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Assume that $f:(X,d_{1})\to (Y,d_{2})$ is a continuous surjective map between compact metric spaces. We define another metric $d_{f}$ on $Y$ With $$ d_{f}(y_{1},y_{2})=Hd(f^{-1}(y_{1}), f^{-1}(y_{2}))$$ where $Hd$ is the Hausdorff distance. This metric is used in this post,too

  1. Is $(Y,d_{f})$ a locally compact space? can this topological space be topologically embedded in $(Y, d_{2})$?

  2. If the answer to the first part of the above question is affirmative, then we have the following situation in the context of commutative $C^{*}$ algebras:

We have an embedding of a (commutative) separable $C^{*}$ algebra $A=C(Y)$ into another separable (commutative) $C^{*}$ algebra $B=C(X)$ and finally we obtain a new (not necessarilly unital $C^{*}$ algebra. What is a possible non commutativization of this processes?

  1. Is there an example of this situation such that $f:(X,d_{1})\to (Y, d_{f})$ would be discontinuous at all points.

Edit: According to the counterexample of Eric, I would like to add some extra conditions to the first question as follows:

What about if each of the following conditions are assumed :

i)$f:X\to Y$ is a covering map (a covering space structure)

Or

ii) $X$ and $Y$ are smooth compact manifold and $f$ is an smooth map

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  • $\begingroup$ Your notation is confused. I suggest writing $x_1,x_2\in X$ and $y_1.y_2\in Y$. Then this gives a metric on $X$, not $Y$. $\endgroup$ – Paul Taylor Oct 23 '14 at 10:05
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Here's an example where the $d_f$-topology is not locally compact. Let $Y=[0,1]$ and let $$X=[0,1]\times \{0\}\cup \{(q,1/n):q\in[0,1]\cap\mathbb{Q}\text{ and $n$ is the minimal denominator of $q$}\}.$$

Take $f:X\to Y$ to be the projection $f(a,b)=a$. Then the $d_f$-topology is the refinement of the usual topology on $[0,1]$ obtained by making every rational point isolated. This is not locally compact, since any neighborhood of an irrational point will contain sequences that ought to converge to a rational point but have no $d_f$-limit.

More generally, let $Y$ be arbitrary and let $(A_n)$ be a sequence of closed subsets of $Y$. Taking $X=Y\times \{0\}\cup \bigcup A_n\times \{1/n\}$ and $f$ to be the projection, the $d_f$-topology will be the $d_2$-topology refined by making each $A_n$ open. This should be a rich source of counterexamples to properties you might want $d_f$ to have.

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Here is an application of a similar but simpler idea. By allowing infinite distances the function does not have to be surjective.

For any continuous function $f:X\to Y$ and points $x\in X$ and $y\in Y$, let $$ d_y(x) = \inf \{ d(x,x') | f(x')=y \}. $$ Then $d_y(x)$ is a continuous function of two variables iff $f$ is an open map.

This result belongs in undergraduate courses on metric spaces and topology, but whenever I have asked, nobody has seen it before.

In my draft paper on Overt Subspaces of ${\mathbb R}^n$ I took this result as the starting point to try to explain overtness (which is often said to be invisible in classical topology) to the ordinary mathematician.

In particular, I point out the similarity with the Newton-Raphson algorithm for finding zeroes of continuously differentiable functions.

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  • $\begingroup$ Thank you for the interesting answer and the link paper. $\endgroup$ – Ali Taghavi Oct 23 '14 at 21:16

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