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I'm curious about the following: Is every real $n$-manifold isomorphic to a quotient of $\mathbb{R}^n$?


EDIT: As Tilman points out, the manifold should be connected. Also, yes, I'm thinking about topological quotients. Specifically, is there a surjective map $\mathbb{R}^n\to M$ such that $M$ has the quotient topology?

EDIT': I guess an interesting addendum to the question is "when is it true?"

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No. Most manifolds are not. Explicit examples include the $n$-dimensional sphere, where $n \ge 2$ and $R^n$ minus the origin. – Deane Yang Nov 13 '10 at 20:23
You mean: is there a surjective map from $\mathbf{R}^n$ to the manifold such that the manifold has the quotient topology? Then $S^n$ and $\mathbf{R}^n-\{0\}$ are not counterexamples. – Tilman Nov 13 '10 at 20:36
You'll also want to assume that the manifold is connected, otherwise it's obviously false. – Tilman Nov 13 '10 at 20:37
Every connected 2nd countable Hausdorff manifold is a quotient of $\mathbb R$. This follows from the fact that every compact connected manifold is a quotient of $[0,1]$ and the fact that non-compact manifolds have proper functions so they're a countable union of compact manifolds. – Ryan Budney Nov 13 '10 at 21:44
There is a loosely related open question due to Gromov which asks whether or not every manifold can be realised as the quotient of hyperbolic space by a discrete group of isometries. (He mentions essentially this question on page 12 of a recent paper. See ) – Joel Fine Nov 13 '10 at 21:46
up vote 25 down vote accepted

Hahn–Mazurkiewicz Theorem: Suppose $X$ is a nonempty Hausdorff topological space. Then the following are equivalent:

  1. there is a surjection $[0,1]\to X$,
  2. $X$ is compact, connected, locally connected and second-countable.

It follows that a Hausdorff space satisfying the conditions of (2) is a quotient of $I = [0,1]$.

Cor: Every connected compact manifold is a quotient of $I$.

Since $I$ is a quotient of $\mathbb{R}^n$, we have your answer.

Cor: Every compact connected $m$-manifold is a quotient of $\mathbb{R}^n$ for any $n\geq 1$.

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As I mentioned in my comment above this generalizes to quotient maps $\mathbb R \to M$ for $M$ non-compact and connected by using the existence of a proper functions $M \to \mathbb R$, for which you only need partitions of unity. – Ryan Budney Nov 13 '10 at 21:54
This answer while completely correct strikes me as funny. – Spice the Bird Nov 9 '12 at 3:16

Note that any continuous surjection from a compact space to a Hausdorff space is automatically a quotient map. Also, there are 'space-filling curves', which are continuous surjections from $[0,1]$ to $[0,1]^2$. This means it is not very difficult for a space to be a quotient of $\mathbb{R}^n$ or even of $\mathbb{R}$. In particular, any connected Hausdorff second countable manifold will be such a quotient.

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Uhm, perhaps I should have picked this answer instead. – Eivind Dahl Nov 13 '10 at 22:33

Every closed (that is, compact without boundary) smooth connected manifold $M$ is a quotient of some $\mathbb R^n$. Put on $M$ an arbitrary riemannian metric. A metric on a compact manifold is necessarily complete, hence for every point $x$ in $M$ there is an exponential map $T_x \to M$ from the tangent space on $x$ to $M$. This map is smooth and surjective (because every point of $M$ may be connected to $x$ by a geodesic). Since $T_x$ is homeomorphic to $\mathbb R^n$ and a surjective map from a compact space to a Hausdorff space is a quotient, we are done.

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Yet the tangent space is not compact! :) – Mariano Suárez-Alvarez Nov 14 '10 at 15:19
Ah ah you are right! :-D We must restrict the exponential map to a disc centered in the origin of radius $R$: since $M$ is compact there is a $R$ such that this restriction is still surjective. – Bruno Martelli Nov 14 '10 at 21:15
(simply take $R$ to be the diameter of $M$) – Bruno Martelli Nov 14 '10 at 21:17

That should only be true for things which have $\mathbb{R}^n$ as their universal covering space. In particular I think it fails for things like Lens spaces.

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Hmm, and I interpreted the question to mean quotient wrt an equivalence relation. I guess the question is more than a little ambiguous and the OP should edit it to say precisely what he means by "quotient". – Dick Palais Nov 13 '10 at 20:33
You're right, I was thinking it was more restrictive, but on second thought there's no real reason for it to be, especially given these kinds of questions would really be much more trivial if that were the case. – Adam Hughes Nov 14 '10 at 2:24

It's worth mentioning, even though is a trivial observation, if you weaken your question to ask whether a manifold is the quotient of disjoint unions of copies of $\mathbb{R}^n$, then EVERY manifold is such a quotient, even as a differentiable quotient. Indeed, given an $n$-manifold $M$, choose an atlas $\mathcal{U}$ such that every element of the covering is diffeomorphic to $\mathbb{R}^n$. Consider the Cech-groupoid $M_{\mathcal{U}}$. This is a Lie-groupoid and its orbit space is $M$. This is nothing but spelling out that every manifold is the colimit of its atlas. In particular, your question is equivalent to asking whether or not every differentiable stack admits an atlas from a Cartesian space $\mathbb{R}^n$. (One direction of this being equivalent is clear, and conversely, if every manifold is such a quotient, then every Lie groupoid can be replaced by a Morita-equivalent one which has a Cartesian-space as an atlas.)

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I should mention that this isn't quite an equivalence- you'd need every manifold to be a quotient in a "natural" way- i.e. the colimit expressing this qoutient needs to be preserved by the Yoneda embedding into sheaves. – David Carchedi Nov 13 '10 at 21:32
Nice! Can't wait to figure out what that actually means :-) – Eivind Dahl Nov 13 '10 at 21:42
I admit the statement about stacks may be technical, however, the first part of what I said, namely that every manifold is a quotient of COPIES of $\mathbb{R}^n$, is nothing more than saying that manifolds are obtained by glueing together patches that look like $\mathbb{R}^n$. – David Carchedi Nov 13 '10 at 21:45
Technical $\neq$ bad. Stacks is something I'll eventually want to understand, and now I actually have a context where I may have something to think about. Thanks! – Eivind Dahl Nov 13 '10 at 21:59

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