# Injective:Surjective :: Isometric:?

I used this sort of title before, in “Continuous:Lipschitz :: Open:?“, but the topics are related anyway.

In some sense (formal or informal) the following classes of maps are dual to each other.

• Injective : Surjective
• Immersion : Submersion
• Monomorphism : Epimorphism

An isometric embedding of metric space $X$ into a metric space $Y$ is a map $f\colon X\to Y$ such that $d_Y(f(a),f(b))=d_X(a,b)$ for all $a,b\in X$. This concept belongs to the left column of the table above. What should be its counterpart?

Candidate 1. Observe that a 1-Lipschitz map $f\colon X\to Y$ is an isometric embedding iff it does not factor through any 1-Lipschitz surjection $g\colon X\to Z$ (for any space $Z$) unless $g$ is an isometric isomorphism. Reversing the order of arrows, we arrive at the following concept:

A 1-Lipschitz map $f\colon Y\to X$ is a metric quotient map if it does not factor through any 1-Lipschitz injection $g\colon Z\to X$ (for any space $Z$) unless $g$ is an isometric isomorphism.

This can be reformulated as follows: $\rho(a,b):=d_{X}(f(a),f(b))$ is the greatest pseudo-metric on $Y$ subject to

1. $\rho(a,b)\le d_{Y}(a,b)$
2. $\rho(a,b)=0$ if $f(a)=f(b)$

This seems reasonable: $f$ does as little damage as possible, given the structure of its fibers. There is also a natural way to construct $\rho$ for any reasonable fibering of $Y$: begin by defining $\widetilde{d_Y}=d_Y$ for points in different fibers and $0$ otherwise. Then force the triangle inequality by letting $\rho(a,b)=\inf \sum_{j=1}^n \widetilde{d_Y}(y_j,y_{j-1})$ subject to $y_0=a$ and $y_n=b$. As long as the fibers stay at positive distance from one another, this $\rho$ will be a metric. The corresponding metric quotient map sends each point of $Y$ onto its fiber.

Here is a simple example, in which a part of an interval is mapped to a point.

However, the above example made me unhappy. The only nontrivial fiber is the interval $[1,2]$. Both points $0$ and $3$ belong to trivial fibers, but the distance between them decreases from 3 to 2. This looks like a wrong kind of quotient to me.

Candidate 2 already appeared in my post on Lipschitz quotient, but wasn’t recognized at the time. It could be called $(1,1)$-Lipschitz quotient, but a better name is available. A map $f\colon Y\to X$ is a submetry if $f(B_Y(y,r))=B_X(f(y),r)$ where the balls are closed (using open balls yields something almost equivalent, but generally weaker). Such $f$ need not be an isometry: consider orthogonal projections in a Euclidean space. It does have to be 1-Lipschitz. The additional property that distinguishes it from general 1-Lipschitz maps is the 2-point lifting property: for every $x_0,x_1\in X$ and every $y_0\in f^{-1}(x_0)$ there is $y_1\in f^{-1}(x_1)$ such that $d_{Y}(y_0,y_1)=d_X(x_0,x_1)$. Incidentally, this shows that $x\mapsto f^{-1}(x)$ is an isometric embedding of $X$ into the hyperspace $\mathcal{H}(Y)$ which I covered earlier (“This ain’t like dusting crops, boy“).

The concept and the name were introduced by В.Н. Берестовский (V.N. Berestovskii) in his paper “Submetries of space-forms of nonnegative curvature” published in 1987 in Siberian Math. J. Among other things, he proved that a submetry between spheres (of any dimension) is an isometry. Of course, there are many submetries of $S^{n-1}$ onto other spaces: take the quotient by a subgroup of $O(n)$, which can be either discrete or continuous. Are there any submetries that are not quotients by isometries?

Yes, there are. I’ll describe a (modified) example given by Berestovskii and Guijarro (2000). Let $\mathbb{H}$ be the hyperbolic plane realized as the upper half-plane with the metric $\frac{dx^2+dy^2}{y^2}$. Define $f\colon \mathbb{H}\to\mathbb{R}$ by $\displaystyle f(x,y)=\begin{cases} \log y, \qquad x\le 0 \\ \log\frac{x^2+y^2}{y}, \qquad x\ge 0 \end{cases}$

Don’t panic; this is just the signed distance function (in the metric of $\mathbb H$) to the fat green curve below. I also drew two other level sets of $f$, to the best of my Friday night line-drawing ability.

To convince yourself that $f$ is a submetry, first consider $y\mapsto \log y$ for which the submetry property is clear (it’s the quotient by horizontal translation), and then note that the inversion in the unit circle exchanges horizontal lines (horocycles at infinity) with horocycles at 0. An interesting feature of this submetry that it is not very smooth: $C^{1,1}$ but not $C^2$.