Is the preimage of the closure the closure of the preimage under a quotient map?

Question

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.

Answer 1

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 $G$.

Answer 2

To supplement Todd's answer, I'd first emphasize that it's better not to place the quotient variety construction in algebraic geometry in the setting of general topology. In traditional algebraic geometry the natural (Zariski) topology on affine algebraic groups and their quotients by closed subgroups fails to be Hausdorff, so one can't automatically carry over classical ideas about quotients or quotient maps without further discussion. In scheme theory deciding on the correct notion of "quotient" gets even more delicate.

Here you are basically working over an algebraically closed field of arbitrary characteristic, within the Borel-Chevalley structure theory as developed in several textbooks by Borel, Springer, and myself using approximately the language of Mumford's old lecture notes. (More sophisticated treatments involving schemes occur in SGA3, Demazure-Gabriel, and the notes by Jim Milne.) In the standard development one shows that the natural map from $G$ to $G/H$ is open (section 12 of my book, for example). In the special case where $H$ is a Borel subgroup $B$, it's then easy to go back and forth between the double cosets $BwB$ or their closures in $G$ and the Bruhat cells or their closures (Schubert varieties) in the projective variety $G/B$.

By the way, it's worth looking further at the classic paper by Chevalley (circa 1958) where the "Bruhat ordering" of the Weyl group is first defined relative to the inclusions of Schubert varieties. This unpublished paper was edited by Borel and appears at the beginning of the two volume AMS Summer Institute proceedings (Penn State, 1991) published in 1994 as PSPM 56.

ADDED: Leaving quotient maps aside, the question raised about the Bruhat decomposition can be addressed directly inside $G$ by viewing the finitely many double cosets $BwB$ as orbits of the algebraic group $B×B$ acting on $G$ via left and right translations. Then the basic theory of algebraic group actions implies that the closure of such an orbit consists of the orbit together with possibly other orbits of lower dimension (ensuring that an orbit of smallest dimension is already closed, but here that's just $B$).

Answer 3

The answer seems to be yes in the case of interest to the OP. Nevertheless, the answer to the question as stated is no, and this is the chance for me to pull out my favorite example of badly behaved quotient map.

Let $\mathcal{S} = [0,1]^{\mathbb{N}}$, and define for all $\mathbf{x} \in \mathcal{S}$ the support of $\mathbf{x}$ to be $\mathrm{supp}(\mathbf{x})=\bigcup_{i\in \mathbb{N}} \mathbf{x}_i .$ Let $\mathcal{X} \subset [0,1] \times \mathcal{S}$ be the set defined by $$ \mathcal{X} =\lbrace (x,\mathbf{x}) \mid x\in \mathrm{supp}(\mathbf{x})\rbrace.$$

Fact 1: $\mathcal{X}$ is a metric space.

We can define a distance $d$ by: $d[(x,\mathbf{x}), (y,\mathbf{y})]:=1$ if $\mathbf{x}\neq \mathbf{y}$, and $d[(x,\mathbf{x}),(y,\mathbf{x})]:=|x-y|.$

Fact 2: The map $q:\mathcal{X} \to [0,1]$ defined by $q(x,\mathbf{x})=x$ is a quotient map.

The map $q$ satisfies $|q(x,\mathbf{x})-q(y,\mathbf{y})| \leq d[(x,\mathbf{x}), (y,\mathbf{y})]$, so $q$ must be continuous. So it's enough to check that for any $F \subseteq [0,1]$, $q^{-1}(F)$ closed implies that $F$ is closed. Consider a convergent sequence $(x_n)$ in $F$ whose limit is $\ell$: it can be lifted to the sequence $(x_n, \mathbf{x})\in q^{-1}(F)$ (where the second factor is constant). This sequence converges too, so if we assume that $q^{-1}(F)$ is closed, the limit $(\ell, \mathbf{x})$ must be an element of $q^{-1}(F)$, and thus $\ell$ is in $F$; it follows that $F$ is closed.

Fact 3: The map $q$ fails to verify $q^{-1}(\overline{S})=\overline{q^{-1}(S)}$ for all $S$.

Consider $S=(0,1)$. Then $q^{-1}(\overline{S})=\mathcal{X}$, but $\overline{q^{-1}(S)}$ does not contain the element that corresponds to the constant sequence 1.

Answer 4

We have $\overline{f^{-1}(S)}\subset f^{-1}(\overline S)$ since $f^{-1}(\overline S)$ is a closed set that contains $f^{-1}(S)$ but the reverse inclusion does not hold in general. Partition the reals under the usual topology into two equivalence classes $(-\infty, 0]$ and $(0,\infty)$ to get a two-point quotient space $F$. The closure of the singleton $\{(0,\infty)\}$ is $F$, hence its preimage is $ℝ$, but the closure of the preimage of $\{(0,\infty)\}$ is $[0,\infty)$. (Note that $F$ is the Sierpiński space.)

Answer 5

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 $1$.