When are simple foliations strictly simple? Any submersion $f: M → N$ defines a foliation of M whose
leaves are the connected components of the fibres of $f$. Foliations
associated to the submersions are called simple foliations. The foliations
associated to submersions with connected fibres are called strictly
simple. A simple foliation is strictly simple precisely when its space of
leaves is Hausdorff.
 Any idea for a proof?
 A: If the fibres of the submersion $f:M\rightarrow N$ are connected, the space of leaves is the image of $f$ which separated since it is an open subset of $N$ which is separated.
A: By Tsemo's answer, only one direction is still open. Let $B$ be the leaf space with the quotient topology, and let $g\colon M\to B$ and $p\colon B\to N$ be the natural continuous maps with $f=p\circ g$. Then $B$ is second countable. Assume that $B$ is Hausdorff, too. If we show that $p$ is a local covering, then $B$ inherits a manifold structure from $N$, and $g$ is the desired submersion.
Fix $y\in B$ and a preimage $x\in M$, let $z=p(y)$. We identify a sufficiently small transversal slice $\bar U$ around $x$ in $M$ with a coordinate neighbourhood $U\subset N$ of $z$. Let $W$ denote the connected component of $f^{-1}(U)$, and let $V\subset B$ be its image under $g$. Claim: $p\colon V\to U$ is a local homeomorphism. Because id$_U$ factorises as $U\cong\bar U\to V\to U$, it suffices to show that $p\colon V\to U$ is a bijection.
Suppose not. Then $V$ contains a point $q\notin\mathrm{im}(g|_{\bar U})$. Let $r\in g^{-1}(q)$, and join $x$ to $r$ with a path $\bar\gamma\colon[0,1]\to W$. Let $\gamma=g\circ\bar\gamma$ be the image in $V\subset B$. Let $\bar\gamma'\colon[0,1]\to\bar U\subset M$ and $\gamma'=g\circ\gamma'$ be lifts of $p\circ\gamma$. Then $\gamma'(0)=\gamma(0)=y$, but $\gamma'(1)\ne\gamma(1)=g(q)$. If $\gamma'(s)\ne\gamma(s)$, separate both points using that $V\subset B$ is Hausdorff. In fact, the two separating open sets can be chosen to have the same image in $N$. It follows that $J=\{s\in[0,1]\mid\gamma'(s)\ne\gamma(s)\}\subset[0,1]$ is open with $1\in J$, $0\notin J$.
Now fix $t\notin J$, then $\bar\gamma(t)$ can be joined to $\bar\gamma'(t)$ by a path $\delta\colon[a,b]\to W$ in $g^{-1}(\gamma(t))$ because by construction of $B$, both lie in the same leaf. Consider the holonomy of the foliation along $\delta$, which locally maps $\bar\gamma'$ to $\bar\gamma$, because both paths have the same image in $N$. Then there is a neighbourhood of $t$ where $\bar\gamma'$ and $\bar\gamma$ can be joined by a path within one leaf that is close to $\delta$. So $I\setminus J$ is open, too. This gives a contradiction, which proves the claim.
