Note that for each $x$ , $x=x\cdot e\in xH$ , and $xH=yH$ iff $y=x\cdot z$ for some $z\in H$ .

(a) Let $p:G\rightarrow G/H$ be the quotient map. Then, $p_{\alpha}=p\circ f_{\alpha}:G\rightarrow G/H$ is a quotient map as the composite of a homeomorphism with a quotient map. It maps every $x\in G$ to $\alpha xH\in G/H$ . Now, $\alpha xH=\alpha yH$ iff $\alpha\cdot x\cdot z=\alpha\cdot y$ for some $z\in H$ iff $y=x\cdot z$ for some $z\in H$ iff $xH=yH$ , therefore, $p_{\alpha}$ is constant on sets $p^{-1}(xH)$ . By Corollary 22.3, $p_{\alpha}$ induces a homeomorphism $h_{\alpha}:G/H\rightarrow G/H$ carrying each $xH$ to $\alpha xH$ . Moreover, $h_{y\cdot x^{-1}}(xH)=yH$ .

(b) $H$ is closed, $f_{\alpha}$ is a homeomorphism, therefore, $xH=f_{x}(H)$ is closed in $G$ , and $\{xH\}$ is closed in $G/H$ .

(c) Let $U$ be open in $G$ . Since $g_{\alpha}$ is a homeomorphism, $p(U)=$ $\cup_{x\in U}xH$ consists of elements $\cup_{x\in U}\cup_{h\in H}x\cdot h=$ $\cup_{h\in H}g_{h}(U)$ , which is open in $G$ , hence, $p(U)$ is open in $G/H$ .

(d) The operation is defined on $G/H$ as follows: $xH\cdot yH=(x\cdot y)H$ which we denote simply as $xyH$ . This operation is well-defined as if $xH=x'H$ and $yH=y'H$ , then for some $z,w\in H$ , $x\cdot y=x'\cdot z\cdot y'\cdot w$ , and, since $H$ is normal, $z\cdot y'=y'\cdot z'$ for some $z'\in H$ , therefore, $x\cdot y=x'\cdot y'\cdot(z'\cdot w)$ and $xyH=x'y'H$ . Note, that $eH$ is the identity element of $G/H$ , $\cdot$ is associative, and $(xH)^{-1}=x^{-1}H$ , hence, $(G/H,\cdot)$ is a group. Using (b), $G/H$ is also a $T_{1}$ -space. Using Exercise 1, it is sufficient to show that $h(xH,yH)=(xH)\cdot(yH)^{-1}$ is continuous. Now, let $f(x,y)=x\cdot y^{-1}$ , which is continuous by Exercise 1 again. $h(xH,yH)=f(x,y)H$ , therefore, $h\circ(p\times p)=p\circ f$ . $p\circ f$ is continuous as a composite of continuous maps, moreover, by (c), $p$ is an open quotient map so that $p\times p$ is an open quotient map as well, therefore, using Theorem 22.2, $p\circ f$ induces the continuous function $h$ via the quotient $p\times p$ .