Closure in Quotient Space: If $\pi : X \to Y$ is a quotient map and $A \subseteq Y$, is it true that $ \overline{A}=\pi(\overline{\pi^{-1}(A)})$? 
If $\pi : X \to Y$ is a quotient map and $A \subseteq Y$, is it true that $ \overline{A}=\pi(\overline{\pi^{-1}(A)})$?

What I have thought is that $\pi$ is a closed map so $\pi(\overline{\pi^{-1}(A)})$ is a closed set, and $ \pi(\overline{\pi^{-1}(A)}) \supseteq A$. I am trying to prove that $\pi(\overline{\pi^{-1}(A)})$ is indeed the smallest closed set containing $A$. I have found more than one example of that being true, and no counterexample.
Any help is appreciated. Thanks in advance.
 A: The answer is No. Instead of writing down a counterexample (which one can do directly, see also the end of this answer) I will explain how I came up with a class of counterexamples.
As already explained in my comment, the inclusions $\pi(\overline{\pi^{-1}(B)}) \subseteq \overline{B}$ and $B \subseteq \pi(\overline{\pi^{-1}(B)})$ are clear for every subset $B \subseteq Y$, so that the claim is equivalent to the statement that $\pi(\overline{\pi^{-1}(B)})$ is closed. To disprove this, it is not enough to find a closed subset $A \subseteq X$ such that $\pi(A)$ is not closed, since it may happen that $A$ cannot be written as $\overline{\pi^{-1}(B)}$.
Since $\pi : X \to Y$ is surjective, the subsets of $Y$ correspond bijectively to those subsets $A \subseteq X$ which are saturated, meaning that $A = \pi^{-1}(\pi(A))$ holds. The correspondence is given by $A \mapsto \pi(A)$ and $B \mapsto \pi^{-1}(B)$. So the claim is equivalent to: For every saturated $A \subseteq X$ the set $\pi(\overline{A})$ is closed. And this means, by the definition of the quotient topology, that the saturation $\pi^{-1}(\pi(\overline{A}))$ is closed.
Now $\pi$ corresponds to an equivalence relation $\sim$ on $X$. The saturated subsets are the unions of equivalence classes, and the saturation is given by taking the union of all equivalence classes of elements in the set. We have thus completely transformed the statement to something internal to $X$. The question becomes: is the saturation of the closure of a saturated subset a closed subset? But the process of taking the saturation (and hence the notion of saturatedness) is completely unrelated to the topology on $X$. This makes the claim very unlikely to be true.
Let us look at a simple class of examples of topological spaces. Any partial order $(X,\leq)$ gives rise to a topological space with the same underlying set: the closed subsets are the lower sets. The closure $\overline{A}$ of a subset $A \subseteq X$ contains those elements which are $\leq$ some element of $A$. So the saturation of $\overline{A}$ is given by $\{x \in X : \exists x' \in X, a \in A ~ (x \sim x' \leq a)\}$, and the question is if this is a lower set. But since $\sim$ can be defined completely independently from $\leq$, this cannot be true in general.
Here is a concrete example: Consider the partial order $\{0 < 1 < 2 < 3\}$. Define $\sim$ so that the equivalence classes are $\{0,3\}$, $\{1\}$, $\{2\}$. Let $A = \{1\}$. Then $\overline{A} = \{0,1\}$, and its saturation is $\{0,1,3\}$, which is not closed.
