A metric associated with a continuous surjective map $f:X\to Y$ 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

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

  
*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

 A: 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.
A: 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.
