Continuous homorphisms between topological groups. 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. 
 A: 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.
A: 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 $x\in U$,
$$
\rho(\lambda_g(x))=\rho(gx)=\rho(g)\rho(x)\in\lambda_{\rho(g)}(V)
$$
A: 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 $(*)$.
