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Let $G,H$ be topological groups and let $\rho: G \rightarrow H$ be a homomorphism of topological groups. I understand this to mean that $\rho$ preserves group structure and is a map between the topologies.

If we say $\rho$ is continuous I understand that this means the map between the two topologies must be a continuous map i.e. the preimage of open sets will be open etc.

However I have come across the following claim, which in my eyes using the definitions above seems trivial. It goes as follows:

Let $G,H$ be topological groups and let $\rho: G \rightarrow H$ be a homomorphism of topological groups. Then $\rho$ is continuous if and only if $\rho^{-1} (V)$ is open for each $V$ in a basis of neighbourhoods $\mathcal{V}$ of the identity $1_h$.

Maybe I have not fully grasped one of the definitions.

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  • $\begingroup$ The main difference with topological continuity is that this (the one you expose) only requires to be so for neighbourhoods of the identity of the group. That is a big difference. $\endgroup$ – MyUserIsThis Jun 9 '15 at 18:17
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That result actually does say something. It's parallel to the result that a linear transformation $T:V \to W$ between two normed vector spaces is continuous if and only if it is continuous a $0$.

Usually, to check that $\rho$ is continuous, you would have to check that for any $V \subset H$ open, $\rho ^{-1}(V)$ is open. However, in topological groups, you only have to check this for what is potentially a very small subset of the open sets: a neighborhood basis of the identity.

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  • $\begingroup$ Ok. So if I understand correctly continuous implies RHS is simple. But for the RHS implies continuous I need to show that it is sufficient for open neighbourhoods of the identity have open preimages to imply continuous (possibly be showing open neighbourhoods of the identity have open preimages implies that all open neighbourhoods have open preimages). $\endgroup$ – Slimeonline Jun 9 '15 at 18:24
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    $\begingroup$ Yes, that's exactly right. $\endgroup$ – Noah Olander Jun 9 '15 at 18:26
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The ordinary definition of continuity requires that

$(*)$ $\rho^{-1}(V)$ be continuous for all elements $V$ of a basis for the topology on $H$.

The point of this claim is that it is not necessary to check all basis elements $V$, you need only to check it for those elements $V$ in a neighborhood basis of $1_H$, because once that's done then you can use it to prove $(*)$.

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In a topological group $G$, the map $\lambda_g\colon x\to gx$ (for $g\in G$) is a homeomorphism, because multiplication is continuous, so $\lambda_g$ is continuous, and the inverse map is $\lambda_{g^{-1}}$, which is continuous as well.

Thus, in order to verify that a homomorphism between topological groups is continuous it's sufficient to check it's continuous at one element, the identity is as good as any other one.

Indeed, the fact that $\lambda_g$ is a homeomorphism implies that the neighborhoods of $g$ are exactly the subsets of the form $\lambda_g(U)$, where $U$ is a neighborhood of the identity.

So, suppose you know that $\rho$ is continuous at the identity $1_G$ of $G$ and let $g\in G$. Take a neighborhood of $\rho(g)$, that we can see as $\lambda_{\rho(g)}(V)$, where $V$ is a neighborhood of $1_H$.

By assumption, there is a neighborhood $U$ of $1_G$ such that $\rho(U)\subseteq V$. But then $$ \rho(\lambda_g(U))\subseteq \lambda_{\rho(g)}(V) $$ Indeed, for $x\in U$, $$ \rho(\lambda_g(x))=\rho(gx)=\rho(g)\rho(x)\in\lambda_{\rho(g)}(V) $$

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