Is every real n-manifold isomorphic to a quotient of $\mathbb{R}^n$? I'm curious about the following:
Is every real $n$-manifold isomorphic to a quotient of $\mathbb{R}^n$?
Thanks.
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?"
 A: Hahn–Mazurkiewicz Theorem:  Suppose $X$ is a nonempty Hausdorff topological space.
Then the following are equivalent:


*

*there is a surjection $[0,1]\to X$, 

*$X$ 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 $n\geq 1$.
A: 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.
A: 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.
A: 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.
A: 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.)
