Topology problem on Lie Transformation groups.

Question

I'm reading Kobayashi's book Transformation Groups in Differential Geometry and at the page 14 is this lemma:

My question is why the uniqueness of that topology is trivial? I just know that i have to use the fact that $A_\varphi$ continuous but I don't know how..

Answer 1

Assume $G$ is given a topology compatible with the group structure, with the given restriction to $G^*$, and that $G^*$ is open.

It is clear that $\varphi V$ is an open neighborhood of $\varphi\in G$ if $V\subseteq G^*$ is an open neighborhood of the identity.
Conversely, if $U\subseteq G$ is open and $\varphi\in U$, then $V:=\varphi^{-1}U\cap G^*$ is an open neighborhood of the identity within $G^*$, and $\varphi\in\varphi V\subseteq U$.
So the sets $\varphi V$ indeed form a basis for the topology of $G$: every open set is a union of these.