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.

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