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:

  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.