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](https://mathoverflow.net/questions/184820/a-metric-on-s2)



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> 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:
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> 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?
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> 3. Is there an example of this situation such that $f:(X,d\_{1})\to (Y, d\_{f})$ would be discontinuous at all points.
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**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 :



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



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> ii) $X$ and $Y$ are smooth compact manifold and $f$ is an smooth map
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A metric associated with a continuous surjective map $f:X\to Y$
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.


