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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

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    $\begingroup$ 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 $g$. $\endgroup$ Commented Jun 18, 2017 at 4:32
  • $\begingroup$ ... and a set is open ifff it is a neighbourhood for all its elements $\endgroup$ Commented Jun 18, 2017 at 5:38

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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 $G$.

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