Multiplication by group elements is an open mapping

Question

In a topological group $G$, let $L_x(u)=xu$. In much of the literature on topological groups, the fact that this is an open mapping (so, if $U \subset G$ is open then $L_x(U)$ is open) is used without proof. Any good proofs, given that we can assume it is a continuous map?

Answer 1

It is an homeomorphism, so it is open, $L_{x^{-1}}$ is the inverse of $L_x$, if $h$ is an homeomorphism, $({h}^{-1})^{-1}(U)=h(U)$ is open since $h^{-1}$ is continuous.