## Continuous:Lipschitz :: Open:?

A map is continuous if the preimage of every open set is open. If the topology is defined by a metric, we can reformulate this as: the inverse image of an open ball $B_R(f(x))$ contains an open ball $B_r(x)$. Like this:

But bringing these radii $R$ and $r$ into the picture will not serve any purpose unless we use them to quantify continuity. For example, if we insist that $r\ge cR$ for a fixed constant $c>0$, we arrive at the definition of a Lipschitz map.

But why do we look at the inverse image; what happens if we take the direct image instead? Then we get the definition of an open map: the image of every open set is open. Recasting this in metric terms: the image of an open ball $B_R(x)$ contains an open ball $B_r(f(x))$. Like this:

If we quantify openness by requiring $r\ge cR$ for a fixed $c>0$, we arrive at the definition of a co-Lipschitz map. [Caution: some people use “co-Lipschitz” to mean $|f(a)-f(b)|\ge c|a-b|$, which is a different condition. They coincide if $f$ is bijective.]

I don’t know if openness without continuity is good for anything other than torturing students with exercises such as: “Construct an open discontinuous map from $\mathbb R$ to $\mathbb R$.” We probably want both. At first one can hope that open continuous maps will have reasonable fibers $f^{-1}(x)$: something $(m-n)$-dimensional when going from $m$ dimensions to $n$, with $m\ge n$. The hope is futile: an open continuous map $f\colon \mathbb R^2\to\mathbb R^2$ can squeeze a line segment to a point (construction left as an exercise).

A map that is both Lipschitz and co-Lipschitz is called a Lipschitz quotient; this is a quantitative analog of “open continuous”. It turns out that for any Lipschitz quotient $f\colon \mathbb R^2\to\mathbb R^2$ the preimage of every point is a finite set. Moreover, $f$ factors as $f=g\circ h$ where $g$ is a complex polynomial and $h$ is a homeomorphism.

This is encouraging… but going even one dimension higher, it remains unknown whether a Lipschitz quotient $f\colon \mathbb R^3\to\mathbb R^3$ must have discrete fibers. For an overview of the subject, see Bill Johnson’s slides.