## 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 beopenif, 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 here, *UH* is open in *G*. Since is open, so is *p*(*U*) by definition of quotient topology. ♦

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

–

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
*f*:*X*→*Y*and*g*:*Y*→*Z*are open, then so is*gf*:*X*→*Z*. - If
*f*:*X*→*Y*is open and , then restricting to*f*:*X*→*Z*still gives an open map since if*f*(*U*) is open in*Y*, then*f*(*U*) ∩*Z*=*f*(*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 is open whether we use the product or box topology. To prove that, use the fact that for any map we have
- If each is open, then the resulting which takes is also open. Again, its proof requires

**Exercises**

Generally, open continuous maps preserve local properties. For example, prove the following for an open continuous *f* : *X* → *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*).

**Answers (Highlight to Read)**

- 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.

- G/H is T2.
- G/H is T1.
- H is closed in G.

**Proof**.

Let *p* : *G* → *G*/*H* be the projection map, which is open by proposition 1. Now (1→2) is obvious. (2→3) follows from 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 is closed in *G*/*H* × *G*/*H*. Now its complement is:

Let *q* = (*p*, *p*) : *G* × *G* → *G*/*H* × *G*/*H*, which is open because *p* is. Then is open since the continuous map gives us

Since *q* is surjective, we have 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 and inverse map are continuous.

**Proof**.

First, we prove a general result:

Algebraic Quotient Lemma. Suppose is a collection of quotient maps which are open. Let be the projection map which takes .Then q is a quotient map. [ Compare this result with the warning after example 2 here. ]

**Proof of AQL**.

We need to show that is open if and only if is open. (→) is obvious since *q* is continuous. For (←), suppose is open; we have since *q* is surjective. Since each *p _{i}* is open, so is

*q*. And since

*q*is surjective 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 *q*: *G* × *G* → *G*/*N* × *G*/*N* is a quotient map too. Thus, showing *m’* is continuous is equivalent to showing *m’q* : *G* × *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 is a continuous homomorphism of topological groups, this induces a continuous injective map

**Proof**. The only non-trivial part is continuity. If *p*: *G* → *G*/ker(*f*) is the projection map, then *gp* = *f* is continuous, so by the universal property of quotient topology, *g* 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 where

Next, recall (proposition 7 here) that the connected component of 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 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 which contains points from more than one connected component, so it is disconnected, i.e. for some non-empty disjoint open subsets *U*, *U’* 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 as a disjoint union; since <*x*> is connected and we must have Thus *U* and *U’* are both unions of connected components of *X*.

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

- ;
- : indeed, if it contains
*y*, then pick such that*p*(*x*) =*p*(*x’*) =*y*. Then so they’re in*U*,*U’*respectively. Since*p*(*x*) =*p*(*x’*),*x*and*x’*must belong to the same connected component, which contradicts

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

- Consider
**Q**as a topological subgroup of**R**. The quotient**R**/**Q**has the coarsest topology. - Take . The map which takes has kernel equal to
**Z**. By theorem 5 above, this gives a bijective continuous homomorphism**R**/**Z**→*S*^{1}. Since the image is Hausdorff, if we could prove**R**/**Z**is compact, then we’d have shown that**R**/**Z**and*S*^{1}are isomorphic topological groups. But then**R**/**Z**is the continuous image of the composition [0, 1] →**R**→**R**/**Z**. Case closed. - 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. - Take . The same reasoning holds as before and one can show that
**R**^{2}/**Z**^{2 }is isomorphic to*S*^{1}×*S*^{1}. - Finally, take The determinant map GL
_{n}(**R**) →**R*** induces a bijective continuous homomorphism GL_{n}(**R**)/SL_{n}(**R**) →**R***. Now we can’t use the same trick since the groups aren’t compact. Instead, we compose:**R*** → GL_{n}(**R**) → GL_{n}(**R**)/SL_{n}(**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.