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(Updated to include effective epimorphism.)

This question is prompted by the recent discussion of why analysts don't use category theory. It demonstrates what happens when an analyst tries to use category theory.

Consider the category CpltMet in which the objects are complete metric spaces and morphisms are 1-Lipschitz maps; the maps $f:X\to Y$ such that $d_Y(f(a),f(b))\le d_X(a,b)$ for all $a,b\in X$. Note that in this category monomorphisms are injective 1-Lipschitz maps, and epimorphisms are 1-Lipschitz maps with dense range, but not necessarily surjective.

An isometric embedding is a map $f:X\to Y$ such that $d_Y(f(a),f(b)) = d_X(a,b)$ for all $a,b\in X$. I can describe such maps in the arrow-speak as follows. $f:X\to Y$ is an isometric embedding iff the following holds: whenever $f$ factors through an epimorphism $g:X\to Z$ (meaning $f=h\circ g$ for some $h:Z\to Y$), $g$ is an isomorphism. (Proof is given in Note 1.) If there is a better categorical description of isometric embeddings, I'd like to see it.

A submetry is a map $f:X\to Y$ such that for every $a\in X$ and every $r\ge 0$ we have $f(B_X(a,r))=B_Y(f(a),r)$ where $B$ denotes a closed ball. (See Note 2 about the definition). To appreciate this definition, consider the following.

  • isometric embeddings are characterized by the condition $f^{-1}(B_Y(f(a),r))=B_X(a,r)$, mirroring the definition of submetry.
  • among 1-Lipschitz maps, submetries are characterized by the 2-point lifting property:
    for every $y_0,y_1\in Y$ and every $x_0\in f^{-1}(y_0)$ there exists $x_1\in f^{-1}(y_1)$ such that $d_X(x_0,x_1)=d_Y(y_0,y_1)$.
  • for linear operators between Banach spaces, the adjoint of an isometric embedding is a submetry (the proof is an exercise with Hahn-Banach).

My question is: can the submetries be defined categorically, preferably in a way that makes them a dual class to isometric embeddings?

The problem is that reversing the arrows in the above definition of an isometric embedding gives a wider class of maps than submetries. Indeed, the reversed definition is: $f:Y\to X$ does not factor through any monomorphism $g:Z\to X$ unless $g$ is an isomorphism. But this holds, for example, for the function $f:\mathbb R\to \mathbb R$ defined by $$ (*)\qquad \qquad f(x)=\begin{cases} x+1,\quad &x\le -1 \\ 0,\quad &|x|\le 1 \\ x-1,\quad &x\ge 1 \end{cases}$$ which is not a submetry in the standard metric of the real line. (See Note 3 for the proof.) Maybe I'm reversing arrows in a "wrong" description of isometric embeddings.


Note 1. If $f$ preserves distances, then so does $g$; having dense range, $g$ must be onto because $X$ is complete; hence, $g$ is an isomorphism. Conversely, if $f$ decreases distances somewhere, let $Z$ be the same set as $X$ with the metric $(d_X(a,b)+d_Y(f(a),f(b)))/2$. The identity map $g:X\to Z$ is an epimorphism, $f$ factors through it, but $g$ is not an isomorphism.

Note 2. I am following the original definition of submetry given by Berestovskii ("Submetries of space-forms of nonnegative curvature", 1987). If one uses open balls instead of closed, the class is enlarged to weak submetries. In Riemannian Geometry by Petersen the term submetry is used for more general maps, which I would call weak local submetries.

Note 3. Proof: Suppose $f=g\circ h$ where $g: Z\to \mathbb R$ is a monomorphism. Then $h$ maps $[-1,1]$ into a single point $z\in Z$. When $a\le -1$ and $b\ge 1$, the triangle inequality yields $d_Z(h(a),h(b))\le |a-b|-2=|f(a)-f(b)|$. Hence, $g$ must be an isomorphism.

Note 4. Following the immersion:submersion terminology of differential geometry, I'd like to call an isometric embedding an immetry, but I'm not sure that the neologism would catch on.


Following the suggestion by @t.b., I considered the concept of an effective epimorphism. Unfortunately, the undesirable map defined by (*) appears to be effective. Indeed, let $R=\{(x,y)\in\mathbb R^2: f(x)=f(y)\}$. The orthogonal projections $\pi_x,\pi_y : R\to \mathbb R$ are 1-Lipschitz and satisfy $f\circ \pi_x=f\circ \pi_y$ by construction. As far as I can tell, $\pi_x$ and $\pi_y$ qualify as a kernel pair for which $f$ is a coequalizer.

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Very nice question (as usual). Just a small remark: you characterize isometries as extremal monomorphisms in category-speak and you dualize correctly. Here's the nlab page on extremal epimorphisms where you can find further notions of epimorphisms, maybe one of them fits submetries. –  t.b. Jul 11 '12 at 16:28
    
I've been thinking about the problem a bit, but with no satisfactory outcome so far. Of course, the two-point lifting property of submetries can be phrased categorically, but I don't think it's what you're looking for (because it looks somewhat arbitrary): submetries are precisely the maps that are right orthogonal to all inclusions of the one point space in a two point space. I've played with factorization systems a bit, but haven't managed to find a duality submetry vs immetry. –  t.b. Jul 13 '12 at 2:19
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