# Topological groups - Actions and Hausdorffness

I guess this problem is widely-known, but I couldn't finish it.

If $X$ is a topological group (and compact), and $G$ a closed subgroup acting on $X$ by left translation, show that $X/G$ is Hausdorff.

I was trying to use that famous lemma about an equivalence relation on $X$ being open, but I couldn't do it yet. Is this the way? Do you have any suggestion?

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Lemma 1 Let $f\colon X \rightarrow X'$ and $g\colon Y \rightarrow Y'$ be open maps of topological spaces. Then $f\times g \colon X\times Y \rightarrow X'\times Y'$ is open.

Proof: Clear.

Lemma 2 Let $X, Y$ be topological spaces. Let $R$(resp. $S$) be an equivalence relation on $X$(resp. $Y$). Suppose the canonical map $X \rightarrow X/R$(resp. $Y \rightarrow Y/S$) is open. Let $R\otimes S$ be the equivalence relation on $X\times Y$ such that $(x, y) \equiv (x', y')$ (mod $R\otimes S)$ if and only $x \equiv x'$ (mod $R)$ and $y \equiv y'$ (mod $S)$. Then the canonical map $(X\times Y)/(R\otimes S) \rightarrow (X/R)\times (Y/S)$ is an isomorphism.

Proof: Let $f\colon X\times Y \rightarrow (X/R)\times (Y/S)$ be the canonical map. By Lemma 1, $f$ is open. Hence $f$ induces an isomorphism $(X\times Y)/(R\otimes S) \rightarrow (X/R)\times (Y/S)$. QED

Lemma 3 Le $X$ be a topological space. Let $\Delta$ be the diagonal of $X\times X$, i.e, $\Delta = \{(x, x) \in X \times X\colon x \in X\}$. Then $X$ is Hausdorff if and only if $\Delta$ is closed.

Lemma 4 Le $X$ be a topological space. Let $R$ be an equivalence relation on $X$. Suppose the canonical map $X \rightarrow X/R$ is open. Let $C$ be the graph of $R$, i.e. $C = \{(x, y) \in X\times X\colon x \equiv y$ (mod $R)\}$. Suppose $C$ is closed. Then $X/R$ is Hausdorff.

Proof: Let $f\colon X\times X \rightarrow (X/R)\times (X/R)$ be the canonical map. By Lemma 2, $f$ induces an isomorphism $(X\times X)/(R \otimes R) \rightarrow (X/R)\times (X/R)$. Let $\Delta$ be the diagonal of $(X/R)\times (X/R)$. Since $f^{-1}(\Delta) = C$, $\Delta$ is closed. Hence $X/R$ is Hausdorff by Lemma 3. QED

Proposition Let $X$ be a topological group. Let $G$ be a closed subgroup of $X$. Let $X/G$ be the space of left cosets of $X$ by $G$. Then $X/G$ is Hausdorff.

Proof: Let $R$ be the equivalence relation on $X$ defined by $x \equiv y$ (mod $R)$ if and if $x^{-1}y \in G$. Then $X/G = X/R$. Let $C$ be the graph of $R$. Let $f\colon X\times X \rightarrow X$ be the map definded by $f(x, y) = x^{-1}y$. Then $f$ is continuous and $f^{-1}(G) = C$. Since $G$ is closed, $C$ is closed. Clearly the canonical map $X \rightarrow X/G$ is open. Hence $X/G$ is Hausdorff by Lemma 4. QED

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The relation that $R$ and $S$ naturally induce on $X\times Y$ isn’t actually $R\times S$: it’s $$\left\{\big\langle\langle x_0,y_0\rangle,\langle x_1,y_1\rangle\big\rangle:\big\langle\langle x_0,x_1\rangle,\langle y_0,y_1\rangle\big\rangle\in R\times S\right\}\;.$$ – Brian M. Scott Oct 9 '12 at 10:59
@BrianM.Scott Thanks. I corrected the answer. – Makoto Kato Oct 9 '12 at 12:59
The following lemma may make the proof of Lemma 4 clearer. Lemma 1.5 Let $f\colon X \rightarrow Y$ be a surjective open map of topological spaces. Let $U, F$ be subests of $Y$. Then If $f^{-1}(U)$ is open, $U$ is open. If $f^{-1}(F)$ is closed, $F$ is closed. Proof: Suppose $f^{-1}(U)$ is open. Since $f$ is surjective $f(f^{-1}(U)) = U$. Since $f$ is open, $U$ is open. Suppose $f^{-1}(F)$ is closed. Since $X - f^{-1}(F) = f^{-1}(Y - F)$ is open, $Y - F$ is open by the above. Hence $F$ is closed QED – Makoto Kato Oct 9 '12 at 21:31
I read this proof a dozen times and I really can't find any problem with it. But it bothers me that the OP said $X$ is compact yet that was never used - particularly because this is exercise 30 (of section 4.4) of Armstrong Basic Topology and he also assumes $X$ is compact. – Gregory Grant May 1 '15 at 18:29