Intuition for an open mapping What is an intuitive picture of an open mapping?
The definition of an open mapping (a function which maps open sets to open sets) is simple sounding, but it's really not as easy to picture as the simple language would suggest. When I think of fields, for example, I immediately think of the rational numbers, the real numbers, the complex numbers, the integers modulo a prime, etc. When I think of continuous functions, I can picture common examples like polynomials, the absolute value function, etc., or nastier "artificial" examples like the Weierstrass function. What are the functions I should think of, nasty and nice, when I think of open mappings?
 A: The characterizations below should be useful. The notation and definitions mean what you'd expect them to mean and so are given after the theorems. Throughout, $f : X\to Y$ is any map (not assumed to be continuous or surjective) and by default $y_{\bullet}=(y_i)_{i\in I}$ denotes a net of points in $Y.$
Closed maps: $f$ is closed $\iff$ whenever $y_{\bullet}\to y$ in $Y$ then $f^{-1}(y_{\bullet})\to f^{-1}(y)$ in $X.$

*

*In comparison, $f$ is continuous $\iff$ whenever $x_{\bullet} \to x$ in $X$ then $f\left(x_{\bullet}\right) \to f(x)$ in $Y.$
Open maps 1: $f$ is open $\iff$ whenever $x \in X$ and $y_{\bullet}$ is a net that clusters at $f(x)$ in $Y,$ then $f^{-1}\left(y_{\bullet}\right)$ clusters at $x$ in $X.$

*

*This can be reworded as: $f$ is open $\iff$ whenever $y_{\bullet}$ is a net that clusters at a point $y$ in $Y,$ then $f^{-1}\left(y_{\bullet}\right)$ clusters at every point $x \in f^{-1}(y).$

*In comparison, $f$ is continuous $\iff$ whenever $x_{\bullet}$ is a net that clusters at a point $x$ in $X,$ then $f\left(x_{\bullet}\right)$ clusters at $f(x)$ in $Y.$
Open maps 2: $f$ is open $\iff$ whenever $y \in Y$ and $y_{\bullet} \to y$ in $Y$ and $C$ is a closed subset of $X$ that eventually contains $f^{-1}(y_{\bullet}),$ then $C$ contains $f^{-1}(y).$

*

*In comparison, by the closure characterization of continuity, $f$ is continuous $\iff$ whenever $x_{\bullet} \to x$ in $X$ then any closed subset of $Y$ that contains $f\left(x_{\bullet}\right)$ also contains $f(x).$
The first two characterizations remain true if $y_{\bullet},$ $y,$ and $x$ are allowed to be sets (instead of necessarily being points). The last remains true if $y_{\bullet}$ is allowed to be a net of sets that is not eventually empty.
Definitions used
$f : X \to Y$ is an open map if it sends open subsets of $X$ to open subsets of $Y$. Closed map is defined by replacing "open" with "closed".
If $y_{\bullet}=(y_i)_{i\in I}$ is a net in $Y$ (i.e. a function $I \to Y$ from a directed set) then the notation $f^{-1}(y_{\bullet}):=(f^{-1}(y_i))_{i\in I}$ denotes the net of set $I \to \wp(X)$ defined by sending the index $i \in I$ to the set $f^{-1}(y_i).$ Similarly, $f\left(y_{\bullet}\right):=\left(f\left(y_i\right)\right)_{i\in I}.$

*

*A set $C$ contains $f^{-1}(y)$ means $f^{-1}(y)\subseteq C.$ Similarly, a set $C$ contains $f^{-1}\left(y_{\bullet}\right)$ means that $f^{-1}\left(y_{i}\right) \subseteq C$ for every index $i.$ And $C$ eventually contains the net of sets $f^{-1}(y_{\bullet})$ if there exists some index $i \in I$ such that $f^{-1}(y_j)\subseteq C$ for every $j\geq i.$

*$f^{-1}(y_{\bullet})\to f^{-1}(y)$ means the following: whenever $U$ is an open subset of $X$ such that $f^{-1}(y) \subseteq U,$ then $U$ eventually contains $f^{-1}(y_{\bullet}).$ Side note: This is used in characterization Closed maps. In contrast, in characterization Open maps 2, "eventually contains $f^{-1}(y_{\bullet})$" is a hypothesis (rather than a conclusion) and $f^{-1}(y)\subseteq \text{ (set)}$ is the conclusion (rather than a hypothesis).

*$f^{-1}(y_{\bullet})$ clusters at $x$ means: whenever $U$ is an open subset of $X$ that contains $x$ and $i \in I$ is an index then there exists some index $j \geq i$ such that $U \cap f^{-1}(y_j) \neq \varnothing.$
Open maps 1 intuition and example
I think that characterization Open maps 1 gives the best intuition for what it means for a map to be open while characterization Closed maps gives the best intuition for what it means for a map to be closed.
For example, consider the canonical projection $\pi : \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ from $X := \mathbb{R} \times \mathbb{R}$ to $Y := \mathbb{R}$ defined by $(x, y) \mapsto y.$ Every fiber is of the form $\pi^{-1}(y) = \mathbb{R} \times \{y\}$. Characterizations Closed maps and Open maps 1 become:
$\pi$ is closed $\iff$ whenever $y_{\bullet}$ is a net of real numbers that converges to a point $y$ in $\mathbb{R}$ and $U$ is an open subset of $\mathbb{R} \times \mathbb{R}$ that contains $\mathbb{R} \times \{y\},$ then $U$ eventually contains all fibers $\mathbb{R} \times \{y_i\}$ (explicitly, there exists an index $i_0 \in I$ such that $\mathbb{R} \times \{y_i\} \subseteq U$ for every $i \geq i_0$).
$\pi$ is open $\iff$ whenever $y_{\bullet}$ is a net of real numbers that clusters at a point $y$ in $\mathbb{R}$ and $U$ is an open subset of $\mathbb{R} \times \mathbb{R}$ that contains a point $x \in \mathbb{R} \times \{y\},$ then $U$ intersects cofinally many fibers $\mathbb{R} \times \{y_i\}$ (explicitly, for every $i_0 \in I,$ there exists a larger index $i \geq i_0$ such that $\mathbb{R} \times \{y_i\} \cap U \neq \varnothing$).
$\pi$ is NOT a closed map because the sequence $\tfrac{1}{1}, \tfrac{1}{2}, \tfrac{1}{3}, \ldots$ converges to $0$ in $Y := \mathbb{R}$ but the set $U := \left\{(x, y) : |x y| < 1\right\}$ (consisting of the axes and everything "inside" the hyperbolas $x = \pm 1/y;$ see a graph here) is an open neighborhood of $\pi^{-1}(0) = \mathbb{R} \times \{0\}$ in $X := \mathbb{R} \times \mathbb{R}$ that does NOT contain even a single one of the fibers
$$\pi^{-1}\left(\tfrac{1}{1}\right) = \mathbb{R} \times \left\{1\right\}$$
$$\pi^{-1}\left(\tfrac{1}{2}\right) = \mathbb{R} \times \left\{\tfrac{1}{2}\right\}$$
$$\pi^{-1}\left(\tfrac{1}{3}\right) = \mathbb{R} \times \left\{\tfrac{1}{3}\right\}$$
$$\vdots$$
However, consistent with $\pi$ being an open map, $U$ does intersect cofinally many fibers (in fact, this particular open set happens to intersect every one of these fibers, although this is not true of other open neighborhoods, such as $U \setminus (\mathbb{R} \times \{1\})$ for instance).
Open maps 2 intuition
These next definitions are non-standard (i.e. my own made up definitions) but they make make Open maps 2's statement more intuition friendly:
A subset (or point) $R$ of a topological space is close to another subset $S$ if $R$ is contained in the closure of $S$ (i.e. $R \subseteq \overline{S}$ if $R$ is a set while if $R$ is a point then this means $R \in \overline{S}$). With these definitions, a subset is closed if and only if it contains every point/subset that is close to it.
The set of all fibers of a map $f : X \to Y$ (i.e. sets of the form $f^{-1}(y)$ for some $y \in Y$) form a partition of $X$. 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 $S$.
In my opinion, characterization Open maps 2 also gives a good intuitive understanding of what it means for a map $f : X \to Y$ to be open.
Open maps 2 tells us that the property of $f$ being open expresses a relationship between (1) points in $f$'s codomain $Y$ and (2) the fibers of f (rather than the points in $f$'s domain $X$).
If $f$ is open then intuitively, Open maps 2 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 (i.e. image under $f$) 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").
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".
Continuous and open
Now let's investigate what it intuitively means for a continuous map to be open:
Lemma: 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.
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 results to determine whether or not they're open.
A: 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.
