# Quotient of a manifold

Suppose we have a manifold $M$, and a connected submanifold $N$. We can make the quotient $\frac{M}{N}$, which send $N$ to a single point. Now, there are known restrictions on $N$ such that $\frac{M}{N}$ is also a manifold?

I can see that in the Euclidean space, as a trivial example, $\frac{\mathbb{R}^n}{\mathbb{R}^m}$ is $\mathbb{R}^{n-m}$ which is a manifold too ($m<n$). I guess this is closely related to the fact that $\mathbb{R}^m$ foliates $\mathbb{R}^n$ for any $m<n$. Is it necessary that $N$ foliates $M$? Is it sufficient?

I'm trying to answer the second question, which I think has a possitive answer, for which I'm using the definition of the foliation. Since any point $p$ in $M$ has a coordinate neighborhood and chart $(\phi, U)$ such that the connected components of $N \cap U$ are sets of the form {$q\in U | x^{m+1}(q)=a^{m+1}, \dots, x^{n}(q)=a^{n}$} for $a^{m+1}, \dots, a^{n}$ fixed real numbers, $m=$dim $N$ and $\phi=(x^1,\dots, x^n)$. So, locally our quotient looks like $R^{n-m}$, but I'm stuck with the Hausdorff property. In the first question, I've tried a little, but I don't even know whether it's possitive or not. Any help?

Be attentive when quotienting: the quotient of $\mathbb{R}^n$ by its subspace $\mathbb{R}^m$ by sending the second one to a point is absolutely not the space $\mathbb{R}^{n-m}$, which is obtained by taking the quotient of the two as (topological, if you want) vector spaces!
If you have a foliation of $M^n$ of constant dimension $k$ by leaves $F_m$, one necessary and sufficient condition to have that the moduli space of leaves is a smooth manifold with smooth projection map is the following:
The quotient is Haudorff, and for each point $m_0\in M$ there exists a smooth submanifold $S$ of $M$ of dimension $n-k$ such that $m_0\in S$ and such that at each point $m\in S$, $S$ and the leaf $F_m$ through $m$ intersect transversally and at most once.
• I think of it as $\mathbb{R}^n$ as foliated by all the sets $(a, {\mathbb{R}^m})$, for $a\in \mathbb{R}^{n-m}$so when I say that we quotient, I'm actually sending every one of this set to a single poin $a$. – Paco Pacotilla Jan 17 '16 at 18:33
• So, this quotient actually it's $\mathbb{R}^{n-m}$ as a manifold. – Paco Pacotilla Jan 17 '16 at 18:35