TITLE: Is the preimage of the closure the closure of the preimage under a quotient map?
QUESTION [11 upvotes]: Let $f : X \to X/\sim$ be a quotient map from a topological space $X$ to the quotient space $X/\sim$ for $\sim$ some equivalence relation. Let $S \subseteq X/\sim$. Is it true that $f^{-1}(\overline{S}) = \overline{f^{-1}(S)}$? 
The specific case I have in mind is a Borel subgroup $B$ of a Chevalley group $G$ acting on $G$. The Bruhat decomposition decomposes $G$ into the disjoint union of Bruhat cells $BwB$ over representatives $w$ of the elements of the Weyl group $N(T)/T$ where $T$ is a maximal torus in $B$ in $N(T)$ is it's normalizer in $G$. Taking the quotient under the action of $B$ on $G$ induces a decomposition of $G/B$ into the disjoint union of $Bw.B$ where $Bw.B$ is the set of cosets of the form $bwB$. The closures of the Bruhat cells $Bw.B$ in $G/B$ are unions of Bruhat cells and I want to know if this therefore implies that the closures of the corresponding Bruhat cells in $G$ are also unions of Bruhat cells.

REPLY [4 votes]: Since the question has been bumped to the front page, I won't feel guilty about adding another answer. It's a bit more complicated that mathmatrucker's answer, but it involves nicer spaces (metric, whereas mathematrucker's quotient isn't even $\text{T}_1$); it may also be related to the "fairly simple semilinear set" mentioned by Thierry Zell in his comment on his answer.
Let $X$ be the subspace $\{\frac1n:n=1,2,\dots\}\cup\{0\}$ of $\mathbb R$, i.e., a convergent sequence together with its limit. Form a quotient by identifying $1$ with $0$, call the resulting point $p$, and let $S$ be the set of all points except $p$ in the quotient.  (So $S$ consists of the images of the points $\frac12,\frac13,\dots$ of $X$.) Then the closure of $S$ is the entire quotient space, so $f^{-1}(\overline S)$ is all of $X$.  But $f^{-1}(S)$ is $\{\frac12,\frac13,\dots\}$, whose closure in $X$ does not contain the point $1$.