# Under what conditions the quotient space of a manifold is a manifold?

There are many operations we can do with topological spaces that when we apply to topological manifolds gives us back topological manifolds. The disjoint union and the product are examples of that.

Another common operation is to take the quotient by some equivalence relation. Now, the definition of manifold I’m used to is the following: a topological manifold is a topological space $M$ such that $M$ is Hausdorff, has an enumerable basis for the topology and is locally euclidean of dimension $n$.

Now, we know that the quotient of a Hausdorff space is not necessarily Hausdorff. So, under what conditions if we have a manifold $M$ and an equivalence relation $\sim$ on $M$ the quotient space $M/\sim$ will be a manifold?

As I’m sure you know, the category of smooth (topological?) manifolds is one of those categories where the objects are very nice but the category itself is terrible. I cannot describe the number of times I’ve heard the algebraic geometers curse the smooth category.

I am not certain this is a total classification, but From Lee’s Introduction to Smooth Manifolds, Theorem 9.19:

If $\tilde M$ is a connected smooth manifold and $\Gamma$ is a discrete group acting smoothly, freely, and properly on $\tilde M$, then the quotient $\tilde M/\Gamma$ is a topological manifold and has a unique smooth structure such that $\pi: \tilde M \to \tilde M/\Gamma$ is a smooth covering map.

The manifold portion of this comes from the Quotient Manifold Theorem:

If $G$ is a Lie group acting smoothly, freely, and properly on a smooth manifold $M$, then the quotient space $M/G$ is a topological manifold with a unique smooth structure such that the quotient map $M \to M/G$ is a smooth submersion.

And then applying this to the (zero-dimensional) Lie group of deck transformations.

Edit: The proof of the Hausdorff property is very similar to @useruser43208’s response, and uses the properness of the action. Take the orbit set

which is closed under the properness assumption. Any two distinct points $\pi(p)$ and $\pi(q)$ in the image of the quotient map $\pi: M \to M/G$ must have arisen from distinct orbits, so $(p,q) \notin \mathcal O$. Hence we may find a product neighbourhood $U_p\times U_q \subseteq M \times M$ of $(p,q)$ disjoint from $\mathcal O$, hence $\pi(U_p)$ and $\pi(U_q)$ are separating open neighbourhoods (since $\pi$ is open).