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The construction of turning a metric space $(X,d)$ into a topological space by inducing the topology generated by the open balls gives rise to a functor $Met\to Top$ for any reasonable category $Met$ of metric spaces. Of course not every topological space is metrizable and thus there can't possibly be an inverse functor. My question is whether one can enlarge the category $Met$ to some category $M$ in a sensible way and so that there is an embedding $Met\to M$ together with a functor $M\to Top$ which extends the open balls topology construction $Met\to Top$, such that $M\to Top$ has some sort of pseudo-inverse functor $Top\to M$. In particular, is it possible for $Met \to Top$ to have a (left or right) adjoint with nice properties (e.g., being a (co)reflection).

The context for the question is curiosity as to the following situation. The problem of classical metrizability is very well studied. There are some variants on what one means by metrizability (i.e., do we demand the metric to be symmetric or not, separated or not, etc.) that can slightly enlarge the class of metrizable spaces but still far from allows a pseudo-inverse. Other possibilities for enlarging $Met$ is to consider probabilistic metric spaces. The category of probabilistic metric spaces is a very natural category to consider and ordinary metric spaces embed in it. It seems that the problem of when a topological space is probabilistically metrizable is not so well-studied, and I wonder why. I also wonder then if one can further weaken the notion of metrizability to finally come up with a single category as in my question above.

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    $\begingroup$ I think that requiring the inverse to come from ${\rm Top}$ is a bit too much. Perhaps restrict to topological metric spaces which stand a chance to even become metric-like, e.g. Hausdorff/normal/first countable? $\endgroup$
    – Asaf Karagila ♦
    Nov 16, 2012 at 9:52
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    $\begingroup$ Not sure if this answers your question but you can embed the category of metric spaces into that of uniform spaces. For the pseuo-inverse, you can take the functor which assigns the finest uniform structure compatible with the topology. Of course, this only works for completely regular spaces but this usually suffices for analysts. $\endgroup$
    – jbc
    Nov 16, 2012 at 9:57
  • $\begingroup$ Of course the game can be played both ways: Either enlarge Met or shrink Top. I'd like to see some evidence though that answering my original question is hopeless before we start shrinking Top. Of course, everybody sort of 'knows' that Met and Top are very different. But the question of how different is kind of in the air. Answering my question, and then measuring how different Met and M are, will then make clearer between topology and metric space theory. $\endgroup$
    – Ittay Weiss
    Nov 16, 2012 at 9:58
  • $\begingroup$ jbc, how do you embed the category of metric spaces into that of uniform spaces? Two different metric structures on the same set can yield the same uniform structure, so the straightforward way of turning a metric space into a uniform space is not an embedding. Is there another way? $\endgroup$
    – Ittay Weiss
    Nov 16, 2012 at 10:00

2 Answers 2

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So, here is what I found out. The Bing-Nagata-Smirnov theorem can be seen as answering the question by restricting $Top$ to a subcategory which is metric like. A construction of Flagg can be seen as answering the question by enlarging $Met$ to a category that is topology like. I collected these observations into a short note.

Edit: The note will appear in the short communications of Algebra Universalis.

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It's just a possible attack direction:

I would consider the profunctors $J^*:Top\not\to Met$ and $J_*:Met\not\to Top$ where $J:Met\to Top$ is the 'open ball' functor you want to start with. I prefer to view profunctors $A\not\to B$ as their collages, ie. 'bipartite' categories themselves, containing disjoint copies of $A$ and $B$ and with additional arrows ('heteromorphisms') from objects of $A$ to objects of $B$.

In this sense, each metric space $X$ will have $J(X)$ as its coreflection to $Top$ in $J^*$, and its reflection to $Top$ in $J_*$.

For $J$ having a left/right adjoint means exactly that $J^*$ [resp. $J_*$] also has the (co-)reflection property in the other direction. I.e., I would look at exactly which top.spaces have (co)-reflections to $Met$ in $J^*$ and $J_*$.

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  • $\begingroup$ But this would lead to an answer of identifying a subcategory of Top closer to Met. I'm looking for an enlargement of Met closer to Top. Unless I'm misunderstanding your suggestion. $\endgroup$
    – Ittay Weiss
    Nov 16, 2012 at 11:19
  • $\begingroup$ Aha, well.. yes. But.. perhaps we can adjoin the missing coreflections formally somehow. I'm not really sure, was just a feeling. $\endgroup$
    – Berci
    Nov 16, 2012 at 11:25
  • $\begingroup$ I'm not sure about formally adjoining coreflections while making sure the larger category exists (i.e., set-theoretic problems) and that it embeds the original category. I do believe that some solution using formally adding solutions should be possible, but it's a hunch. $\endgroup$
    – Ittay Weiss
    Nov 16, 2012 at 11:28

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