Let $X$ be a topological space on which a topological group $G$ acts transitively. Given $x\in X$ let $$stab(x)=\{g\in G\;|\; gx=x\}.$$
I want to show that $X$ is homeomorphic to $G/stab(x)$ for any given $x\in X$. I Know that when $G$ acts transitively then there exists only one orbit so that the quotient $X/G$ has only one element $[x]$ which is the class of any element $x\in X$ but i can't find a one to one correspondance between $X$ and $G/stab(x)$. Thank you for your help!
Let $x\in X$. Define a map $f:G\to X$ by $f(g)=g\cdot x$ for $g\in G$. Note that if $g_1,g_2\in G$ are in the same left coset of $stab(x)$, then $$g_1^{-1}g_2(x)=x,$$and so $$g_2(x)=g_1(x),$$in other words,$$f(g_1)=f(g_2).$$Hence, $f$ induces a well defined map $\tilde{f}:G/stab(x)\to X$. The verification that $\tilde{f}$ is a continuous bijection is standard. However, as studiosus kindly suggests, it is not necessarily an open map. For example, one can take $X$ to have the trivial topology $\tau=\{\phi,X\}$. In order to obtain a homeomorphism, further assumptions are needed.