# The topology of a topological group is determined by the collection of neighborhoods of the identity element of the group

In Jacques Faraut's book "Analysis of Lie Groups", the author says that the topology of a topological group $G$ is determined by the collection of neighborhoods of the identity element of the group, i.e. is determined by the collection

$\Omega = \lbrace{U\subseteq{G} : U\,\, \mbox{is a neighborhood of}\,\,{e} }\rbrace$

Could someone explain to me what this means?

Note: The neighborhoods of $e$ are not necessarily open

• Given $g\in G$, the map $h\mapsto h\cdot g$ is a homeomorphism of $G$ to itself, so once you know all the neighborhoods of $1$, you know all the neighborhoods of any $g$. Commented Jun 18, 2017 at 4:32
• ... and a set is open ifff it is a neighbourhood for all its elements Commented Jun 18, 2017 at 5:38

Suppose that we have a topological group $(G, \ast, e, \mathcal{T})$. Then for any $g \in G$, the translation $t_g: G \to G$ defined by $t_g(x) = g\ast x$ is continuous (as the operation $m: G \times G \to G, m(g,h) = g \ast h$ is continuous by definition of a topological group and $t_g = m \circ e_g$, where $e_g: G \to G \times G, e_g(x) = (g,x)$ is also continuous).
As $t_g$ and $t_{g^{-1}}$ are each other's continuous inverse, translations are homeomorphisms of $G$.
This implies that $U$ is a neighourhood of $g$ iff $t_{g^{-1}}[U]$ is a neighbourhood of $e$, so we only have to specify the neighbourhoods of $e$ to know all neighbourhoods of all points of $G$.