# Showing that an $\mathbb{R}$-action is proper

Let $$\mathbb{R}$$ act freely and smoothly on a manifold $$M$$, such that the space of orbits $$M/\mathbb{R}$$ is a smooth manifold and the projection $$M \to M/\mathbb{R}$$ is a smooth submersion.

Is it then true that the $$\mathbb{R}$$-action is proper? I suspect that it is, but have trouble explicitly showing it. Any advice?

I see that this is some kind of converse statement to the Quotient Manifold Theorem, however that didn't help me so far.

• Does this help? math.stackexchange.com/questions/3233/… Nov 7, 2021 at 12:23
• Thank you, that would be a way to do it. I am just wondering if for this simple example $G=\mathbb{R}$ there is a direct way, without first proving that $M$ is a principal bundle. Nov 7, 2021 at 14:13
• I am not at all convinced that the argument in the link is a valid proof. Details are missing. Nov 8, 2021 at 4:07
• I wrote details for the linked answer. You can avoid using the terminology "principal fiber bundle" but, even in the case when $G={\mathbb R}$, the proof will go through the same steps. Nov 8, 2021 at 19:22
• Thank you very much! Nov 11, 2021 at 15:07