Let $(X,d)$ be a metric space and $R\subseteq X \times X$ an equivalence relation.
There is a condition for which $X/R$ with the usual quotient topology is a metric space?
Thanks!
Let $(X,d)$ be a metric space and $R\subseteq X \times X$ an equivalence relation. There is a condition for which $X/R$ with the usual quotient topology is a metric space? Thanks! 

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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 $D[x]=\{y\in X(x,y)\in D\}$.


