quotient topology on $A^{*}$ equals subspace topology on $\pi(A)$ for $A^{*}:= \pi^{-1}(\pi(A))$ open or closed. Let $\pi:X \to X/\sim$ be a quotient map. Let $A \subset X$ and define $A^{*} := \pi^{-1}(\pi(A)) \subset X$. If $A^{*}$ is open or closed in X, then the subspace topology on $\pi(A)$ is the same as the quotient topology on $A^{*}/\sim$.
My ideas so far:
A set $U \in \pi(A)$ is open in the subspace topology iff $\exists \tilde{U} \subset X/\sim$
open such that $U = \pi(A) \cap \tilde{U}$. I want to show that it is open in the quotient-topology, i.e. that $\pi^{-1}(U) \subset X$ is open. But $\pi^{-1}(U) = A^{*} \cap \:\pi^{-1}(\tilde{U})$. Now if $A^{*}$ is open in X that works. But not if it is closed. And I have no idea for the other inclusion.
Any help would be appreciated.
 A: See the proof of Theorem 22.1 in Munkres. Using the notation that you have provided, the theorem is written as follows:

Let $\pi : X \rightarrow X /\sim$ be a quotient map; let $A^*$ be a subspace of $X$ that is saturated with respect to $\pi$; let $\pi^* : A^* \rightarrow \pi(A^*)$ be the map obtained by restricting $\pi$ to $A^*$. If $A^*$ is either open or closed in $X$, then $\pi^*$ is a quotient map.

In the context of your question, the subspace $A^{*}$ is clearly saturated since it is the inverse image of a subset of the codomain $X /\sim$. Moreover, $\pi^*(A^*) = \pi(\pi^{-1}(\pi(A)) = \pi(A)$, so the restriction $\pi^*$ of $\pi$ to $A^*$ can in fact be written as $\pi^* : A^* \rightarrow \pi(A)$. Thus, the theorem tells you that $\pi^*$ is a quotient map; in particular, a subset $V \subseteq \pi(A)$ is open if and only if it is open in ${\pi^*}^{-1}(V)$ if and only if it is open in $\pi^*(A^*)$. By definition, this implies equality of the subspace topology on $\pi(A)$ and the quotient topology on $\pi^*(A^*)$.
