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. 



Is the preimage of the closure the closure of the preimage under a quotient map?
Yes, this is true for any open continuous map $f: X \to Y$. 


Since $\bar{B} = \neg int (\neg B)$ where $\neg$ denotes complementation and $int$ denotes interior, and since the preimage map $f^{-1}(-): P(Y) \to P(X)$ preserves complementation, we just need to check that $f^{-1}$ preserves the interior operation. The inclusion 


$$f^{-1}(int (B)) \subseteq int(f^{-1}(B))$$ 


is clear since the left side is an open contained in $f^{-1}(B)$ and the right side is the largest open contained in $f^{-1}(B)$. The other containment, 


$$int(f^{-1}(B)) \subseteq f^{-1}(int (B)),$$ 


is equivalent to the containment $f(int f^{-1}(B)) \subseteq int (B)$. This last containment holds because the left side is open since $f$ is open, and the left side is contained in $B$ since $f(f^{-1}(B)) \subseteq B$. Thus it is contained in $int (B)$, as required. 


**Edit:** As mentioned in the comment below, it is not always the case the quotient maps are open, so this does not completely answer the general question. But it does answer the specific case mentioned by the OP, because a quotient map $p: G \to G/B$ to a coset space *is* open. Indeed, for $U$ open in $G$, to check $p(U)$ is open, we must check that $p^{-1}(p(U))$ is open. But it is not hard to see that $p^{-1}(p(U)) = UB$, and the right side is clearly open in $G$. 


