Why are morphisms in the usual category of metric spaces short maps rather than Lipshitz maps?

Question

In the usual category of metric spaces, morphisms are short maps (aka metric maps); that is, a morphism $f$ from $X$ to $Y$ is a map satisfying $d_X(x,y)\leq d_Y(f(x),f(y))$ for all $x,y\in X$. This choice surprises me. Surely choosing morphisms to be Lipshitz maps (functions satisfying $d_X(x,y)\leq Bd_Y(f(x),f(y))$ for all $x,y\in X$ and fixed constant $B$) would give a more useful category since the short map condition seems so restrictive?

I am aware that short maps as morphisms is natural if one views metric spaces themselves as enriched categories (Lawvere metric spaces) and is interested in functors between Lawvere metric spaces. I am much more interested in knowing whether short maps perhaps ensure the category of metric spaces has better properties then Lipschitz maps? If so, how specifically?

Answer 1

One reason to care is that the usual category of metric spaces and short maps (provided you allow two points to have infinite distance from one another) is very well behaved -- it's complete, cocomplete, and symmetric monoidal closed. See, for instance the introduction to Rosicky and Tholen's Approximate Injectivity. It's possible that the category of metric spaces with lipschitz maps doesn't enjoy these nice properties, and that's why people don't consider it as often. If we restrict attention to compact metric spaces with lipschitz maps, then we definitely don't enjoy these properties, since they're not closed under colimits (basically because of compactness) or limits (this I think would continue to be an issue even for the category of all metric spaces with lipschitz maps). See this related question for more details.

Another reason is that the category of metric spaces with lipschitz maps "doesn't remember the metric" in the sense that two metric spaces with different metrics can still be isomorphic in this category (compare with two homeomorphic spaces equipped with different metrics). In this sense, the category "isn't actually seeing the metric structure", it's only seeing that part of the metric structure that we can detect via lipschitz functions (in much the same way that topological spaces only see the part of a metric we can detect via continuous functions)! This is closely related to @N.Virgo's comments about universal properties being only defined up to scaling. Of course, it's easy to see that the category of metric spaces and short maps doesn't have this issue -- it correctly "remembers" the metric structure in the sense that two isomorphic objects must have exactly the same metric.

Obviously this isn't a historically accurate picture, and I would love for someone with more experience with various categories of metric spaces to chime in!

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I hope this helps ^_^

Answer 2

In the other direction, the category of metric spaces (with real valued metrics) and Lipschitz maps has a nice property which is not shared by the category of metric spaces and short maps, namely, that it has (finite) coproducts.

Indeed, for two nonempty metric spaces $(X,d_X)$ and $(Y,d_Y)$ with points $x_0\in X$ and $y_0\in Y$ let $S$ be the disjoint (or disjointified) union and define $d_S$ so that the restrictions to $X$ and $Y$ are $d_X$ and $d_Y$, respectively, and $$d_S(x,y)=d_S(y,x)=d_X(x,x_0)+1+d(y_0,y)$$ for $x\in X$ and $y\in Y$. Together with the inclusions $i_X:X\to S$ and $i_Y:Y\to S$ (which are even short) we get a coproduct. For, if $f:X\to Z$ and $g:Y\to Z$ are Lipschitz with Lipschitz constants $L_f$ and $L_g$, the obvious map $h:S\to Z$ defined as $f$ on $X$ and as $g$ on $Y$ is Lipschitz with Lipschitz constant $$L_h\le \max\{L_f,L_g,d_Z(f(x_0),g(x_0))\}.$$

Answer 3

I assume that, within the context of metric spaces themselves, taking the morphisms between two metric spaces to be the maps which do not increase distances is the "correct" notion of morphism because it gives a natural class of morphisms within which isomorphic spaces are isometric. This seems like a reasonable desideratum in a category whose objects are metric spaces! It means, more informally, that in this category there is a notion of "scale", whereas that notion disappears if you allow Lipschitz morphisms.

The category of metric spaces and distance non-increasing maps also has good "homological" properties, every metric space $X$ has an injective envelope $\epsilon(X)$ and thus Met has "enough injectives". (It is even the case that $\epsilon(X)$ is compact if $X$ is compact, so the subcategory of compact metric spaces inherits this property.)

Also, to clarify a tension between the assertions in the previous answers, the category of metric spaces Met does not have coproducts, i.e. is not (even finitely) cocomplete. However, if one extends the notion of a metric to allow functions $d\colon X \times X \to [0,\infty]$ (i.e. allowing the distance function to take the value $\infty$ which is assumed to be obey $\infty \geq \infty+\infty\geq \infty$ etc.) then this category $\textbf{Met}_{\infty}$ (with the same notion of morphism as in Met) is complete and cocomplete.

For example, if $X$ and $Y$ are metric spaces then $X \bigsqcup Y$, the coproduct of $X$ and $Y$, is their disjoint union with metric $d$ where $$ d(a,b) = \left\{\begin{array}{cc} d_X(a,b), & \text{ if } a,b \in X \\ d_Y(a,b), & \text{if } a,b \in Y\\ \infty & \text{ otherwise}.\end{array}\right. $$