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 $B = f(A)$. One can also think of the inverse as a function defined only on the subset $f(A)$ of the space $B$.
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.
PS: If you can do a diagonal arrow, please edit. :-)
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 $\{y \in Y : f^{-1}(y) \subseteq C\}$ is closed in $Y$.
Remark: 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 $Y$.
Proof: For any subset $U \subseteq X$, let $S_U := \{y \in Y : f^{-1}(y) \subseteq X \setminus U\}$. It is an exercise in set theory to show that $f(U) = Y \setminus S_U$ for any subset $U \subseteq X$ (even when $f$ is not surjective). First assume that $f$ is open and let $C \subseteq X$ be closed. Using $U = X \setminus C$, we see that that $Y \setminus S_U = f(U)$ is open so that $S_U = \{y \in Y : f^{-1}(y) \subseteq C\}$ is closed in $Y$. Now assume that for all closed $C \subseteq X$ the set $S_{X \setminus C} = \{y \in Y : f^{-1}(y) \subseteq C\}$ is closed in $Y$. Let $U \subseteq X$ be open and define $C = X \setminus U$. Then our assumption gives us that $S_{U} = \{y \in Y : f^{-1}(y) \subseteq C = X \setminus U\}$ is closed in $Y$ so that $Y \setminus S_{U} = f(U)$ is open in $Y$. 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:
- the topology of $Y$ and the $Y$-values of $f$ (i.e. points in $f$'s codomain $Y$), and
- 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:
- 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. $R \subseteq \overline{S}$).
- Say that a point $y$ in a topological space is close to a subset $S$ if $\{ y \}$ is close to $S$ (i.e. if $y \in \overline{S}$).
- 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.
- 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.
- 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.
- If, however, the range of $f$ is an open subset of $Y$ then these two definitions of open map are equivalent.