed it with the quotient topology. We have the commutative diagram: $$ \require{AMScd} \begin{CD} A @>{f}>> B \\ @V{\pi}VV @| \\ A/\sim @>{\tilde{f}}>> B \end{CD} $$

What happens here is that $\tilde{f}^{-1}$ is continuous.

In fact, if a set $\tilde{X} \subset A/\sim$ is open, then, $X = \pi^{-1}(\tilde{X})$ is open. Notice that $\tilde{X} = \pi(X)$ and let $\tilde{g} = \tilde{f}^{-1}$. Now, $$ \tilde{g}^{-1}(\tilde{X}) = \tilde{f}(\tilde{X}) = \tilde{f}(\pi(X)) $$ is open, and therefore, $\tilde{g}$ is continuous.

So, being an open mapping is a very easy and intuitive condition to warrant that $\tilde{f}^{-1}$ is continuous. As @MatthewK. points out in his comments, it is not a necessary condition.

PS: If you can do a diagonal arrow, please edit. :-)

OBS: Edited to comply with the very pertinent comments of @MatthewK.

PPS: @MatthewK. seems very upset because my answer is not good enough according to his standards. So, please, read his post as well!!! And also, please, consider giving him an up vote.

(Added Oct. 7, 2021) Both of the following characterizations are technically correct (i.e. true/provable) and I think give pretty good intuition. I included a characterization of "closed map" for contrast.

Open maps: A map $f : X\to Y$ is (strongly) open if and only if whenever $y_{\bullet}=(y_i)_{i\in I} \to y$ in $Y$ and $C$ is a closed subset of $X$ that eventually contains $f^{-1}(y_{\bullet})=(f^{-1}(y_i))_{i\in I},$ then $f^{-1}(y)\subseteq C.$

• $C$ eventually containing the net of sets $f^{-1}(y_{\bullet})$ means that there exists some index $i \in I$ such that $f^{-1}(y_j)\subseteq C$ for every $j\geq i.$
• This statement is in a way, slightly reminiscent of the following characterization of continuity: A function $f$ is continuous if and only if $f\Big(\overline{S}\Big)\subseteq\overline{f(S)}$ for every $S \subseteq X,$ where the right hand side can be restated as: whenever $x_{\bullet}\to x$ in $X$ and $S\subseteq X$ is an arbitrary subset that eventually contains $x_{\bullet},$ then $f(x)\in\overline{f(S)}.$

Closed maps: A map $f : X\to Y$ is (strongly) closed if and only if whenever $y_{\bullet}\to y$ in $Y$ then $f^{-1}(y_{\bullet})\to f^{-1}(y)$ in $X.$

• Here, $f^{-1}(y_{\bullet})\to f^{-1}(y)$ means the following: whenever $U$ is an open subset of $X$ such that $f^{-1}(y) \subseteq U,$ then $U$ eventually contains $f^{-1}(y_{\bullet}).$ Contrast this to the "open map" characterization above, where "eventually contains $f^{-1}(y_{\bullet})$" is a hypothesis (rather than conclusion) and $f^{-1}(y)\subseteq \text{ (set)}$ is the conclusion (rather than a hypothesis).
• A function $f$ is continuous if and only if whenever $x_{\bullet}\to x$ in $X$ then $f(x_{\bullet})\to f(x)$ in $Y.$ So with respect to this characterization, being a "closed map" is actually closer to "continuity in reverse" than is being an "open map."

Remark on continuity vs. openness (added Oct., 2021): The intuitive description of "openness" here is different from that of continuity. Is it reasonable to expect this or a sign that that something's afoul? After all, to get the definition of openness, all you do to continuity's definition is just reverse the direction the open sets are being sent (by replacing $f^{-1}$ with $f$). However, in category theory, reversing arrows usually results in drastically different objects. Compare: inverse limits vs. direct limits, injections vs. surjections, empty set vs. singleton sets, products vs. coproducts, weak topology (e.g. pointwise convergence) vs. final topology (e.g. quotient topology), unions vs. intersections, etc. So it actually might be reasonable to expect that intuition about "continuity" does not transfer over very well to "openness".

In my opinion, the best intuitive understanding of what it means for a map $f : X \to Y$ to be open actually follows from the following characterization of "open map" in terms of closed sets${}^{1}$ and fibers (i.e. sets of the form $f^{-1}(y)$ for some $y \in Y$) that I discovered a while ago (I'm not sure if I'm the first to discover this, but I couldn't find this result anywhere).

Throughout, $f : X \to Y$ will be any map (not necessarily continuous or surjective) between topological spaces and it will be called open if $U \subseteq X$ being open in $X$ implies $f(U)$ is open in $Y$ (WARNING: there exists a competing definition of "open map" that merely requires that $f(U)$ be open in the image of $f$, which weaker than the definition that we're using${}^{2}$).

Lemma 1: A (not necessarily continuous) map $f : X \to Y$ is open if and only if whenever $C \subseteq X$ is closed in $X$ then the set $f_C := \{y \in Y : f^{-1}(y) \subseteq C\}$ is closed in $Y$.

Remark 1: Said differently, $f : X \to Y$ is open if and only if for all closed subsets $C \subseteq X$, the image under $f$ of all fibers contained in $C$ (i.e. the image of $\bigcup\limits_{\substack{y \in Y,\\ f^{-1}(y) \subseteq C}} f^{-1}(y)$) will necessarily be closed in $Y$.

Remark 2: The proof below generalizes to show that a map $f : X \to Y$ is closed (meaning that it maps closed subsets of $X$ to closed subsets of $Y$) if and only if whenever $C \subseteq X$ is open in $X$ then the set $\{y \in Y : f^{-1}(y) \subseteq C\}$ is open in $Y$.

Proof: For any subset $S \subseteq X$, let $f_S = \{y \in Y : f^{-1}(y) \subseteq S\}$. It is an exercise in set theory to show that $f(S) = Y \setminus f_{X \setminus S}$ holds for any subset $S \subseteq X$ (even when $f$ is not surjective) where $X \setminus (X \setminus S) = S$ implies that $$f(X \setminus S) = Y \setminus f_S \quad \text{ for every } S \subseteq X.$$ In particular, $f(X \setminus S)$ is open (resp. closed) in $Y$ if and only if $f_S$ is closed (resp. open) in $Y$. Thus to prove the lemma, it suffices to show:

$f$ is open if and only if $f(X \setminus C)$ is open in $Y$ for every closed subset $C \subseteq X$.

Because an arbitrary subset is closed if and only if its complement is open, using $U := X \setminus C$ we can conclude that:

$f(X \setminus C)$ is open in $Y$ for every closed subset $C \subseteq X$ if and only if $f(U)$ is open in $Y$ for every open subset $U \subseteq X$

But the right hand side condition is the very definition of "$f$ is open", which completes the proof of lemma 1.

The proof that $f$ is closed if and only if $f_C$ is open in $Y$ for every open subset $C \subseteq X$ is obtained from the proof above by replacing every instance of the word "open" with the word "closed" while also replacing every instance of "closed" with "open". Q.E.D.

Recall that the set of all fibers of a map $f : X \to Y$ form a partition of $X$. Lemma 1 tells us that the property of $f$ being open expresses a relationship between:

1. the topology of $Y$ and the $Y$-values of $f$ (i.e. points in $f$'s codomain $Y$), and
2. the topology of $X$ and the fibers of f (rather than the points in $f$'s domain $X$).

We now introduce some non-standard (i.e. my own made up) definitions to make lemma 1's statement more intuition friendly:

1. Given two subsets $R$ and $S$ of a topological space, say that $R$ is close to $S$ if $R$ is contained in the closure of $S$ (i.e. $R \subseteq \overline{S}$).
2. Say that a point $y$ in a topological space is close to a subset $S$ if $\{ y \}$ is close to $S$ (i.e. if $y \in \overline{S}$).
3. If $\emptyset \neq F \subseteq X$ is a single fiber of $f$ then we will call the (unique) element $y \in Y$ such that $F = f^{-1}(y)$ the fiber's $f$-value or its $Y$-value.
4. Given $S \subseteq X$, by the (set of) $f$-fibers in $S$ we mean the union of all fibers of $f$ that are entirely contained in $S$.

Note that with these definitions, a subset is closed if and only if it contains every point/subset that is close to it.

If $f$ is open then intuitively, lemma 1 says that each closed subset $C$ of $X$ contains a unique maximal "fiber-sucking black-hole like" subset $B$ with the property that if the $Y$-value of any fiber is close to $f(B)$ then the entire fiber is "sucked into $B$" (i.e. is then necessarily contained in $B$). (To be clear the set $B$ here is the union of all fibers that are contained in $C$; note that this set $B$ has the property that if the $Y$-values of a fiber is close to $f(B)$ then the fiber is necessarily entirely contained in $B$).

Now, normally you'd expect that with a "fiber-sucking black-hole like" subset $B$ of $X$, whether or not some fiber gets sucked into $B$ would be determined by how near the fiber is to $B$ using $X$'s topology (i.e. using $X$'s notion of "nearness"). However, openness of $f$ means that this is determined instead by how close the fiber's $Y$-value is to the image $f(B)$ of the "black-hole" $B$ using $Y$'s topology (i.e. using $Y$'s notion of "nearness").

Now let's investigate what it intuitively means for a continuous map to be open:

Lemma $2$: If a continuous map $f : X \to Y$ is open then whenever any single point of any fiber is close to the $f$-fibers in a closed set $C \subseteq X$, then the entire fiber belongs to $C$. Furthermore, the set of fibers in any closed subset of $X$ is also closed.

proof: Assume that $f$ is open, let $C \subseteq X$ be closed, and let $S$ be the set of all fibers contained in $C$. Then $f\left( \overline{S} \right) \subseteq \overline{f(S)} = f(S)$ by lemma $1$. If $x \in \overline{S}$ then $y := f(x) \in f(S)$ and since by definition of $S$, we have $f^{-1}(f(S)) = S$, it follows that $f^{-1}(y) \subseteq f^{-1}(f(S)) = S$ and $x \in S$. This also shows that $S$ is closed. Q.E.D.

Thus, if a continuous map is open then for any closed $C \subseteq X$, if a point $x \in X$ is close to the set of $f$-fibers in $C$ then this set "sucks in" the entire fiber containing $x$ (i.e. $f^{-1}\left(f(x)\right)$ into $C$. This property, if lacking in map, prevents it from being an open mapping.

The intuition of what it means for a map to be open will hopefully become clear after imagining some random maps between Euclidean spaces and using the above lemmas $1$ and $2$ to determine whether or not they are open mappings.

1. You can actually essentially define the category of topological spaces using only the closure operator: see Kuratowski closure axioms (technically, this category is equivalent to the category of topological spaces). This justifies thinking of topological spaces in terms of "closeness" rather than open subsets.
2. If, however, the range of $f$ is an open subset of $Y$ then these two definitions of open map are equivalent.