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:

Continuous map

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:

Open map

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.

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