Lets $X$ be a compact metrizable space and $f:X\to Y$ be a quotient map such that $Y$ equipped with the quotient topology is Hausdorff. Thus $Y$ is metrizable. Lets $\sim$ be an equivalence relation on $X$ such that $x\sim y$ if $f(x)=f(y)$. Then for any metric $d$ compatible with the topology of $X$ one can build a (pseudo)metric $d_\sim$ on $Y$ with: \begin{equation} d_\sim(a,b) = \inf\{d(p_1,q_1) + \cdots+ d(p_n,q_n);[p_1] = a,[q_i] = [p_{i+1}],[q_n] = b\}.(1) \end{equation} Where the $\inf$ is taken over all finite chains of points $\{p_i\}_{i=1}^n$, $\{q_i\}_{i=1}^n$ between $a$ and $b$.

As in this question which has not been fully answered (Quotient of metric spaces) can we show that $d_\sim$ is a metric compatible with the quotient topology in $Y$ ?

If not, what would be a sufficient condition on the quotient map in order to have the result ?

Here is an attempt:

In Herman 1968, Quotient of metric spaces, in theorem 4.8, is stated the following :

THEOREM: Let $f$ be a function from a pseudo-metrizable space $X$ to a topological space $Y$, and suppose that $Y$ has the quotient topology relative to $f$, then the following are equivalent:
1) $Y$ is pseudo-metrizable
2) There exists a pseudo-metric $\rho$ compatible with the topology in $X$ such that the quotient pseudo-metric $\rho_\sim$, defined as in (1), is in fact compatible with the quotient topology of $Y$ (The definition of the quotient pseudo-metric by Herman should be equivalent to the one introduced earlier).

If $X$ is in fact metrizable, then it is pseudo-metrizable and $Y$ is also pseudo-metrizable. Therefore, from the theorem there exists a pseudo-metric $d^*$ compatible with the topology in $X$ (it is a metric as $X$ is Hausdorff) such that the quotient pseudo-metric $d^*_\sim$ is compatible with the topology in $Y$ (it is also a metric because $Y$ is Hausdorff too).

Is it possible to show that any quotient (pseudo)metric from an arbitrary metric $d$ is topologically equivalent to $d^*_\sim$ ?

We know that $X$ is metrizable and compact, thus there is a unique uniform structure in it and all metrics compatible with the topology are uniformly equivalent. Therefore any metric $d$ compatible with the topology of $X$ is uniformly equivalent to the metric $d^*$.

From uniform equivalent metrics, maybe there is a relation between their corresponding quotient pseudo-metrics but I am stucked here, do you have any idea/theorem/reference that would help me ?


  • $\begingroup$ According to the first line of your post, I think $Y$ is always metrizable, provided it is Haussdorf. (No quotion topology is needed for its metrizability). am I mistaken? $\endgroup$ Commented Apr 5, 2019 at 13:38

1 Answer 1


For the Cantor starcase function $f:C\to[0,1]$ from the standard ternary Cantor set $C$ onto the interval $[0,1]$ and for the standard Euclidean metric $d$ on $C$ the quotient pseudometric $d_\sim$ is constant zero (this follows from the fact that the Cantor set $C$ has length zero). So, the pseudometric $d_\sim$ is not necessarily a metric.

This example should be known but I cannot mention a suitable reference at the moment.

Added in Edit. Essentially the same counterexample is discussed in the answer of Wlodzimierz Holsztynski to this MO-question.

  • $\begingroup$ Thank for the answer ! Indeed it is the same counter-example than in the question I have quoted. However, I have realised that I need to deal with path-connected spaces so that quotient space is path-connected in the quotient pseudo-metric. Is there a known example that does not use the cantor set ? $\endgroup$
    – VMrcel
    Commented Jun 25, 2018 at 9:44
  • 1
    $\begingroup$ @VMrcel You can extend the Cantor starcase function to a continuous function on the closed interval and then you will get a continuous function between closed intervals, for which the quotient pseudometric still is zero. By the way, the quotient space is path-connected in the quotient metric (since it determines the anti-discrete topology). So, maybe some more precise question should be asked (but a good question is a half of an answer). $\endgroup$ Commented Jun 25, 2018 at 10:53
  • $\begingroup$ Oh you are right, I'll think about it, thank you ! $\endgroup$
    – VMrcel
    Commented Jun 26, 2018 at 12:08

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