Lipschitz duality: immetries and submetries

Continuation of an earlier post that considered submetries, namely the maps f\colon Y\to X between metric spaces such that f(B_r(y))=B_r(f(y)) for all y\in Y and all r\ge 0. (Here and in what follows B_r is a closed ball.) The dual notion is an immetry: a map f\colon X\to Y such that f^{-1}(B_r(f(x)))=B_r(x) for all x\in X and all r\ge 0. Immetries can be characterized by the condition d(f(a),f(b))=d(a,b), which means they are nothing but isometric embeddings. I just made up the word to introduce symmetry between submetry and immetry. And then tried to look it up.

This has got me stumped

In what sense are these dual? Recall that reversal of arrows in a “categorical” definition of an immetry did not produce a submetry. Let’s try another approach. Let our metric spaces be pointed: they all contain the point 0 which is fixed by all maps under consideration. The Lipschitz dual X^\sharp of a metric space X consists of all Lipschitz maps \varphi\colon X\to \mathbb R. This is naturally a vector space. Moreover, it is a Banach space with the norm \displaystyle \|\varphi\|=\sup\frac{|\varphi(a)-\varphi(b)|}{d(a,b)}. A Lipschitz map f\colon X\to Y induces a bounded linear operator f^\sharp\colon Y^\sharp\to X^\sharp. If f is 1-Lipschitz, then so is f^\sharp (i.e., its operator norm is at most 1).

It would be nice to have the following:

  • f is an immetry iff f^\sharp is a submetry
  • f is a submetry iff f^\sharp is an immetry

but that’s too much to hope for. For example, the inclusion of \mathbb Q into \mathbb R is not a submetry, but it induces the identity map \mathbb R^\sharp\to\mathbb Q^\sharp. Let’s go through these one by one.

Suppose f\colon X\to Y is an immetry. Then we think of X as a subset of Y, and f^\sharp is simply the restriction operator. To prove that it’s a submetry, we should verify the 2-point lifting property: given any \varphi,\psi\in X^\sharp and any \Phi\in Y^\sharp that extends \varphi, we must find \Psi\in Y^\sharp that extends \psi and satisfies \|\Psi-\Phi\|=\|\psi-\varphi\|. This is easy: extend \psi-\varphi in a norm-preserving way (by McShane-Whitney) and add \Phi.

I also wrote down an (easy) proof that f being a submetry implies that f^\sharp is an immetry, but WP ate it. Specifically, having pressed “Save Draft”, I was asked to re-login (the cookie expired). Having done so, I was presented with a 10 min old draft.

We already know that f^\sharp being an immetry does not imply that f is a submetry. Whether f^\sharp being an submetry implies that f is an immetry is left as an exercise for the reader.

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