Continuation of an earlier post that considered *submetries*, namely the maps between metric spaces such that for all and all . (Here and in what follows is a closed ball.) The dual notion is an *immetry*: a map such that for all and all . Immetries can be characterized by the condition , 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.

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 which is fixed by all maps under consideration. The *Lipschitz dual* of a metric space consists of all Lipschitz maps . This is naturally a vector space. Moreover, it is a Banach space with the norm . A Lipschitz map induces a bounded linear operator . If is 1-Lipschitz, then so is (i.e., its operator norm is at most 1).

It would be nice to have the following:

- is an immetry iff is a submetry
- is a submetry iff is an immetry

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

Suppose is an immetry. Then we think of as a subset of , and is simply the restriction operator. To prove that it’s a submetry, we should verify the 2-point lifting property: given any and any that extends , we must find that extends and satisfies . This is easy: extend in a norm-preserving way (by McShane-Whitney) and add .

I also wrote down an (easy) proof that being a submetry implies that 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 being an immetry does not imply that is a submetry. Whether being an submetry implies that is an immetry is left as an exercise for the reader.

The arrangement of five things is a quincunx, so it must be the other one.