Let $(X,d)$ be a compact metric space and $\sim$ an equivalence relation on $X$ such that the quotient space $X/\sim$ is Hausdorff. It is well known that in this case the quotient is metrizable. My question is, can we choose a compatible metric on $X/\sim$ so that the quotient map does not increase distances?


As in this question 


[Is there a conceptual reason why topological spaces have quotient structures while metric spaces don't?](https://mathoverflow.net/questions/123739/is-there-a-conceptual-reason-why-topological-spaces-have-quotient-structures-whi)


we can define a pseudometric $d'$ on the quotient $X/\sim$ by letting \begin{equation\*} d'([x],[y]) = \inf\{ d(p\_1,q\_1) + \cdots+ d(p\_n,q\_n):p\_1 = x,q\_i \sim p\_{i+1},q\_n=y\} \end{equation\*} where $[x]$ is the equivalence class of $X$. With this $d'$ the quotient map clearly does not increase distances. So do we know when $d'$ is a metric and when it is compatible with the quotient topology? Wikipedia says that $X$ being compact implies that $d'$ is compatible with the quotient topology; I would appreciate a reference or a quick proof.



Quotient of metric spaces
The following counter-example leaves no doubts :-) (this will answer also some other - more basic - potential similar questions too). This example is related to the *Cantor set*:


Let $\ X:=[0;1]\ $ with the ordinary Euclidean metric $\ |\,.\,|\ \ $ (absolute value). Let $\ f:X\rightarrow Y\ $ be an arbitrary continuous surjection onto an arbitrary (compact) metric space $\ (Y\ d),\ $ such that


$$f(0)\ne f(1)$$


Furthermore, let $\ f\ $ be constant on every interval $\ [a;b]\subseteq [0;1]\ $ such that there exits a non-empty finite sequence $\ T\subseteq\{1\ 2\ \ldots\}\ $ and $\ t:=\max T,\ $ such that


$$b:=2\cdot\sum\_{j\in T}3^{-j}\qquad and\qquad a:=b-3^{-t}$$


Then:


**THEOREM**   Function $\ f\ $ is not Lipschitz, i.e. for every $\ C>0\ $ there exist $\ x\ y\in[0;1]\ $ such that $\ d(f(x)\ f(y)\,>\,C\cdot|x-y|$.


**PROOF** *Mood improver*.


**COROLLARY**   Function $\ f\ $ is not Lipschitz with respect to any metric $\ d'\ $ in $\ Y,\ $ which is topologically equivalent to $\ d\ $ (see the Theorem above).


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**EXAMPLE**   Function $\ f\ $ can be $\ f:[0;1]\rightarrow[0;1]\ $ which is a monotone extension from the Cantor set onto the whole interval, as follows:


$$f(2\cdot\sum\_{j\in S}3^{-j})\ :=\ \sum\_{j\in S}2^{-j}$$


for arbitrary $\ S\subseteq\{1\ 2\ \ldots\}\,\ $ (when $\ S=\emptyset\ $ then we obtain $\ f(0)=0$). One can see directly that this specific function is a quotient function. See also the general remark below.



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> **REMARK**   Every(!) continuous surjection of Hausdorff compact spaces $\ f:X\rightarrow Y\ $ is a quotient map. Indeed, such mappings are closed.
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