# differ between $M$ and $G/G_{x_0}$

I have a Lie group $G$ acting transitively on $M$.

Why is $M$ diffeomorphic to $G/G_{x_0}$?

Here is the function. It remains to see that it is differential both ways. $$\phi:G/G_{x_0}\rightarrow M : g G_{x_0}\rightarrow g\cdot x_0$$

Will inverse function theorem do the trick?

You have to be careful. In general, $M$ does not even have to be homeomorphic to the orbit space. For a counterexample, consider $M=\mathbb{R}$ and $G$ the group of real numbers with the discrete topology. Note that $G$ is still a Lie group with this topology, a $0$-dimensional Lie group. Then there is a smooth action of $G$ on $M$ by left multiplication, which is clearly free and transitive. However, $G$ is not homeomorphic to $M$, since $M$ is connected and $G$ is disconnected.

The good news is that if the map $G/G_{x_0} \to M$ that you describe (which is already continuous and bijective) happens to be a homeomorphism, then it is a diffeomorphism. You can check this for instance in Proposition 4.3 of the book "Differential geometry, Lie groups and symmetric spaces" of S. Helgason.

This happens for example if the action of $G$ on $M$ is proper, that is, if the map

$$A \colon G \times M \to M \times M$$

$$(g,m) \mapsto (m,gm)$$

is proper. This means the inverse image of a compact set is compact. Let us see that in this case the map $G/G_{x_0} \to M$ is a homeomorphism. Consider first the map

$$a \colon G \to M$$ that takes $g$ to $gx_0$. If $K$ a compact set in $M$, then $\{x_0\} \times K$ is compact in $M \times M$ and therefore $A^{-1}(\{x_0\} \times K)$ is compact in $G \times M$. Then $a^{-1}(K) = \text{pr}_2(A^{-1}(\{x_0\} \times K))$ is compact and so $a$ is proper.

By the argument here, the map $a$ is closed and since $G/G_{x_0}$ has the quotient topology from the quotient $G \to G/G_{x_0}$, the map $G/G_{x_0} \to M$ is also closed. Therefore, it is a homeomorphism.

The condition of properness happens very often, for example, an action of a compact Lie group on a manifold is always proper. In any case, if $G$ is compact and $M$ is Hausdorff, then $G/G_{x_0}$ is also compact and then you can conclude that the map $G/G_{x_0} \to M$ is a homeomorphism.