Topology for Coset Space

This is really a continuation from the previous article. Let G be a topological group and H a subgroup of G. The collection of left cosets G/H is then given the quotient topology. This quotient space, however, satisfies an additional property.

Definition. A map f : X → Y between two topological spaces is said to be open if, for any open subset U of X, f(U) is open in Y.

Proposition 1. The map p : G → G/H is open.

Proof. Let U be an open subset of G. From proposition 2 hereUH is open in G. Since $p^{-1}(p(U)) = UH$ is open, so is p(U) by definition of quotient topology. ♦

In general, not every quotient map pX → X/~ is open. For example, glue the endpoints of I = [0, 1] together and form the quotient map $p:I\to S^1.$ Then U = (1/2, 1] is open in I but p(U) is not open in S1.

We should say something about open maps since this is our first encounter with them. Don’t worry too much for now – they don’t appear all that often.

Properties of Open Maps

• The identity map on any X is open.
• If fX → Y and gY → Z are open, then so is gfX → Z.
• If f : X → Y is open and $f(X)\subseteq Z\subseteq Y$, then restricting to f : X → Z still gives an open map since if f(U) is open in Y, then f(U) ∩ Zf(U) is open in Z.
• Restricting the domain doesn’t give an open map in general: e.g. if we restrict id: R → R to the closed interval [0, 1], the image of [0, 1] is not open. However, restricting f : X → Y to an open subset U of X gives an open map U → Y.
• The projection map $\prod_i X_i \to X_i$ is open whether we use the product or box topology. To prove that, use the fact that for any map $f:A\to B,$ we have $f(\cup_i A_i) = \cup_i f(A_i).$
• If each $f_i : X_i \to Y_i$ is open, then the resulting $f:\prod_i X_i \to \prod_i Y_i$ which takes $(x_i) \mapsto (f_i(x_i))$ is also open. Again, its proof requires $f(\cup_i A_i) = \cup_i f(A_i).$

Exercises

Generally, open continuous maps preserve local properties. For example, prove the following for an open continuous fX → Y.

• If X is locally connected, then so is f(X).
• If X is locally path-connected, then so is f(X).
• Let X and Y be Hausdorff. If X is locally compact, then so is f(X).

• If y is in f(X), pick x in X such that f(x)=y. Then x is contained in some connected open subset U of X. So f(x) lies in f(U) which is an open and connected subset of Y and hence f(X).
• Same as connected.
• If y is in f(X), pick x in X such that f(x)=y. Then x is contained in some open subset U of X whose closure cl(U) is compact. So f(x) is contained in f(U), which is open in Y and hence f(X). Also f(cl(U)) is compact and hence closed in f(X), so the closure of f(U) in f(X) is contained in f(cl(U)) and thus compact also.

Properties of Coset Space

Let H be a subgroup of the topological group G. The first result is straightforward.

Theorem 2. G/H is discrete if and only if H is open in G.

Proof.

G/H is discrete iff every singleton subset is open; by the definition of quotient topology, this holds iff every left coset gH is open in G, which holds iff H is open in G. ♦

The next result is trickier.

Theorem 3. The following are equivalent.

1. G/H is T2.
2. G/H is T1.
3. H is closed in G.

Proof.

Let pG → G/H be the projection map, which is open by proposition 1. Now (1→2) is obvious. (2→3) follows from $H = p^{-1}(\{e\});$ if G/H is T1, then {e} is closed in G/H, hence H is closed in G.

Finally, suppose H is closed in G; to show that G/H is Hausdorff, recall that it suffices to show that the image of the diagonal map $\Delta :G/H \to G/H \times G/H$ is closed in G/H × G/H. Now its complement is:

$U := (G/H \times G/H) - \text{im}(\Delta) = \{(p(g), p(g')) : g^{-1}g' \not\in H\}.$

Let q = (p, p) : G × G → G/H × G/H, which is open because p is. Then $q^{-1}(U) = \{(g, g')\in G\times G : g^{-1}g'\not \in H\}$ is open since the continuous map $f: G\times G\to G, (x, y)\mapsto x^{-1}y$ gives us $q^{-1}(U) = f^{-1}(G-H).$

Since q is surjective, we have $U = q(q^{-1}(U))$ which is open since q is open. ♦

Group Quotient

Now suppose N is a normal subgroup of the topological group G. It turns out the group operations for G/N are continuous with respect to the quotient topology, i.e. G/N is also a topological group.

Theorem 4. The product map $m':G/N \times G/N \to G/N$ and inverse map $i':G/N \to G/N$ are continuous.

Proof.

First, we prove a general result:

Algebraic Quotient Lemma. Suppose $p_i : X_i\to Y_i$ is a collection of quotient maps which are open. Let $q = (p_i) : \prod_i X_i \to \prod_i Y_i$ be the projection map which takes $(x_i) \mapsto (p_i(x_i))$.

Then q is a quotient map. [ Compare this result with the warning after example 2 here. ]

Proof of AQL.

We need to show that $V\subseteq \prod_i Y_i$ is open if and only if $q^{-1}(V)$ is open. (→) is obvious since q is continuous. For (←), suppose $q^{-1}(V)$ is open; we have $q(q^{-1}(V)) = V$ since q is surjective. Since each pi is open, so is q. And since q is surjective $q(q^{-1}(V))=V,$ which must be open. ♦

We resume our proof of theorem 4 to show m’ is continuous (i’ is easy). Applying it to G → G/N, we see that the projection qG × G → G/N × G/N is a quotient map too. Thus, showing m’ is continuous is equivalent to showing m’qG × G → G/N is continuous. But m’q is obtained via composing the product map G × G → G with projection G → G/N. Thus, we’re done. ♦

Theorem 5. The quotient group G/N is discrete if and only if N is open in G. Also, G/N is Hausdorff if and only if N is closed in G.

Proof.

Follows immediately from theorem 3 above. ♦

Corresponding to the first isomorphism theorem for groups, we have:

Theorem 6. If $f:G\to H$ is a continuous homomorphism of topological groups, this induces a continuous injective map $g:G/\text{ker}(f)\to H.$

Proof. The only non-trivial part is continuity. If pG → G/ker(f) is the projection map, then gpf is continuous, so by the universal property of quotient topologyg is continuous. ♦

In general, G/ker(f) doesn’t have the subspace topology from H. For example, f can be an injective continuous map which does not provide G with the subspace topology from H.

E.g. take $\mathbf{Z}\to S^1,$ where $m \mapsto (\cos(m\sqrt 2), \sin(m\sqrt 2)).$

Next, recall (proposition 7 here) that the connected component of $e\in G$ is a closed normal subgroup N of G.

Theorem 7. If N is the connected component of e, then G/N is totally disconnected.

Proof.

For each g in G, left-multiplication $l_g:G\to G, x \mapsto gx$ is a homeomorphism. Thus, the connected components of G are the cosets gN, for various g. More generally, we’ll prove:

Lemma. In a topological space X, denote x ~ y if they belong to the same connected component. If the projection map p : X → X/~ is open, then X/~ is totally disconnected.

Proof of Lemma.

Suppose there’s a connected subset Y of X/~ comprising of more than a point. Let $Z := p^{-1}(Y)$ which contains points from more than one connected component, so it is disconnected, i.e. $Z=U\cup U'$ for some non-empty disjoint open subsets UU’ of Z. Now, if Z contains x, then it contains the entire connected component <x> of x. We claim this holds for U as well. Indeed, we have $\left = (U\cap \left) \cup (U'\cap \left)$ as a disjoint union; since <x> is connected and $U\cap \left\ne\emptyset$ we must have $U\cap \left = \left \implies U\supseteq \left.$ Thus U and U’ are both unions of connected components of X.

Write UZ ∩ V and U’Z ∩ V’ for open subsets VV’ of X. Since p is open, p(V) and p(V’) are open subsets of X/~ satisfying:

• $p(V) \cup p(V') = p(V\cup V') \supseteq p(U\cup U') = p(p^{-1}(Y)) = Y$;
• $p(V) \cap p(V')\cap Y=\emptyset$: indeed, if it contains y, then pick $x\in V, x'\in V'$ such that p(x) = p(x’) = y. Then $x, x'\in p^{-1}(Y) = Z$ so they’re in UU’ respectively. Since p(x) = p(x’), x and x’ must belong to the same connected component, which contradicts $x\in V, x'\in V'.$

Hence, p(V) ∩ Y and p(V’) ∩ Y are disjoint non-empty open subsets of Y with union Y, which contradicts our assumption that Y is connected. ♦

Examples of Group Quotients

1. Consider Q as a topological subgroup of R. The quotient R/Q has the coarsest topology.
2. Take $\mathbf{Z}\subset \mathbf{R}$. The map $\exp: \mathbf{R}\to S^1$ which takes $t\mapsto (\cos(2\pi t), \sin(2\pi t))$ has kernel equal to Z. By theorem 5 above, this gives a bijective continuous homomorphism R/Z → S1. Since the image is Hausdorff, if we could prove R/Z is compact, then we’d have shown that R/Z and S1 are isomorphic topological groups. But then R/Z is the continuous image of the composition [0, 1] → R → R/Z. Case closed.
3. If G and H are topological groups, then so is G × H (with the product topology). The continuous bijective group homomorphism (G × H)/H → G is a homeomorphism since it’s open. Thus (G × H)/H and G are isomorphic topological groups.
4. Take $\mathbf{Z}^2\subset \mathbf{R}^2$. The same reasoning holds as before and one can show that R2/Zis isomorphic to S1 × S1.
5. Finally, take $\text{SL}_n(\mathbf{R}) \subset \text{GL}_n(\mathbf{R}).$ The determinant map GLn(R) → R* induces a bijective continuous homomorphism GLn(R)/SLn(R) → R*. Now we can’t use the same trick since the groups aren’t compact. Instead, we compose: R* → GLn(R) → GLn(R)/SLn(R) → R*, where the first map takes c to the diagonal matrix with entries (c, 1, …, 1). This gives the identity map on R*; since the first two maps are continuous, the last map is open.
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