## 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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