Exercise TG.5

Proof. Now, for $\alpha \in G$, define the function $h_{\alpha }:G∕H\to G∕H$ by mapping the left coset $\mathit{xH}\in G∕H$ to $f_{\alpha }(x)H=(\alpha \cdot x)H$. We note that if $\mathit{xH}=\mathit{yH}$ for $x,y\in G$ then, of course, $y\in \mathit{yH}=\mathit{xH}$ so that $y=x\cdot h$ for some $h\in H$. Then

$$\begin{array}{cc}{f}_{\alpha }(y)=\alpha \cdot y=\alpha \cdot (x\cdot h)=(\alpha \cdot x)\cdot h=f_{\alpha }(x)\cdot h& \end{array}$$

so that $f_{\alpha }(y)\in f_{\alpha }(x)H$, which suffices to show that $f_{\alpha }(x)H=f_{\alpha }(y)H$ since $G∕H$ is a partition. Hence $h_{\alpha }(\mathit{yH})=f_{\alpha }(y)H=f_{\alpha }(x)H=h_{\alpha }(\mathit{xH})$ so that the mapping $h_{\alpha }$ is a well-defined function.

To show that $h_{\alpha }$ is a homeomorphism, we first show that it is a bijection. Suppose that $\mathit{xH}$ and $\mathit{yH}$ are left cosets where $h_{\alpha }(\mathit{xH})=h_{\alpha }(\mathit{yH})$. Then we have $h_{\alpha }(\mathit{xH})=f_{\alpha }(x)H=f_{\alpha }(y)H=h_{\alpha }(\mathit{yH})$, and hence $f_{\alpha }(x)\in f_{\alpha }(y)H$. From this it follows that

$$\begin{array}{cc}{f}_{\alpha }(x)=f_{\alpha }(y)\cdot h=(\alpha \cdot y)\cdot h=\alpha \cdot (y\cdot h)=f_{\alpha }(y\cdot h)& \end{array}$$

for some $h\in H$. Since $f_{\alpha }$ is injective (since it was shown in Exercise .4 to be bijective), it has to be that $x=y\cdot h$ so that $x\in \mathit{yH}$. Since also $x\in \mathit{xH}$ and $G∕H$ is a partition, it must be that $\mathit{xH}=\mathit{yH}$, which shows that $h_{\alpha }$ is injective. Now consider any coset $\mathit{yH}\in G∕H$. Since $f_{\alpha }$ is a surjection we have that there is an $x\in G$ where $y=f_{\alpha }(x)$. Thus it immediately follows that $h_{\alpha }(\mathit{xH})=f_{\alpha }(x)H=\mathit{yH}$, which shows that $h_{\alpha }$ is surjective since $\mathit{yH}$ was arbitrary. This completes the proof that $h_{\alpha }$ is a bijection.

Next we digress a bit and show that $\bigcup <!--\; FUNCTION\; APPLICATION\; -->h_{\alpha }(H)=f_{\alpha }\left(\bigcup H\right)$ for any subset $H\subset G∕H$, where we use the notation $\bigcup <!--\; FUNCTION\; APPLICATION\; -->A={\bigcup }_{A\in A}A$ for a collection of sets $A$. So first consider any $x_0\in \bigcup <!--\; FUNCTION\; APPLICATION\; -->h_{\alpha }(H)$ so that there is a coset $\mathit{yH}\in h_{\alpha }(H)$ where $x_0\in \mathit{yH}$. Then, since $\mathit{yH}\in h_{\alpha }(H)$, there is another coset $\mathit{xH}\in H$ where $x_0\in \mathit{yH}=h_{\alpha }(\mathit{xH})=f_{\alpha }(x)H$. We then have that

$$\begin{array}{cc}{x}_0=f_{\alpha }(x)\cdot h=(\alpha \cdot x)\cdot h=\alpha \cdot (x\cdot h)=f_{\alpha }(x\cdot h)& \end{array}$$

for some $h\in H$. Since clearly $x\cdot h\in \mathit{xH}$ and $\mathit{xH}\in H$, we have that $x\cdot h\in \bigcup <!--\; FUNCTION\; APPLICATION\; -->H$. Then of course $x_0\in f_{\alpha }(\bigcup <!--\; FUNCTION\; APPLICATION\; -->H)$ since $x_0=f_{\alpha }(x\cdot h)$, which shows that $\bigcup <!--\; FUNCTION\; APPLICATION\; -->h_{\alpha }(H)\subset f_{\alpha }\left(\bigcup H\right)$ since $x_0$ was arbitrary. Now consider $x_0\in f_{\alpha }(\bigcup <!--\; FUNCTION\; APPLICATION\; -->H)$ so that there is a $y_0\in \bigcup <!--\; FUNCTION\; APPLICATION\; -->H$ where $x_0=f_{\alpha }(y_0)$. Then also there is a coset $\mathit{xH}\in H$ where $y_0\in \mathit{xH}$ since $y_0\in \bigcup <!--\; FUNCTION\; APPLICATION\; -->H$. Hence $y_0=x\cdot h$ for some $h\in H$. Define the coset $\mathit{yH}=h_{\alpha }(\mathit{xH})=f_{\alpha }(x)H$ and we have

$$\begin{array}{cc}{x}_0=f_{\alpha }(y_0)=\alpha \cdot y_0=\alpha \cdot (x\cdot h)=(\alpha \cdot x)\cdot h=f_{\alpha }(x)\cdot h& \end{array}$$

so that $x_0\in f_{\alpha }(x)H=\mathit{yH}$. Then, since $\mathit{yH}=h_{\alpha }(\mathit{xH})$ and $\mathit{xH}\in H$, we have that $\mathit{yH}\in h_{\alpha }(H)$. As we also have $x_0\in \mathit{yH}$, it follows that $x_0\in \bigcup <!--\; FUNCTION\; APPLICATION\; -->h_{\alpha }(H)$. This shows that $\bigcup <!--\; FUNCTION\; APPLICATION\; -->h_{\alpha }(H)\supset f_{\alpha }\left(\bigcup H\right)$ since $x_0$ was arbitrary, which completes the proof that $\bigcup <!--\; FUNCTION\; APPLICATION\; -->h_{\alpha }(H)=f_{\alpha }\left(\bigcup H\right)$.

To return to the main goal, we, therefore, have that

$$\begin{array}{lllllll}\hfill U\text{ is open in }G∕H& \iff \bigcup U\text{ is open in }G& \hfill \text{(by the definition of the quotient space)}& & \hfill & & \hfill \\ \hfill & \iff f_{\alpha }\left(\bigcup U\right)=\bigcup h_{\alpha }(U)\text{ is open in }G& \hfill \text{(since }{f}_{\alpha }\text{ is a homeomorphism)}& & \hfill & & \hfill \\ \hfill & \iff h_{\alpha }(U)\text{ is open in }G∕H,& \hfill \text{(by the definition of the quotient space)}& & \hfill & & \hfill \end{array}$$

noting that $f_{\alpha }$ was shown to be a homeomorphism in Exercise .4. This shows that $h_{\alpha }$ is a homeomorphism as desired.

Lastly, to show that $G∕H$ is homogeneous, consider two cosets $\mathit{xH},\mathit{yH}\in G∕H$. We know from what was shown in Exercise TG.4 that there is an $f_{\alpha }$ such that $y=f_{\alpha }(x)$. If $h_{\alpha }$ is the homeomorphism on $G∕H$ induced by $f_{\alpha }$ as defined above then we have $h_{\alpha }(\mathit{xH})=f_{\alpha }(x)H=\mathit{yH}$. This suffices to show that $G∕H$ is homogeneous since $\mathit{xH}$ and $\mathit{yH}$ were arbitrary. □

Proof. Define the single-point subset $H_0=\left\{H\right\}$ of $G∕H$, noting that we showed above why $H\in G∕H$. We have that $H_0$ is closed in $G∕H$ since we know that $p^{-1}(H_0)=\bigcup <!--\; FUNCTION\; APPLICATION\; -->H_0=H$ is closed in $G$, which follows from the alternative definition of a quotient map. Now consider any arbitrary one-point subset $H=\left\{\mathit{xH}\right\}\subset G∕H$. Since it was shown in part (a) that $G∕H$ is a homogeneous space, there is a homeomorphism $h_{\alpha }:G∕H\to G∕H$ that maps $H$ to $\mathit{xH}$. Then we clearly have that $h_{\alpha }(H_0)=h_{\alpha }(\left\{H\right\})=\left\{\mathit{xH}\right\}=H$. Since $H_0$ is closed in $G∕H$ and $h_{\alpha }$ is a homeomorphism, it follows that $h_{\alpha }(H_0)=H$ is also closed in $G∕H$. This shows the desired result since $H$ was an arbitrary single-point subset. □

Proof. Let $g_{\alpha }:G\to G$ be the function defined by $g_{\alpha }(x)=x\cdot \alpha$ for $\alpha ,x\in G$, which we know is a homeomorphism for any $\alpha \in G$ by Exercise .4. Consider any open set $U$ of $G$.

We first show that $p^{-1}(p(U))={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{h\in H}{g}_h(U)$.

$\text{(⊂)}$ Consider arbitrary $x\in p^{-1}(p(U))$ so that $p(x)\in p(U)$. Of course, $p(x)=\mathit{xH}$ so that $\mathit{xH}\in p(U)$ so that there is a $y\in U$ where $\mathit{xH}=p(y)=\mathit{yH}$. From this it follows that $x\in \mathit{yH}$ so that $x=y\cdot h_0$ for some $h_0\in H$, and so $x=g_{h_0}(y)$. Since $y\in U$ we have that $x\in g_{h_0}(U)$, and thus of course $x\in {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{h\in H}{g}_h(U)$ since $h_0\in H$. This shows that $p^{-1}(p(U))\subset {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{h\in H}{g}_h(U)$ since $x$ was arbitrary.

$\text{(⊃)}$ Now consider $x\in {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{h\in H}{g}_h(U)$ so that there is an $h_0\in H$ where $x\in g_{h_0}(U)$. Hence $x=g_{h_0}(y)$ for some $y\in U$, and so $x=y\cdot h_0$. This shows that $x\in \mathit{yH}$ since $h_0\in H$, and thus it must be that $\mathit{xH}=\mathit{yH}$. However, we have that $\mathit{xH}=\mathit{yH}=p(y)$ and $y\in U$ so that $\mathit{xH}\in p(U)$. Moreover $\mathit{xH}=p(x)$ so that $p(x)\in p(U)$ and hence $x\in p^{-1}(p(U))$. This shows that $p^{-1}(p(U))\supset {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{h\in H}{g}_h(U)$, which shows the desired result.

Now, since each $g_h$ is a homeomorphism for $h\in H$ and $U$ is open in $G$, it follows that each $g_h(U)$ is also open in $G$. Then of course their union ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{h\in H}{g}_h(U)=p^{-1}(p(U))$ is open in $G$ by the definition of a topology. Since $p^{-1}(p(U))$ is open in $G$, it follows that $p(U)$ is open in $G∕H$ since $p$ is a quotient map. Then, since $U$ was an arbitrary open set of $G$, this proves that $p$ is an open map. □

Proof. First, we need to show that we can define an operation on $G∕H$ that makes it into a group. This is done in an expected way: for $\mathit{xH},\mathit{yH}\in G∕H$ define $\mathit{xH}\cdot \mathit{yH}=(x\cdot y)H$. To show that this operation is well-defined, suppose that $x_{0}H=x_{1}H$ and $y_{0}H=y_{1}H$ are elements of $G∕H$. Then of course $x_1\in x_{0}H$ so that $x_1=x_0\cdot h_x$ for some $h_x\in H$. Similarly $y_1\in y_{0}H$ so that $y_1=y_0\cdot h_y$ for some $h_y\in H$. Freely utilizing the associativity of the operation of $G$ and suppressing the $\text{⋅}$ by using multiplication notation, we then have that

$$\begin{array}{llllll}\hfill x_1{y}_1& =(x_0{h}_x)(y_0{h}_y)& \hfill & & \hfill & \\ \hfill (x_1{y}_1){y_0}^{-1}& =(x_0{h}_x)(y_0{h}_y){y_0}^{-1}& \hfill & & \hfill & \\ \hfill x_1{y}_1{y_0}^{-1}& =x_0{h}_x(y_0{h}_y{y_0}^{-1})& \hfill & & \hfill & \\ \hfill x_1{y}_1{y_0}^{-1}& =x_0{h}_x{h}_1& \hfill \text{(where }{h}_1=y_0{h}_y{y_0}^{-1}\in H\text{ since }H\text{ is normal)}& & \hfill & & \hfill \\ \hfill x_1{y}_1{y_0}^{-1}& =x_0{h}_2& \hfill \text{(where }{h}_2=h_x{h}_1\in H\text{ since }H\text{ is a group)}& & \hfill & & \hfill \\ \hfill x_1{y}_1{y_0}^{-1}{x_0}^{-1}& =x_0{h}_2{x_0}^{-1}& \hfill & & \hfill & \\ \hfill x_1{y}_1{y_0}^{-1}{x_0}^{-1}& =h_3& \hfill \text{(where }{h}_3=x_0{h}_2{x_0}^{-1}\in H\text{ since }H\text{ is normal)}& & \hfill & & \hfill \end{array}$$

Hence $(x_1{y}_1)({y_0}^{-1}{x_0}^{-1})=x_1{y}_1{y_0}^{-1}{x_0}^{-1}\in H$ so that also $({y_0}^{-1}{x_0}^{-1})(x_1{y}_1)\in H$ by the equivalent definition of a normal subgroup. Therefore, for some $h\in H$, we have

$$\begin{array}{cc}{y_0}^{-1}{x_0}^{-1}{x}_1{y}_1=({x_0}^{-1}{y_0}^{-1})(x_1{y}_1)=h& \\ {x_0}^{-1}{x}_1{y}_1=y_{0}h& \\ x_1{y}_1=x_0{y}_{0}h& \\ x_1{y}_1=(x_0{y}_0)h& \end{array}$$

so that $x_1\cdot y_1\in (x_0\cdot y_0)H$, which of course shows that $(x_1\cdot y_1)H=(x_0\cdot y_0)H$, and hence the operation on $G∕H$ is well-defined.

Now we show that $G∕H$ with this operation satisfies the group axioms. We first note that, for $x,y\in G$, we have $\mathit{xH},\mathit{yH}\in G∕H$ and $x\cdot y\in G$ since $G$ is a group so that $\mathit{xH}\cdot \mathit{yH}=(x\cdot y)H\in G∕H$. Hence the operation is closed in $G∕H$. Next, clearly $\mathit{eH}=H$ itself is the identity element for $G∕H$ since we have $\mathit{eH}\cdot \mathit{xH}=(e\cdot x)H=\mathit{xH}$ and $\mathit{xH}\cdot \mathit{eH}=(x\cdot e)H=\mathit{xH}$ for any $\mathit{xH}\in G∕H$. We also have that the inverse element of $\mathit{xH}$ is $x^{-1}H$ since $\mathit{xH}\cdot x^{-1}H=(x\cdot x^{-1})H=\mathit{eH}$ and $x^{-1}H\cdot \mathit{xH}=(x^{-1}\cdot x)H=\mathit{eH}$. Lastly, we have that

$$\begin{array}{llll}\hfill (\mathit{xH}\cdot \mathit{yH})\cdot \mathit{zH}& =(x\cdot y)H\cdot \mathit{zH}=((x\cdot y)\cdot z)H=(x\cdot (y\cdot z))H& \hfill & \\ \hfill & =\mathit{xH}\cdot (y\cdot z)H=\mathit{xH}\cdot (\mathit{yH}\cdot \mathit{zH})& \hfill & \end{array}$$

since of course, the operation on $G$ is associative. This shows that the operation on $G∕H$ is associative as well, which completes the proof that $G∕H$ is a group.

To show that $G∕H$ is in fact a topological group, first, it was shown in part (b) that one-point subsets of $G∕H$ are closed in $G∕H$ since $H$ is closed in $G$. From this is follows that $G∕H$ satisfies the $T_1$ axiom since any finite subset of $G∕H$ is a finite union of one-point sets and so is also closed in $G∕H$.

At this point we take a short digression and show that if $f:G\times G\to G$ is continuous, then the function $h:G∕H\times G∕H\to G∕H$ defined by $h(\mathit{xH},\mathit{yH})=f(x,y)H$ is also continuous. To see this, we first claim that $h\circ (p\times p)=p\circ f$, where the function $p\times p:G\times G\to G∕H\times G∕H$ is defined as $(p\times p)(x,y)=(p(x),p(y))$ as in Exercise 18.10. This is easy to show as we have

$$\begin{array}{llll}\hfill (h\circ (p\times p))(x,y)& =h((p\times p)(x,y))=h(p(x),p(y))=h(\mathit{xH},\mathit{yH})& \hfill & \\ \hfill & =f(x,y)H=p(f(x,y))=(p\circ f)(x,y)& \hfill & \end{array}$$

for any $x,y\in G$. Since both $p$ and $f$ are continuous, it follows that $p\circ f=h\circ (p\times p)$ is also continuous by Theorem 18.2 part (c). We also have that $p\times p$ is an open quotient map by the remarks in the text since $p$ is an open quotient map by part (c).

At this point, we use Theorem 22.2 to show that $h$ is continuous. As this can be confusing, we include the following table, which shows how the sets and functions in the statement of Theorem 22.2 map to the sets and functions we are working with:

Now we show that the conditions of the theorem are met. We have already shown that $p\times p$ is a quotient map. Now let $P=\left\{(\mathit{xH},\mathit{yH})\right\}$ be a one-point subset of $G∕H\times G∕H$. Since $p\times p$ is a quotient map, it is surjective so that $(p\times p)({(p\times p)}^{-1}(P))=P$ by Exercise 2.1. Then clearly

$$\begin{array}{llll}\hfill (p\circ f)({(p\times p)}^{-1}(P))& =(h\circ (p\times p))({(p\times p)}^{-1}(P))& \hfill & \\ \hfill & =h((p\times p)({(p\times p)}^{-1}(P)))=h(P)& \hfill & \\ \hfill & =\left\{h(\mathit{xH},\mathit{yH})\right\},& \hfill & \end{array}$$

which shows that $p\circ f$ is constant on the set ${(p\times p)}^{-1}(P)$. Thus $p\circ f$ induces the function $h$ per Theorem 22.2 since $h\circ (p\times p)=p\circ f$ as shown above. It then follows by the theorem that $h$ is continuous since we have shown above that $p\circ f$ is.

Returning to the main problem, since $G$ is a topological group the function $f:G\times G\to G$ defined by $f(x,y)=x\cdot y^{-1}$ is continuous by Exercise .1. It then follows by what was just shown that $h:G∕H\times G∕H\to G∕H$ defined by

$$\begin{array}{cc}h(\mathit{xH},\mathit{yH})=\mathit{xH}\cdot {(\mathit{yH})}^{-1}=\mathit{xH}\cdot y^{-1}H=(x\cdot y^{-1})H=f(x,y)H& \end{array}$$

is also continuous. This suffices to show that $G∕H$ is also a topological group as desired, again by Exercise TG.1. □