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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?

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    $\begingroup$ Holomorphic maps are locally open on an open connected set $U \subset \mathbb{C}$. Homeomorphisms are also open maps. These are just the first ones that I thought of. $\endgroup$ – Mustafa Said Feb 13 '14 at 11:35
  • $\begingroup$ Projections $\pi\colon X\times Y\to X$ are nice examples of open maps. $\endgroup$ – Cheerful Parsnip Dec 4 '19 at 20:35
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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. :-)

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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 does exist a competing definition of "open map" that merely requires that $f(U)$ be open in the image of $f$).

Lemma 1: A 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: That is, $f : X \to Y$ 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.

We now introduce some definitions to make the above lemma more intuition friendly.

Non-standard (i.e. my own made up) definitions:

  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 $S$.
  3. Given $S \subseteq X$, by the (set of) $f$-fibers in $S$ we mean the union of all fibers of $f$ that are contained in $S$.
  4. If $F \subseteq X$ 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.

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

Now the set of all fibers of a map $f : X \to Y$ forms a partition of $X$ so if $f$ is open then intuitively, the above lemma means that each closed subset $C$ of $X$ contains a unique maximal "fiber-sucking black-hole like" subset $S$ where if the $Y$-value of any fiber contained in $C$ is close to $f(S)$ then the entire fiber is "sucked into $S$" (i.e. is then necessarily in $S$). The only difference is that whereas in a "black-hole like" subset $S$ of $X$, you'd expect that whether or not some fiber gets sucked into $S$ would be determined by how near the fiber is to $S$ in $X$'s topology (i.e. using $X$'s notion of "nearness"), instead openness means it is determined instead by how close the fiber's $Y$-value is to the image $f(S)$ of the "black-hole" $S$ using $Y$'s notion of "nearness".

If in addition we assume continuity then we get:

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.
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