I was learning about topological groups from Atiyah-Macdonald's chapter on Completions, and I have the following question:

Let $G$ be an abelian topological group. Let $H$ be the intersection of all neighborhoods of $0$. Then in Lemma $1.10$ of Atiyah-Macdonald, it is shown that $H$ is in fact a subgroup of $G$. Now, the quotient group $G/H$ can be given the structure of a topological space via the canonical projection $\pi:G \rightarrow G/H$.

So, then when we want to show that $G/H$ is also a topological group, one of the verifications requires us to check that the binary operation (addition) $G/H \times G/H \rightarrow G/H$ is continuous.

My question is do we regard $G/H \times G/H$ as a topological space under the product topology, or as a topological space which makes the map $\pi \times \pi: G \times G \rightarrow G/H \times G/H$ into a quotient map? Assuming the latter, I have been able to show that addition on $G/H$, i.e., the map from $G/H \times G/H \rightarrow G/H$ is continuous.

I initially thought that the product topology on $G/H \times G/H$ and the quotient topology on $G/H \times G/H$ w.r.t. the map $\pi \times \pi$ would be the same. But, after being unable to prove this statement, I realized that exercise $22.6(b)$ in Munkres shows that this need not be the case in general.

  • $\begingroup$ I see. I couldn't prove continuity with product topology, but I could with quotient topology. Interesting... Thanks. I will give it another try to prove the statement with product topology. $\endgroup$ – Rankeya Nov 7 '11 at 1:29
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    $\begingroup$ One fact that seems useful is that the quotient (in both senses) map $G \to G/H$ is always open. $\endgroup$ – Dylan Moreland Nov 7 '11 at 1:32
  • $\begingroup$ Yes, this is true. So, that map $\pi \times \pi$ is also an identification map. $\endgroup$ – Rankeya Nov 7 '11 at 1:35
  • $\begingroup$ So, I guess both the quotient topology and the product topology coincide then? $\endgroup$ – Rankeya Nov 7 '11 at 1:37
  • $\begingroup$ Hm. It does seem that way! Sometimes it's hard to see the logical implications of what one is saying :) $\endgroup$ – Dylan Moreland Nov 7 '11 at 1:40

More generally, if $N$ is a normal subgroup of a topological group $G$, then $G/N$ (the quotient of the underlying groups, endowed with the quotient topology of $\pi : G \to G/N$) is again a topological group.

Proof: For the inversion map $\iota_{G/N}$; we have that $\iota_{G/N} \circ \pi = \pi \circ \iota_G$ and therefore $\iota_{G/N}$ is continuous. By a very similar argument, all left (or right) translations $\lambda_{\pi(g)} : G/N \to G/N$ are continuous. Now in order to show that the multiplication $\mu_{G/N} : G/N \times G/N \to G/N$ is continuous, it suffices to check continuity at $(1,1)$. So let $1 \in U \subseteq G/N$ be an open neighborhood. Then $1 \in \pi^{-1}(U) \subseteq G$ is an open neighboorhood. Since $\mu_G$ is continuous, there are open neighboorhoods $1 \in V,W \subseteq G$ such that $V \cdot W \subseteq \pi^{-1}(U)$. Now apply $\pi$ to get $1 \in \pi(V) \cdot \pi(W) \subseteq U$. Thus we are done, if we can show that $\pi$ is open. But for $V \subseteq G$ open, we see that $\pi^{-1}(\pi(V)) = \cup_{n \in N} V \cdot n$ is open, thus $\pi(V)$ is open. [qed]


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