Exercise TG.7

Proof. Let $V$ be any neighborhood of $y=f(x,x)$ in $Y$. Then there is a neighborhood $U^{\prime }$ of $(x,x)$ in $X\times X$ such that $f(U^{\prime })\subset V$ by Theorem 18.1 part (4). Now, since $U^{\prime }$ is an open set of $X\times X$ containing $(x,x)$, there is a basis element $B=U_1\times U_2$ of $X\times X$ such that $(x,x)\in U_1\times U_2=B\subset U^{\prime }$, where of course $U_1$ and $U_2$ are open in $X$. Then, being a finite intersection of open sets, $U=U_1\cap U_2$ is also open in $X$ and we have $x\in U$ since $x\in U_1$ and $x\in U_2$. Hence $U$ is a neighborhood of $x$ in $X$, and of course both $U\subset U_1$ and $U\subset U_2$.

Now consider any $z\in f(U\times U)$ so that there is an $(x_1,x_2)\in U\times U$ where $f(x_1,x_2)=z$. Then $x_1\in U\subset U_1$ and $x_2\in U\subset U_2$ so that $(x_1,x_2)\in U_1\times U_2=B\subset U^{\prime }$. Hence $z=f(x_1,x_2)\in f(U^{\prime })$ so that also $z\in V$ since $f(U^{\prime })\subset V$. This shows the desired result that $f(U\times U)\subset V$ since $z$ was arbitrary. □

Proof. Let $e$ be the identity element of the group. Then we have

which shows the desired result. □

Proof. For $\alpha \in G$, of course $\alpha \cdot U$ denotes the set $\left\{\alpha \cdot x\mid x\in U\right\}$, and analogously $U\cdot \alpha =\left\{x\cdot \alpha \mid x\in U\right\}$. The openness of $\alpha \cdot U$ and $U\cdot \alpha$ follow almost immediately from what was shown in Exercise TG.4. It is trivial to show that $\alpha \cdot U=f_{\alpha }(U)$, where $f_{\alpha }:G\to G$ is defined by $f_{\alpha }(x)=\alpha \cdot x$ as in Exercise TG.4. We know from that exercise that $f_{\alpha }$ is a homeomorphism so that $\alpha \cdot U=f_{\alpha }(U)$ is open since $U$ is. Analogously $U\cdot \alpha =g_{\alpha }(U)$ is also open for the same reason, where $g_{\alpha }(x)=x\cdot \alpha$ as in Exercise TG.4. If $C$ is a closed set of $G$ then $\alpha \cdot C=f_{\alpha }(C)$ and $C\cdot \alpha =g_{\alpha }(C)$ are also closed since the homeomorphisms $f_{\alpha }$ and $g_{\alpha }$ also of course preserve closed sets. □

Proof. First we show that $U\cdot U^{-1}$ is indeed a neighborhood of $e$. We claim that

We have

which of course shows the desired result. Now, we know from Lemma 3 that each $U\cdot \alpha$ is open since $U$ is open (being a neighborhood of $e$). Hence their union is open by the definition of a topology, which shows that $U\cdot U^{-1}$ is in fact open. Also, since $e\in U$ and $e^{-1}=e$ (a well-known property of the identity element in any group), we clearly have that $e=e\cdot e=e\cdot e^{-1}\in U\cdot U^{-1}$ and therefore $U\cdot U^{-1}$ is a neighborhood of $e$.

To show that $U\cdot U^{-1}$ is symmetric, we have

which shows that ${(U\cdot U^{-1})}^{-1}=U\cdot U^{-1}$ so that $U\cdot U^{-1}$ is symmetric by definition. □

Proof. Suppose that $U$ is any neighborhood of $e$. Since $G$ is a topological group, we know that the function $f:G\times G\to G$ defined by $f(x,y)=x\cdot y$ is continuous. Then, since $f(e,e)=e\cdot e=e$, it follows from Lemma 1 that there is a neighborhood $V^{\prime }$ of $e$ such that $f(V^{\prime }\times V^{\prime })\subset U$. Now we claim that $f(V^{\prime }\times V^{\prime })=V^{\prime }\cdot V^{\prime }$. For any $z\in f(V^{\prime }\times V^{\prime })$ we have that there is an $(x,y)\in V^{\prime }\times V$ where $f(x,y)=x\cdot y=z$. Since $x,y\in V^{\prime }$ and $z=x\cdot y$, this shows that $z\in V^{\prime }\cdot V^{\prime }$ so that $f(V^{\prime }\times V^{\prime })\subset V^{\prime }\cdot V^{\prime }$. To show the other direction, for any $z\in V^{\prime }\cdot V^{\prime }$ we have that $z=x\cdot y$ for some $x,y\in V^{\prime }$. Then $(x,y)\in V^{\prime }\times V^{\prime }$ and $z=x\cdot y=f(x,y)$ so that $z\in f(V^{\prime }\times V^{\prime })$, which shows that $f(V^{\prime }\times V^{\prime })\supset V^{\prime }\cdot V^{\prime }$. This shows the desired result that $V^{\prime }\cdot V^{\prime }=f(V^{\prime }\times V^{\prime })\subset U$.

Similarly the function $g:G\times G\to G$ defined by $g(x,y)=x\cdot y^{-1}$ is also continuous by Exercise TG.1 since $G$ is a topological group. Then, since $V^{\prime }$ is a neighborhood of $e$ and $g(e,e)=e\cdot e^{-1}=e\cdot e=e$, it follows again from Lemma 1 that there is a neighborhood $W$ of $e$ such that $g(W\times W)\subset V^{\prime }$. We have that $W\cdot W^{-1}=g(W\times W)$ by an argument analogous to that for $f$ above so that $W\cdot W^{-1}=g(W\times W)\subset V^{\prime }$. Let $V=W\cdot W^{-1}\subset V^{\prime }$, which we know is a symmetric neighborhood of $e$ by Lemma 4 and is the neighborhood we seek.

So consider any $z\in V\cdot V$ so that $z=x\cdot y$ for some $x,y\in V$. Then also $x,y\in V^{\prime }$ since $V\subset V^{\prime }$. From this it follows that $z=x\cdot y\in V^{\prime }\cdot V^{\prime }$ so that also $z\in U$ since $V^{\prime }\cdot V^{\prime }\subset U$. This shows the desired result that $V\cdot V\subset U$ since $z$ was arbitrary, which completes the overall proof. □

Proof. Suppose that $x,y\in G$ and that $x\ne y$. Then there are neighborhoods $U_x^{\prime }$ of $x$ and $U_y^{\prime }$ of $y$ such that $y\notin U_x^{\prime }$ and $x\notin U_y^{\prime }$. This follows from Exercise 17.15 since $G$ satisfies the $T_1$ axiom on account of it being a topological group. Then $U_x=U_x^{\prime }\cdot x^{-1}$ and $U_y=U_y^{\prime }\cdot y^{-1}$ are both neighborhoods of $e$ since they are open by Lemma 3 and we have that $e=x\cdot x^{-1}\in U_x^{\prime }\cdot x^{-1}=U_x$ since $x\in U_x^{\prime }$, and analogously $e=y\cdot y^{-1}\in U_y^{\prime }\cdot y^{-1}=U_y$ since $y\in U_y^{\prime }$. Note that we also have that

so that clearly $U_y\cdot y=U_y^{\prime }$.

Now, let $U=U_x\cap U_y$, which is also obviously a neighborhood of $e$. It then follows from part (a) that there is a symmetric neighborhood $V$ of $e$ such that $V\cdot V\subset U$. Suppose that $V\cdot x$ and $V\cdot y$ are not disjoint so that there is a $z\in V\cdot x$ where also $z\in V\cdot y$. Then we have that $z=v_x\cdot x=v_y\cdot y$ for some $v_x,v_y\in V$. It then follows that $x=({v_x}^{-1}\cdot v_y)\cdot y$ and, since ${v_x}^{-1}\in V^{-1}=V$ as $V$ is symmetric, we have ${v_x}^{-1}\cdot v_y\in V\cdot V\subset U\subset U_y$. Thus $x=({v_x}^{-1}\cdot v_y)\cdot y\in U_y\cdot y=U_y^{\prime }$, but we know that $x$ cannot be in $U_y^{\prime }$ by its definition per the $T_1$ axiom! This contradiction means that it must be that $V\cdot x$ and $V\cdot y$ are disjoint, which shows the desired result.

From this the fact that $G$ is Hausdorff readily follows. Clearly $V\cdot x$ is a neighborhood of $x$ since it is open by Lemma 3 and we have $x=e\cdot x\in V\cdot x$. Similarly $V\cdot y$ is a neighborhood of $y$. As we have shown that these are disjoint, this suffices to show that $G$ is a Hausdorff space. □

Proof. A proof of this is similar to the proof of part (b). Since $A$ is closed, we know that $A\cdot x^{-1}$ is also closed by Lemma 3 . Then $G-A\cdot x^{-1}$ is of course open. Moreover if $e$ were in $A\cdot x^{-1}$ then we would have $e=a\cdot x^{-1}$ for some $a\in A$ so that $a=e\cdot x=x$, which is not possible since we know that $x\notin A$. So it must be that $e\notin A\cdot x^{-1}$ so that $e\in G-A\cdot x^{-1}$ since of course $e\in G$. Hence $G-A\cdot x^{-1}$ is a neighborhood of $e$, and thus there is a symmetric neighborhood $V$ of $e$ such that $V\cdot V\subset G-A\cdot x^{-1}$ by what was shown in part (a).

We claim that $V\cdot A$ and $V\cdot x$ are disjoint. To see this, suppose to the contrary that there is a $y\in V\cdot A$ where also $y\in V\cdot x$. Then $y=v_a\cdot a=v_x\cdot x$ for some $v_a,v_x\in V$ and $a\in A$. Then we have

Since $V$ is symmetric we have that ${v_a}^{-1}\in V^{-1}=V$ so that ${v_a}^{-1}\cdot v_x\in V\cdot V\subset G-A\cdot x^{-1}$ and hence ${v_a}^{-1}\cdot v_x\notin A\cdot x^{-1}$. However, clearly ${v_a}^{-1}\cdot v_x=a\cdot x^{-1}$ so that ${v_a}^{-1}\cdot v_x\in A\cdot x^{-1}$ since $a\in A$. This contradiction can only mean that in fact $V\cdot A$ and $V\cdot x$ are disjoint.

Now, for any $a\in A$, we have that $a=e\cdot a\in V\cdot A$ since $e\in V$, which shows that $A$ is contained in $V\cdot A$. Similar to what was done in the beginning of the proof of Lemma 4 , it is easy to show that

which is open since each $V\cdot a$ is open by Lemma 3 . Thus $V\cdot A$ is an open set containing $A$. We also have that $V\cdot x$ is a neighborhood of $x$ since it is open by Lemma 3 and $x=e\cdot x\in V\cdot x$ since $e\in V$. Since we have already shown that $V\cdot A$ and $V\cdot x$ are disjoint, this shows the desired result that $G$ satisfies the regularity axiom. □

Proof. Suppose that $A$ is a closed subset of $G∕H$ and $\mathit{xH}$ is an element of $G∕H$ not contained in $A$. Then $p^{-1}(A)$ is a closed subset of $G$ since $p$ is a quotient map. Moreover $p^{-1}(A)$ cannot contain $x$ for, if it did, then we would have $\mathit{xH}=p(x)\in A$. So let $V$ be a symmetric neighborhood of $e$ such that $V\cdot p^{-1}(A)$ and $V\cdot x$ are disjoint, as shown to exist in part (c).

It was also shown in part (c) that $V\cdot p^{-1}(A)$ is an open set in $G$ containing $p^{-1}(A)$ and $V\cdot x$ is a neighborhood of $x$ in $G$. So, for any $\mathit{yH}\in A$ we have that $p(y)=\mathit{yH}\in A$ so that $y\in p^{-1}(A)$ and hence also $y\in V\cdot p^{-1}(A)$. Then of course $\mathit{yH}=p(y)\in p(V\cdot p^{-1}(A))$, which shows that $A\subset p(V\cdot p^{-1}(A))$ since $\mathit{yH}$ was arbitrary. We also have that $p(V\cdot p^{-1}(A))$ is open in $G∕H$ since $V\cdot p^{-1}(A)$ is open in $G$ and $p$ is an open map by Exercise TG.5 part (c). Similarly, since $x\in V\cdot x$, we have that $\mathit{xH}=p(x)\in p(V\cdot x)$ and $p(V\cdot x)$ is open in $G∕H$ since $p$ is an open map. Thus $p(V\cdot x)$ is a neighborhood of $\mathit{xH}$ in $G∕H$.

Now we show that $p(V\cdot p^{-1}(A))$ and $p(V\cdot x)$ are disjoint subsets of $G∕H$. Suppose to the contrary that they are not so that there is a $\mathit{zH}\in p(V\cdot p^{-1}(A))$ where also $\mathit{zH}\in p(V\cdot x)$. Then we have that there is a $y_a\in V\cdot p^{-1}(A)$ where $z=p(y_a)$ and likewise a $y_x\in V\cdot x$ where $z=p(y_x)$. Thus $y_{a}H=p(y_a)=z=p(y_x)=y_{x}H$ so that $y_x\in y_{a}H$, and hence there is an $h\in H$ where $y_x=y_a\cdot h$. Also since $y_a\in V\cdot p^{-1}(A)$ we have $y_a=v_a\cdot a^{\prime }$ for some $v_a\in V$ and $a^{\prime }\in p^{-1}(A)$. Putting this together we have

Since $a^{\prime }\in p^{-1}(A)$ we have that $a^{\prime }H=p(a^{\prime })\in A$. Also clearly $a^{\prime }\cdot h\in a^{\prime }H$ since $h\in H$. Therefore $a^{\prime }\cdot h\in \bigcup <!--\; FUNCTION\; APPLICATION\; -->A=p^{-1}(A)$. Hence we have that $y_x=v_a\cdot (a^{\prime }\cdot h)\in V\cdot p^{-1}(A)$. Since also $y_x\in V\cdot x$, this violates the fact that $V\cdot p^{-1}(A)$ and $V\cdot x$ are disjoint. This contradiction means that it must be that in fact $p(V\cdot p^{-1}(A))$ and $p(V\cdot x)$ are disjoint, which completes the proof that $G∕H$ satisfies the regularity axiom. □