A topological group is a group that is also a topological space satisfying the axiom, such that the maps sending to and sending to are continuous (note this means is a homeomorphism). We will use to denote an arbitrary topological group.
Theorem 1: If is a subgroup of , and are both topological groups under the subspace topology.
Proof) It is quite clear that is a topological group; the hard part is dealing with ; it suffices to show that is also a subgroup of . First of all, if , as is a homeomorphism that keeps invariant, . Furthermore, as is continuous, , and thus is closed under the group operation. We are done.
Theorem 2: Let be an arbitrary element of . Then, the maps defined by and are homeomorphisms of .
Proof) We show that any function of the form is continuous. The same logic could then be used to show that any function of the form is continuous, and as both functions are clearly bijections with being the inverse of , Theorem 2 is proven.
Consider the function such that . This is clearly a homeomorphism between and with the subspace topology, and the inclusion map is also continuous. If we denote the continuous map as , we can see that must be continuous. We are done.
Theorem 3: Giving a subgroup of , consider the quotient set , composed of left cosets () of . Let be the canonical quotient map into . (i) Show that mapping to is a homeomorphism of . (ii) If is closed in the topology of , show that one-point sets are closed in . (iii) Show that the quotient map is open (iv) Show that if is closed in and is a normal subgroup of , then is a topological group.
Proof) (i) As is a homeomorphism, for any arbitrary subset , is open $\iff
p^{-1}(X) \subset G$ is open $\iff$ $f_\alpha (p^{-1}(X))$ is open $\iff$ $p( f_\alpha(p^{-1}(X)) ) = f'_\alpha(X)$ is open. Thus, $f_\alpha'$ is a homeomorphism of $G/H$. (ii) For any $\alpha \in G$, as $f_\alpha$ is a homeomorphism, if $H$ is closed the left coset $\alpha H$ is also closed. Clearly, this implies one-point sets are closed in $G/H$. (iii) It suffices to show that for any arbitrary open set $X \subset G$, the set $X \cdot H$ is open. This must be true as $X \cdot h$ is open for any $h \in H$ by (i), and thus $X \cdot H = \bigcup_{h \in H} Xh$ must be open as well. (iv) If $H$ is closed, $G/H$ is $T1$ as one-point sets are closed by (ii). The inversion map $xH \mapsto x^{-1}H$ must be continuous as $x^{-1}H = (xH)^{-1}$ and the canonical inversion on $G$ is a homeomorphism. It suffices to show the map $F: G/H \times G/H \to G/H$ such that $xH \times yH \mapsto (xy)H$ is continuous. If we define the quotient map $P: G \times G \to G/H \times G/H$ defined by the equivalence relation $a \times b \sim c \times d \iff aH = cH \text{ and } bH = dH$ and define the map $f: G \times G \to G$ as $x \times y \mapsto x \cdot y$, we can see that $X \subset G/H$ is open $\iff$ $p^{-1}(X) \subset G$ is open $\Rightarrow$ $f^{-1}(p^{-1}(X))$ is open $\iff$ $F^{-1}(X) = P(f^{-1}(p^{-1}(X))) \subset G/H \times G/H$ is open. **Theorem 4:** Let $A$ and $B$ be subsets of $G$. Let $A^{-1}$ denote the set of points $a^{-1}$ for $a \in A$ and $A \cdot B$ denote the points $a \cdot b$ for $a \in A$, $b \in B$. (i) A neighborhood $V$ of the identity element $e$ is said to be symmetric if $V = V^{-1}$. Show that for any neighborhood $U$ of $e$, there is a symmetric neighborhood $V \subset U$ of $e$ such that $V \cdot V \subset U$. (ii) Show that if $x \neq y$, there is a neighborhood $V$ of $e$ such that $V \cdot x$ and $V \cdot y$ are disjoint; this implies $G$ is Hausdorff. (iii) Show that for any point $x$ and closed set $A$ such that $x \notin A$, there is a neighborhood $V$ of $e$ such that $V \cdot x$ and $V \cdot A$ are disjoint. This implies that $G$ satisfies the regularity axiom. (iv) Let $H$ be a subgroup that is closed in the topology of $G$; show that $G/H$ in the quotient topology is regular. *Proof)* (i) Let $F: G \times G \to G$ be the continuous multiplication map; then, the image $F^{-1}(U)$ is open in $G \times G$. As $e \times e \in F^{-1}(U)$, there is some neighborhood $W$ of $0$ such that $W \times W \subset F^{-1}(U)$. Taking, $V = W \cap W^{-1}$, we obtain a symmetric neighborhood $V$ of $0$ such that $V \times V \subset W \times W \subset F^{-1}(U) \Rightarrow V \cdot V \subset U$. (ii) It suffices to show there is a neighborhood $V$ of $e$ such that for any $v_1, v_2 \in V$, $v_1 x \neq v_2 y$, which is equivalent to $xy^{-1} \notin V^{-1} \cdot V$. By (i), it suffices to show there is a neighborhood $U$ of $e$ such that $xy^{-1} \notin U$; this is true as $G$ is a $T1$ topology. (iii) It suffices to show there is a neighborhood $V$ of $e$ such that for any $v_1, v_2 \in V$, $v_1 \cdot x \notin v_2 \cdot A$. Thus, it suffices to show there is a neighborhood $V$ of $e$ such that $V \cdot V^{-1} \cap A \cdot x^{-1} = \emptyset$. As $A$ is closed, $A \cdot x^{-1}$ is also closed and doesn't contain $e$; thus, there exists a neighborhood $U$ of $e$ such that $U \cap A \cdot x^{-1} = \emptyset$. By (i), we are done. (iv) It suffices to show that for any closed saturated set $A$ under the quotient map into $G/H$ and arbitrary element $x \in G$ such that $xH \cap A = \emptyset$, there is a neighborhood $V$ of $e$ such that $A \cdot V \cap xH \cdot V = \emptyset$. To go further, it suffices to show there is a neighborhood $U$ of $e$ such that $U \cap xH \cdot A^{-1} = \emptyset$. Here, as $A$ is saturated under the quotient map, $AH = A$ and thus $H \cdot A^{-1} = (A \cdot H^{-1})^{-1} = (A \cdot H)^{-1} = A^{-1}$, and thus $xH \cdot A^{-1} = x \cdot A^{-1}$ which is closed as $A$ is closed. As $G$ is regular, such an open set $U$ exists and we are done by (i). **Theorem 5:** If $A$ is closed in $G$ and $B$ is compact, show that $A \cdot B$ is closed. *Proof 1)* It suffices to show that for any $c \notin A \cdot B$, there is a neighborhood $W$ of $c$ such that $W \cdot B^{-1} \cap A = \emptyset$. Consider the map $F: G \times G \to G$ such that $x \times y \mapsto x \cdot y$. As $F$ is continuous, $F^{-1}(A)$ is closed; this implies $U = (F^{-1}(A)))^C$ is open. As $c \cdot B^{-1} \cap A = \emptyset$; $c \times B^{-1} \subset U$; and by the tube lemma, this forces there to be a neighborhood $W$ of $c$ such that $W \times B^{-1} \in U \Rightarrow W \cdot B^{-1} \cap A = \emptyset$. We are done. *Proof 2)* It suffices to show that if there is a net in $A \cdot B$ converging to some point $c \in G$, $c$ must be in $A \cdot B$. Let $(a_\alpha \cdot b_\alpha)_{\alpha \in J}$ be this net where $a_\alpha \in A$ and $b_\alpha \in B$. By compactness, some subnet of $b_\alpha$ converges to an element $v \in B$. For notational efficiency, we simply assume $b_\alpha$ converges to $v$. Then, for any neighborhood $V$ of $v$ and $W$ of $c$, there exists some $\beta \in J$ such that $\gamma \geq \beta$ implies $b_\gamma \in V$ and $a_\gamma \cdot b_\gamma \in W \Rightarrow a_\gamma \in W \cdot V^{-1}$. If we let $u = c \cdot v^{-1}$, for any neighborhood $U$ of $u$ we can take a symmetric neighborhood $U' \subset Uu^{-1}$ such that $U' \cdot U' \subset Uu^{-1}$ by Theorem 4(i). Then, if we take the neighborhood $V = vU'$ of $v$ and $W = cU'$ of $c$, $(cU') \cdot (vU')^{-1} \subset cUu^{-1}v^{-1}$, and thus there is some $\beta \in J$ such that whenever $\gamma \geq \beta$, $a_\gamma \in cUu^{-1}v^{-1}$. Now, the map $x \mapsto cxu^{-1}v^{-1}$ is clearly a homeomorphism, and thus for any desired neighborhood $U$ of $u$ we can find some $\beta$ such that $a_\gamma \in U$ for all $\gamma \geq \beta$. This implies the map $a_\alpha$ converges to $u$, which forces $u \in A$ by closure. Thus, $c = u \cdot v \in A \cdot B$. **Theorem 6:** Let $H$ be a compact subgroup of $G$. Then, (i) The quotient map $p: G \to G/H$ is closed. (ii) If $G/H$ is compact, then $G$ is compact. *Proof)* (i) follows immediately from Theorem 5, as it suffices to show for any closed set $X \subset G$, $X \cdot H$ is closed. As for (ii), let $\mathcal{A}$ be an open covering of $G$. For any $g \in G$, $gH$ is compact and thus there exists a finite subcollection $U_1, U_2, \ldots, U_n$ that covers $gH$. Let $U = \bigcup U_i$; then, as $U^C$ is closed, $p(U^C)$ is closed in $G/H$. By Theorem 4(iv), there is some open neighborhood $W$ of $gH$ such that $p^{-1}(W) \subset U$. That is, for any $g \in G$, there is some open subset $W$ of $gH$ such that $p^{-1}(W)$ has a finite subcover in $G$. As $G/H$ is compact, this implies $G$ is compact. **Theorem 7:** Let $G$ be a first-countable topological group. Show that if $G$ is Lindelof or separable, it must be second-countable. *Proof)* We first assume $G$ is first-countable and Lindelof. Let $B_i$ be a countable basis about the identity element $e$. For each $B_i$, $\{gB_i | g \in G\}$ is an open cover of $G$, and thus there exists a countable subcover $g_j B_i$ indexed by $j$ as $G$ is Lindelof. We Claim that $g_j B_i$ is a countable basis of $G$. Let $U$ be an arbitrary open set in $G$ and let $u \in U$ be an arbitrary element; then, $W = u^{-1}U$ is a neighborhood of $e$. There must exist a neighborhood $V$ of $e$ such that $V \cdot V \subset W$, and a $B_i$ such that $B_i \subset V$. Going back to $u$, there is some $g_j$ such that $u \in g_j B_i$ as $g_j B_i$ is a cover of $G$. Then, $g_j \in uB_i^{-1}$ and $g_j B_i \subset u B_i^{-1} B_i \subset u V \cdot V \subset u W = U$. We are done. We now show that if $G$ is first-countable and separable, $G$ is second-countable. Let $B_i$ be a countable basis about the identity $e$ and let $D$ be a countable dense subset of $G$; take the countable collection $\mathcal{A} = d_j B_i$ where $d_j$ is an arbitrary element of $D$; we claim $\mathcal{A}$ is a basis for the topology of $G$. For any $u \in G$ and neighborhood $U$ of $u$, it suffices to show there is some $d_j B_i$ such that $u \in d_j B_i \subset U$. Take some neighborhood $V$ of $e$ such that $V \cdot V \subset u^{-1} U$ and some $B_i$ such that $B_i^{-1} \subset V$. Then, as $D$ is dense, there is some $d_j$ such that $d_j \in u B_i^{-1} \Rightarrow u \in d_j B_i$. Furthermore, $d_j B_i \subset u B_i^{-1} \cdot B_i \subset u u^{-1} U = U$, and we are done.