Quotient Metric Space

Let $(X,d)$ be a metric space and $R\subseteq X \times X$ an equivalence relation.

Is there a condition for which $X/R$ with the usual quotient topology is a metric space?

Thanks!

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The quotient is always a pseudometric space with the quotient pseudometric. On the other hand the quotient topology can certainly fail to be Hausdorff. – Qiaochu Yuan Dec 14 '13 at 20:12
There are certainly a lot of such conditions. If the equivalence classes are compact, you can try to endow the quotient with the Hausdorff metric. What do you want to do with the quotient? What compatibility do you want with the quotient map? Without such precision your question is unanswerable. – Benoît Kloeckner Dec 14 '13 at 20:31
The book Introduction to Uniform Spaces defines compatible and weakly compatible equivalence relations on uniform spaces and it defines quotient uniform spaces by compatible and weakly compatible equivalence relations. – Joseph Van Name Dec 14 '13 at 20:39
I think this question should be opened. I think it is clear that the asker wants conditions for which one can take a quotient metric space which preferably are in some sense necessary and sufficient or at least very general. – Joseph Van Name Dec 22 '13 at 18:55

I am going to answer this question in the more general setting of uniform spaces rather than metric spaces. Most of the content of this answer is taken from the book Introduction to Uniform Spaces by I.M. James.

If $R$ is an equivalence relation on a set $X$, then write $\pi_{R}$ for the projection from $X$ to $X/R$.

Let $R$ be an equivalence relation on a uniform space $(X,\mathcal{U})$. Then we say that $R$ is weakly compatible with the uniform structure on $X$ if for each $D\in\mathcal{U}$ there is an $E\in\mathcal{U}$ such that $E\circ R\circ E\subseteq R\circ D\circ R$. Furthermore, we say that $R$ is compatible with the uniform structure on $X$ if for each $D\in\mathcal{U}$ there is an $E\in\mathcal{U}$ where $R\circ E\subseteq D\circ R$. Clearly every compatible equivalence relation is weakly compatible.

If an equivalence relation $R$ is weakly compatible with a uniformity $(X,\mathcal{U})$, then the images $\pi_{R}\times\pi_{R}[D]=\{(\pi_{R}(x),\pi_{R}(y))|(x,y)\in D\}$ of the entourages $D\in\mathcal{U}$ generate a uniformity $\mathcal{U}/R$ on the quotient space $X/R$. The mapping $\pi_{Z}:(X,\mathcal{U})\rightarrow(X/R,\mathcal{U}/R)$ is clearly uniformly continuous. Clearly the mapping $\pi_{R}:(X,\mathcal{U})\rightarrow(X/R,\mathcal{U}/R)$ is uniformly continuous.

It is well known that a uniform space $(X,\mathcal{U})$ is generated by a metric if and only if $\mathcal{U}$ is generated by countably many elements. Therefore, if $(X,d)$ is a metric space that induces a uniformity $\mathcal{U}$ and $R$ is a weakly compatible equivalence relation on $X$, then the uniformity $\mathcal{U}$ and also the quotient uniformity $\mathcal{U}/R$ are generated by countably many elements. Therefore, the quotient uniformity $\mathcal{U}/R$ is induced by some metric (and I am sure there is an easy explicit description of such a metric).

If we assume that $R$ is a compatible equivalence relation on a uniform space rather than just a weakly compatible equivalence relation, then the quotient uniform spaces are essentially the uniformly continuous uniformly open surjections.

If $(X,\mathcal{U}),(Y,\mathcal{V})$ are uniform spaces, then we say that a map $f:X\rightarrow Y$ is uniformly open if for each entourage $D\in\mathcal{U}$ there is an entourage $E\in\mathcal{V}$ such that $E[f(x)]\subseteq\phi[D[x]]$ for all $x\in X$. Of course, by $D[x]$ we mean $D[x]=\{y\in X|(x,y)\in D\}$.

$\mathbf{Theorem}:$ Suppose that $R$ is an equivalence relation on a uniform space $(X,\mathcal{U})$. Then:

1. If $R$ is compatible with the uniform structure on $(X,\mathcal{U})$, then the mapping $\pi_{R}:(X,\mathcal{U})\rightarrow(X/R,\mathcal{U}/R)$ is uniformly continous and uniformly open.

2. If $\mathcal{V}$ is a uniformity on the quotient set $X/R$ and the projection $\pi_{R}:(X,\mathcal{U})\rightarrow(X/R,\mathcal{V})$ is uniformly continuous and uniformly open, then $R$ is compatible with the uniform structure $(X,\mathcal{U})$ and $\mathcal{V}=\mathcal{U}/R$.

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