Is $g\cdot U$ open for continuous group action and open $U$?

Question

I think given a continuous group action on manifold $M$ and $U\subset M$ open we should have $g\cdot U$ open, but I don't see how to prove this.

Answer 1

For each $g\in G$, the map $T_g(x)=gx$ is a homeomorphism $M\to M$ with inverse $T_{g^{-1}}$, so in particular is an open map.