Does a left group action induce an open continuous map

Question

Let $X$ be a topological space and $G$ a topological group with an action $f:G\times X\to X$ so that $f(g,x)$ is denoted by $g\cdot x$. Let us fix $g\in G$, I want to know if given an open set $U\subset X$ the set $g\cdot U=\{g\cdot x\mid x\in U \}$ is homeomorphic to $U$. My guess is that the answer is positive for the map $x\mapsto g\cdot x$ is clearly bijective, so it suffices to show that it is both continuous and open. I figure that $g$ must be continuous since $f$ already is, but I cannot justify this, how could I prove openness of this map?

Answer 1

In fact, for each $g$, the map $x\mapsto g\cdot x$ is a homeomorphism. Clearly it is continuous and bijective, and its inverse is $x\mapsto g^{-1}\cdot x$, which is also continuous. Therefore in particular the map is open.

To elaborate on why $x\mapsto g\cdot x$ is continuous, the map $f:\{g\}\times X\to X$ is continuous being the restriction of a continuous map, and $x\mapsto g\cdot x$ is $X\to \{g\}\times X\to X$.