HW11

1. Let $G_{1}, G_{2}$ be topological groups and $f : G_{1} \to G_{2}$ be a homomorphisms which is continuous at the point $e \in G_{1}$ . Show that $f$ is continuous.

\begin{proof} For any $y \in G_{2}$ and any open set $U \subseteq G_{2}$ , the set $- y + U$ is a open subset containing $f (e) \in G_{2}$ . Then $f^{- 1} (- y + U)$ is an open set of $G_{1}$ by $f$ continuous at $e \in G_{1}$ . Take $x \in f^{- 1} (y)$ , then we claim that

f^{- 1} (- y + U) = - x + f^{- 1} (U) .

Now we prove the claim. Since $f (x + f^{- 1} (- y + U)) \subseteq U$ , we have $f^{- 1} (- y + U) \subseteq - x + f^{- 1} (U)$ . On the other hand, since $f (- x + f^{- 1} (U)) \subseteq - y + U$ , we have $f^{- 1} (- y + U) \supseteq - x + f^{- 1} (U)$ . Thus $f^{- 1} (U)$ is open. By the arbitrary of $x$ and $U$ , we know $f$ is continuous. \end{proof}

2. Let $G$ be a topological group, and $H$ a normal subgroup. Then $H$ is closed if and only G/H is Hausdorff.

\begin{proof} Define $π : G \to G / H$ be the canonical quotient map, and $G / H$ is a topology space with quotient topology, that is, $V \subseteq G / H$ is open iff $f^{- 1} (V)$ is open.

Assume that $G / H$ is Hausdorff, then for any $\overline{x} \in G / H$ , we claim that ${\overline{x}}$ is closed. For any $\overline{y} \neq = \overline{x}$ , there exist open set $V_{1}, V_{2} \subseteq G / H$ such that $\overline{x} \in V_{1}$ , $\overline{y} \in V_{2}$ and $V_{1} \cap V_{2} = \emptyset$ . Set $V_{y} = V_{2}$ , and define $V = \cup_{\overline{y} \in {\overline{x}}^{c}} V_{y}$ . Then $V$ is an open set and $V = {\overline{x}}^{c}$ . Therefore, ${\overline{x}}$ is closed. Let $\overline{0}$ be the identity of $G / H$ , then ${\overline{0}}$ is a closed set of $G / H$ and $π^{- 1} ({\overline{0}}) = H$ is closed.

Conversely, assume $H$ is closed, then $\overline{0} \in G / H$ is a closed point. Let $\overline{x}, \overline{y} \in G / H$ be distinct points. Then $\overline{x} - \overline{y} \neq = \overline{0}$ , and there exists $U \in N (\overline{x} - \overline{y})$ such that $0 \in / U$ . Also $U = (\overline{x} - \overline{y}) + U_{0}$ for $U_{0} \in N (0)$ . Take $V_{0} \in N (0)$ such that $V_{0} - V_{0} \subseteq U_{0}$ by ^36d8qm. Since $0 \in / U$ , we have $0 \in / (\overline{x} - \overline{y}) + (V_{0} - V_{0})$ . It deduces that $(\overline{x} + V_{0}) \cap (\overline{y} + V_{0}) = \emptyset$ . Therefore, $G / H$ is Hausdorff. \end{proof}

(1) $c l (A^{- 1}) = c l (A)^{- 1}$ and $c l (A) c l (B) \subset c l (A B)$ for every subsets $A, B$ of $G$ .
(2) If $H \leq G$ is a subgroup then $c l (H) \leq G$ is also a subgroup.

此处 $c l$ 是 closure 的意思。

\begin{proof} Remark that $c l (A) = {x \in G : for any open set U containing x, U \cap A \neq = \emptyset}$ .

i) For any $x \in c l (A^{- 1})$ , we aim to show $x^{- 1} \in c l (A)$ . For any open set $U$ containing $x^{- 1}$ , the set $V := U^{- 1} = {u^{- 1} : u \in U}$ is also open by $g \mapsto g^{- 1}$ homeomorphism. Since $x \in V$ and $V \cap A^{- 1} \neq = \emptyset$ , we know $U \cap A \neq = \emptyset$ and then $x^{- 1} \in c l (A)$ . On the other hand, if $x \in c l (A)^{- 1}$ , then $x^{- 1} \in c l (A)$ . For any open set $W$ containing $x$ , $W \cap A^{- 1} \neq = \emptyset$ iff $W^{- 1} \cap A \neq = \emptyset$ , and the latter is guaranteed by $x^{- 1} \in W^{- 1}$ and $x^{- 1} \in c l (A)$ . Now we proved $c l (A^{- 1}) = c l (A)^{- 1}$ .

For any $ab = x \in c l (A) c l (B)$ with $a \in c l (A), b \in c l (B)$ and open set $U$ contains $x$ , since $f : G \times G \to G, (g, h) \mapsto g h$ is continuous, there exist open sets $V, W$ such that $a \in V, b \in W$ and $VW \subseteq U$ (remark that $f^{- 1} (U)$ is open and ${U_{1} \times U_{2} : U_{1}, U_{2} are open sets of G}$ is a set of topology basis of $G \times G$ ). Since $a \in c l (A)$ , $A \cap V \neq = \emptyset$ and we take $a^{'} \in A \cap V$ . Similarly take $b^{'} \in B \cap W$ . Then $a^{'} b^{'} \in A B \cap VW$ is nonempty and so $x \in c l (A B)$ . Now we finish the proof.

ii) Since $H ⩽ G$ is a subgroup, the identity $e \in H \subseteq c l (H)$ . For any $h_{1}, h_{2} \in c l (H)$ , by i) one can check $h_{1} h_{2} \in c l (H) c l (H) \subseteq c l (HH) = c l (H)$ and so $c l (H)$ is multiplicative closed. Furthermore, note that $c l (H) = c l (H^{- 1}) = c l (H)^{- 1}$ , which deduces that $c l (H)$ is a subgroup. \end{proof}

is exact.
Link to original

^9tavuo

\begin{proof} First we prove that $0 \to lim_{\leftarrow} A_{n} \to lim_{\leftarrow} B_{n} \to lim_{\leftarrow} C_{n}$ is exact. Define

0 \to A_{n} \to f_{n} B_{n} \to g_{n} C_{n} \to 0 (*)

and it induces $f : lim_{\leftarrow} A_{n} \to lim_{\leftarrow} B_{n}$ . If there exists $(a_{n})_{n ⩾ 1}$ such that $f ((a_{n})) = 0$ , then by the commutative diagram

we know $f ((a_{n})) = (f_{n} (a_{n}))$ . Since $f_{n}$ is injective for all $n$ , $(a_{n})_{n ⩾ 1} = 0$ and so $f$ is injective. Now we check $im f = ker g$ . For any $f ((a_{n})) \in im f$ , there is

g (f (a_{n})) = g ((f_{n} (a_{n}))_{n ⩾ 1}) = (g_{n} \circ f_{n} (a_{n})) = 0

by $(*)$ exact. So $im f \subseteq ker g$ . Take $(b_{n}) \in ker g$ , then $g ((b_{n})) = (g_{n} (b_{n})) = 0$ yields $b_{n} \in ker g_{n} = im f_{n}$ . Thus there exists $a_{n}$ such that $f_{n} (a_{n}) = b_{n}$ . Define $a = (a_{n})_{n ⩾ 1}$ , then by the commutative diagram it is easy to check $a \in lim_{\leftarrow} A_{n}$ and $f (a) = (b_{n})$ . Therefore, $im f = ker g$ and so $0 \to lim_{\leftarrow} A_{n} \to lim_{\leftarrow} B_{n} \to lim_{\leftarrow} C_{n}$ is exact.

Now assume that ${A_{n}}$ is a surjective system. To show $0 \to lim_{\leftarrow} A_{n} \to lim_{\leftarrow} B_{n} \to lim_{\leftarrow} C_{n} \to 0$ is exact, it is enough to show $g$ is surjective. We prove it by induction. For any $c = (c_{n})_{n ⩾ 1} \in lim_{\leftarrow} C_{n}$ , there exists $b_{1} \in B_{1}$ such that $g_{1} (b_{1}) = c_{1}$ as $g_{1}$ is surjective. Suppose for $k ⩽ n - 1$ , there exists $b_{k} \in B_{k}$ such that $g_{k} (b_{k}) = c_{k}$ and $θ_{k} (b_{k}) = b_{k - 1}$ .

Now we construct $b_{n}$ such that $g_{n} (b_{n}) = c_{n}$ and $θ_{n} (b_{n}) = b_{n - 1}$ . Define $θ_{n} : B_{n} \to B_{n - 1}$ . Since $g_{n}$ is surjective, there exists $b_{n}^{'}$ such that $g_{n} (b_{n}^{'}) = c_{n}$ . If $b_{n - 1}^{'} = θ_{n} (b_{n}^{'}) \neq = b_{n - 1}$ , then $g_{n - 1} (b_{n - 1}^{'}) = g_{n - 1} (b_{n - 1}) = c_{n - 1}$ and so $b_{n - 1} - b_{n - 1}^{'} \in ker g_{n - 1} = im f_{n - 1}$ . Thus there is $a_{n - 1}$ such that $f_{n - 1} (a_{n - 1}) = b_{n - 1} - b_{n - 1}^{'}$ . Since $A_{n}$ is a surjective system and the diagram above commutes, there exists $a_{n} \in A_{n}$ such that $θ_{n} (f_{n} (a_{n})) = b_{n - 1} - b_{n - 1}^{'}$ . Then $b_{n} = f (a_{n}) + b_{n}^{'}$ is what we desired. Repeat this procedure, and we get $b = (b_{n})$ such that $g ((b_{n})) = c$ . Now we finish the proof. \end{proof}

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