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?

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.

up vote 4 down vote accepted

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).


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.

Can we choose a metric on quotient spaces so that the quotient map does not increase distances? This is trivially true, when the metric have an upper bound. But the problem is there is no natural way of doing this.

Consider a sphere $S$ in $\mathbb{R}^3$. We identify two points on the sphere. Maybe we would like that for the quotient map $p$, the tangent map $p_t:T_x S\rightarrow T_{p(x)} (S/\sim)$ is an isometry at smooth points. However, this is not possible. In fact you can prove it using differential geometric arguments.

For example, see this paper. http://www.sciencedirect.com/science/article/pii/S0001870813000923

The singularity can not be classified to any of the three kinds.

  • Why is this trivially true when the metric has an upper bound? – burtonpeterj Feb 4 '15 at 17:16
  • because when the metric $d$ has an upper bound, you can take $\lambda d$ to make the metric arbitrarily small. – user2173168 Feb 5 '15 at 11:27
  • For example, consider a compact space $X$, $X$ has a metric $d$, $d$ therefore has an upper bound. $X/\sim$ is metrizable, it can be endowed with a metric $d'$. It is trivial to prove, there is a $\lambda>0$, such that $\lambda d'$ is smaller than $d$. – user2173168 Feb 5 '15 at 11:33
  • We can find such a $\lambda$ only if the quotient map is Lipschitz. So why can we assume the quotient map is Lipschitz? – burtonpeterj Feb 5 '15 at 20:58

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