Every translation of a topological group is a homeomorphism Definition. A topological group is a group endowed with a topology such that
(1) the multiplication map $m: G\times G\rightarrow G$, $(g,h)\mapsto gh$ is continuous, and
(2) the inversion map $G\rightarrow G$, $g\mapsto g^{-1}$ is continuous.
What I want to do is to prove the following easy fact:
Fact. For any $g\in G$ the left translation $L_g:G\rightarrow G$, $x\mapsto gx$ is continuous.
To prove this, we first take any open subset $U\subseteq G$, and our task is to show that $L_g^{-1}(U)$ is an open subset of $G$. My idea is to use the following relation

$L_g^{-1}(U)=m^{-1}(U)\cap (\{ g \}\times G)$.

By condition (1) of the definition, $m^{-1}(U)$ is open in $G\times G$. If one could show that $\{ g \}\times G$ is open in $G\times G$, then $L_g^{-1}(U)$ would be open because it is the intersection of two open subsets. However, $G$ is generally not discrete, so $\{ g \}\times G$ may not be open in $G\times G$.
It seems that this Fact is easy to prove, but I don't know how to do it.
Any hint or help will be appreciated.

As $m \colon G \times G \to G$, $(g, h) \mapsto g h$ is continuous, so is its restriction $$m|_{G \times \{ h_0 \}} \colon G \times \{ h_0 \} \to G h_0 = \{ g h_0: g \in G \}, \qquad (g, h_0) \mapsto g h_0.$$
for any $h_0 \in G$, which can clearly be identified with $L_{h_0}$.
