# Induced topology on the homomorphic image of a topological group

I would like to do a small sanity check on the following situation:

Let $\pi: G \rightarrow G'$ be a surjective homomorphism of topological groups. Let the topology of $G$ be given by a sequence of subgroups $\left\{G_n\right\}$, i.e. $G=G_0 \supseteq G_1 \supseteq G_2 \supseteq \cdots$ where the $G_n$ form a fundamental system of neighborhoods. Define the topology on $G'$ by letting the open subsets of $G'$ be images of open subsets of $G$ under $\pi$, i.e. for any open set $G_n$ define $G^{'}_n=\pi(G_n)$ to be open.

I want to show that $\pi$ is continuous with respect to this topology. Towards this end, it is enough to show that $\pi^{-1}(G^{'}_n)$ is open in $G$. But $\pi^{-1}(G^{'}_n)=G_n+Ker(g)=\cup_{\xi \in Ker(g)} (G_n + \xi)$ which is open since $G_n + \xi$ is open, because the translation map in $G$ is continuous.

Is this argument sound?

Thanks.

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What do you mean by the topology given by a sequence of groups? – tomasz Sep 22 '12 at 19:29
@tomasz: I added an explanation. – Manos Sep 22 '12 at 19:49

Yes, it's correct. Maybe I would had an explanation of the equality $\pi^{—1}(\pi(G_n))=(\ker\pi)G_n$. $\supset$ is clear, and if $\pi(x)\in \pi(G_n)$, $\pi(x)=\pi(g)$, where $g\in G_n$. Hence $xg^{-1}\in \ker \pi$ and $x\in (\ker \pi)G_n$. (we don't use additive notation unless we know that $G$ is abelian)