Is this identification map(called metric identification) is a quotient map?

Question

First of all. This is not homework. the problem is as follows:

The definition of an open set in a (pseudo)metric space with a metric $\rho$ is:

a set $O$ is an open set if and only if $\forall x \in O, \exists$ an open ball $\mathbf{B}_\rho(x,\epsilon) \subset O$.

We have found that statement 3 of problem 2C is wrong. Here is a counterexample

The right statement should be "a set $A$ in $M$ is closed (open) iff $A$ is saturated and $h(A)$ is closed (open) in $M^*$".

My question is: How to prove the map $h$ is a quotient map?

My attempt:

1. The definition of a quotient map $p: X \rightarrow Y $ is surjectve and provided that a subset $U$ of $Y$ is open iff $p^{-1}(U)$ is open in $X$.
2. $h$ is obviously surjective. It is an identification map!
3. For one direction, given an open set $h^{-1}(U)$ in $M$, since $h$ is surjective, $hh^{-1}(U) = U$, so by above 4b), $U$ is open too. In other words, h is an open map.
4. For the other direction, given an open subset $U$ of $M^*$, how to prove $h^{-1}(U)$ is open? I am stuck here. In other words, how to prove $h$ is a continuous map. Here is my proof: we can first prove $\forall x \in M, \forall \epsilon > 0, h^{-1}(\mathbf{B}_{\rho^*}(h(x),\epsilon)) = \mathbf{B}_{\rho}(x, \epsilon)$. In other words, for any neighbourhood base $B_1$ of $h(x)$ in $M^*$, there is a neighbourhood base $B_2$ of $x$ such that $B_2 \subset h^{-1}(B_1)$. If you can give a more succinct and more illuminative proof than my mechanical proof, it would be deeply appreciated.

Answer 1

3 needs to be corrected to

If $A$ is open in $M$, then $h[A]$ is open in $M^\ast$.

which just says that $h$ is an open map.

To see this let $h(a) \in h[A]$ and find $r>0$ so that $B_\rho(a,r) \subseteq A$. This implies that $B_{\rho^*}(h(a), r) \subseteq h[A]$ as well and so that $h[A]$ is open in $(M^\ast, \rho^\ast)$. Your example shows that the reverse does not hold.

The continuity of $h$ follows from $h^{-1}[B_{\rho^\ast}(h(a), r)] = B_\rho(a,r)$ for all $a \in M$ and $r>0$; a fact you mention in your post and which shows that the inverse images of all open balls of $(M,\rho^\ast)$ are open (being balls again) and this suffices for continuity, as balls form a base in the image space.

It follows that $h$ is a quotient map (because $h$ is continuous, open and onto): if $h^{-1}[A]$ is open, so is $h[h^{-1}[A]] = A$ by ontoness and openness; the other side ($A$ open then $h^{-1}[A]$ open) is just the continuity of $h$, which I did above.