Is a manifold modulo a free action still a manifold?

Question

If $D$ is a manifold, and $G$ is a group that acts freely on $D$, then is it true that $D/G$ is still a manifold? Why? And why is it important that the action of $G$ on $D$ is free?

Answer 1

Theorem. Let $G$ be a Lie group acting freely, smoothly, and properly on a smooth manifold $M$. Then, $M/G$ is a topological manifold and has a unique smooth structure such that the quotient map $M\to M/G$ is a smooth submersion.

This is standard, see e.g. Lee's book Introduction to smooth manifolds. All three assumptions on the action are necessary.

Answer 2

Represent $S^1$ as the quotient of $\mathbb{R}$ by $t(x) =x+1$. Let $ i$ be an irrational number, $f(x) =x+i$ induces a free action on $S^1$ and the quotient space is not a manifold. The quotient topology is not Hausdorff.

Answer 3

The group $\mathbb Z^2 = \mathbb Z \oplus \mathbb Z$ acts on the real line $\mathbb R$ by the formula $(m,n) \cdot t = t + m + n \sqrt{2}$. Every orbit of this action is dense in the real line, and yet there are uncountably many orbits. So the quotient is not a manifold, it is not even Hausdorff.