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?

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    If the quotient is a covering space map, this will imply that you space is a manifold. Their may be more general conditions that work however. – Baby Dragon Sep 17 '13 at 17:23
  • I might be wrong but how about the orbit space of a free action by a finite group? Is that the sort of condition you're looking for? – Dan Rust Sep 17 '13 at 17:28
  • I believe many of the conditions you seek are already satisfied by some definitions of the manifold. Are you asking what will ensure that it is hausdorff? – Dohleman Sep 17 '13 at 17:33
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    @BabyDragon: For topological manifolds (which, of course, is what the OP mentioned), you are right. But I think it's worth pointing out that in the smooth case, it doesn't necessarily follow. See http://math.stackexchange.com/questions/349693/a-covering-map-from-a-differentiable-manifold/350743#350743 – Jason DeVito Sep 17 '13 at 17:34
  • To preserve Hausdorffness it is enough if the quotient map is closed or equivalently if the corresponding decomposition is upper semi-continous. However note that even such simple closed map as gluing two points together may easily destroy the locally euclidean condition. – user87690 Sep 17 '13 at 17:40
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    Ironically, it is user43208 who satisfactorily answered your question, not the user whose answer got accepted and received the most votes so far. – Alex M. Jan 31 '16 at 15:05
  • Even more ironically, the best answer by far is the one by PVAL-inactive that received only two votes. – Moishe Kohan Mar 02 '18 at 19:21

3 Answers3


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 $$ \mathcal O = \{ (g\cdot p,p): g \in G, p \in M \} \subseteq M \times M$$ 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).

Tyler Holden
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My memory is that Bourbaki has a theorem which says: if $E \subseteq M \times M$ is an equivalence relation such that

  • $E$ is a closed submanifold of $M \times M$, and

  • The first projection $\pi_1: E \to M$ is a submersion,

then the quotient space $M/E$ inherits from $M$ a smooth manifold structure. Still trying to track down the exact reference.

Edit: Since there has been some commentary below this answer, let me provide a reference to the original Bourbaki. It's section 5.9.5 in Bourbaki's Variétés différentielles et analytiques, (Hermann, 1967). The statement is as I have it above, except that I added that $E$ should be closed in $M \times M$, in order for the quotient $M/E$ to come out Hausdorff.

Note also that Bourbaki formulates it in the form of an if and only if statement.

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    This is in fact *equivalent* to $M/E$ having a smooth structure compatible with the quotient topology such that $M\to M/E$ is a submersion, according to http://www.math.illinois.edu/~ruiloja/Math519/notes.pdf, theorem 8.3 on page 63. (if by submanifold you mean embedded submanifold) – Daan Michiels Mar 13 '14 at 17:32
  • @DaanMichiels: Why is it *equivalent*? The theorem that you mention requires that $(E, i)$ be a submanifold (where $i$ is the natural inclusion $E \hookrightarrow M \times M$) and that $i$ be a *proper* map. I don't see the answer mentioning anything about the inclusion being proper. – Alex M. Jan 30 '16 at 20:42
  • @AlexM. - I'm not sure I understand your question correctly, but isn't it enough to observe that every closed embedded submanifold is proper? Regardless, I was saying that if $M/E$ inherits a smooth structure from $M$, then $E$ is a closed submanifold of $M\times M$ and the first projection $E\to M$ is a submersion. – Daan Michiels Jan 31 '16 at 05:05
  • Does this criterion imply the above quotient manifold theorem? Is there any example such that the quotient exists but the action is not free? I would like to see more examples such that this criterion applies but the quotient manifold theorem doesn't. – Q. Zhang Sep 06 '16 at 01:45
  • I'm sure that this answer is useful to this question https://math.stackexchange.com/questions/4183490/does-a-continuous-map-from-mathbbrn-into-an-infinite-dimensional-hilbert-s – Rafael Rojas Jun 28 '21 at 10:23

There is quite a lot on this problem already known. The study of the Hausdorff (usually metrizable) spaces gotten from taking quotients of manifolds is called decomposition space theory and has a wonderful history.

A special case of this which has gathered a lot of attention is when is the quotient of a manifold $M$ actually homeomorphic to $M$. Bing in the 1950's and 1960's studied cases of this problem extensively and showed a certain quotient of $S^3$ is actually homeomorphic to $S^3$ and used such a decomposition to construct involutions of $S^3$ not conjugate to any smooth ones. Brown proved the generalized Schoenflies theorem using similar techniques. In the 80's Freedman used this theory to prove to 4-dimensional Poincare conjecture.

In your case, if the quotient map has only one nontrivial set $K$ being identified to a point and $K$ is compact, then the quotient space is homeomorphic to a manifold if and only if $K$ is a countably infinite intersection of nested closed balls ($K$ is cellular according to the literature). Such a characterization also works perfectly if there are only finitely many nontrivial sets identified to a point.

If there are infinitely many such sets, things become more complicated. Realizing when a "generalized manifold" is a manifold to my knowledge is not completely understood though there are a lot of partial results.

Daverman gives a nice account of most of this story in his book here including some of the mentioned partial results near the end.

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