Let us denote by $(\mathrm{CompMet})$ the category of compact metric spaces with Lipschitz maps as morphisms. I'm interested in properties of this category. It seems to me that it has finite products (take the usual cartesian product with the maximum metric) and equalizers (just restrict the metric), thus all finite limits. Is there a rigorous proof that infinite products don't exist? Note that the forgetful functor $(\mathrm{CompMet}) \to (\mathrm{Set})$ is representable (by the point), therefore preserves all limits. In particular, the underlying set of an infinite product has to be usual cartesian product.

As for colimits, I think that finite coproducts exist. The empty coproduct is the empty space. The coproduct of two compact metric spaces $X,Y$ has as underlying set the disjoint union $X \coprod Y$. The distance is defined as follows: For two points in $X$ or in $Y$ use the metrics of $X$ and $Y$. For $x \in X$, $y \in Y$, define $d(x,y):=c$ for a constant $c$ which is at least $\max(diam(X),diam(Y))/2$. Then we get a metric space and the universal property may be verified. Meanwhile I have checked the details; do you know a reference for this construction? (Here it is crucial that we work with Lipschitz maps. I think the notion of Gromov-Hausdorff distance is all about the failure of the category of compact metric spaces with isometries as morphisms having coproducts?)

Now what about infinite coproducts or just coequalizers? Remark that the category of compact Hausdorff spaces has arbitrary colimits, although the forgetful functor to topological spaces doesn't preserve them. Here the Stone-Cech-compactification is crucial. Therefore a related question: Does the forgetful functor $(\mathrm{CompMet}) \to (\mathrm{Set})$ has a left adjoint?

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    $\begingroup$ It has, I think, countable products (the usual sum metric will probably do?), but not higher. Or are you interested in countable products specifically? As to the left adjoint question: any compact metric space has size at most continuum, which will probably kill the idea of a left adjoint. $\endgroup$ – Henno Brandsma Dec 21 '11 at 13:50
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    $\begingroup$ 1) Do you mean the metric $d(x,y) = \sum_n \frac{d(x_n,y_n)}{1 + d(x_n,y_n)} 2^{-n}$? I doubt that you can prove the universal property since you cannot bound the infinitely many Lipschitz constants. 2) If $L(-)$ is the left adjoint, then it is not hard to show that the counit $L(|X|) \to X$ is surjective (first show that it is an isomorphism if $X$ is a point, and then use naturality). But you seem to claim that the unit $Y \to |L(Y)|$ is injective, why? $\endgroup$ – Martin Brandenburg Dec 22 '11 at 13:02
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    $\begingroup$ Although it is not written in categorical terms you might want to check out Burago and Ivanov's book "A Course in Metric Geometry". It has some chapters dedicated to Lipschitz and Gromov-Hausdorff Distances and it puts some of the constructions you mention in easy terms. $\endgroup$ – Sak Feb 17 '12 at 17:34
  • $\begingroup$ To expand on Chu's comment, it is known that the set of isometry classes of compact metrics spaces actually has the structure of a (non compact) metric space. I don't really know what that gains you in terms of categorical properties (since this fact forgets about maps between compact metric spaces), but I figured it might help somehow... $\endgroup$ – Jason DeVito Feb 22 '12 at 15:47

Infinite coproducts do not exist. For instance, suppose $C$ were a coproduct of countably infinitely many $1$-point spaces. The inclusion maps of these $1$-point spaces then give a sequence $(x_n)$ of points of $C$. Since $C$ is compact, some subsequence $(x_{n_k})$ must converge. But now you can just take any compact metric space $Y$ with a sequence $(y_k)$ which does not converge. By the universal property of $C$, there must be a Lipschitz map $C\to Y$ which sends $x_{n_k}$ to $y_k$ (and the $x_n$ which are not of the form $x_{n_k}$ to wherever you want in $Y$). But there does not even exist such a map which is continuous, since $(x_{n_k})$ converges and $(y_k)$ does not. (This example also shows the forgetful functor to Sets has no left adjoint, since such a left adjoint would need to send $\mathbb{N}$ to such a coproduct.)

Coequalizers do exist. Indeed, it is easier to just show that quotients by arbitrary equivalence relations exist. Quotients by arbitrary equivalence relations exist in the category of metric spaces and Lipschitz maps with Lipschitz constant $1$ (for instance, see this answer of mine on MO), and a quotient of a compact metric space is compact since the quotient map is surjective. These quotients have the same universal property when you allow Lipschitz maps with arbitrary Lipschitz constant, since a map $f:X\to Y$ is Lipschitz with constant $K$ iff it is Lipschitz with constant $1$ when you rescale the metric on $Y$ by $1/K$.


A rigorous proof that the infinite product of countably many 2-point spaces $A_n = \{0,1\}$ does not exist: (Really, just an elaboration of Martin's comment.)

Assume that $P$ is such a product, then $P$ must be $\prod_n A_n$ as a set. Let $e_n$ be the distance (in $P$) between the $0$-function and the function $x_n$ that takes the value $1$ only in the $n$-th place. The sequence $(e_n)$ must converge to $0$, because the $x_n$ will converge to the $0$ function. (They must converge by compactness, and the continuity of the projections means that all entries of the limit must be $0$.) [EDIT: As Martin Brandenburg pointed out, compactness is not enough to show convergence of the whole sequence, but only of some subsequence $(e_{n_i})$. For notational simplicity, assome $n_i=i$ for all $i$; otherwise restrict attention from $(x_n)$, $(e_n)$, $(d_n)$ to the respective subsequences $(x_{n_i}$, $(e_{n_i})$, $(d_{n_i})$ from now on.]

Let $(d_n)$ be a sequence of real numbers converging to $0$, but more slowly than $(e_n)$, i.e., $e_n = o(d_n)$. (E.g., $e_n=\sqrt{d_n}$.) Let $Q = \prod A_n$, but with the following metric: $d(x,y) = e_n$, if $x$ and $y$ agree on the first $n$ values, but not on the next one. Project $Q$ to each $A_n$ $-$ these maps are Lipschitz-continuous. So they factor through a map from $Q$ to $P$. Now this map must be the identity (apply the forgetful functor); but points with distance $e_n$ are then mapped to points with distance $d_n$, which is not possible for a Lipschitz function.

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    $\begingroup$ Thanks. Why does $x_n$ converge? A priori only a subsequence converges. $\endgroup$ – Martin Brandenburg Feb 26 '12 at 14:13
  • $\begingroup$ Okay subsequences are ok. Why is $Q$ compact? // By the way, there is a formal argument why your example already shows that no infinite product exists (if there are infinitely many spaces involved which have at least one element). The reason is: a) if $\prod_{i \in I} A_i$ exists, then for every subset $J \subseteq I$ also $\prod_{i \in J} A_j$ exists (realize it as a closed subspace of the whole product, by means of an equalizer). b) If $\prod_{i \in I} A_i$ exists and $B_i \subseteq A_i$ are closed subspaces, then $\prod_i B_i$ also exists (... as above). $\endgroup$ – Martin Brandenburg Feb 27 '12 at 10:57

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