Suppose I have a function $g$ from some open subset $M \subset R^n$ to the reals. I seem to recall that if $g$ is a submersion then, for any $t$ in the range of $g$ the level set $g^{-1}(t)$ is "smooth".
How about the converse: Suppose I only know that the function $g$ is continuous but $g^{-1}(t)$ is smooth for every $t$ in the range of $g$. Must $g$ be differentiable? Concretely, I have a smooth regular codimension one foliation $\mathscr{F}$ of $M$ and have constructed a continuous function $g$ such that, for each $t$ in the range of $g$, $g^{-1}(t)$ is a leaf of $\mathscr{F}$ ($g$ is a variant of the quotient map of $\mathscr{F}$). I've thought about this for a while now trying, among other things, using foliated charts but can't seem to get anywhere. My sense is it's true but I easily could be wrong, of course. I'd appreciate any pointers so I can work this problem out for myself.
Edit: Pursuant to Lee Mosher's comment, I know a few more things about the foliation $\mathscr{F}$. First, $M$ is actually $P^n$, where by $P$ I mean the strictly positive reals. Second, I know that $\mathscr{F}$ arises from a smooth plane distribution via Frobenius' Theorem (hence each leaf is a connected integral manifold which also happens to be proper). Finally, I know that every ray through the origin intersects each leaf once and only once. I still don't think that is imposing any "control" in the transverse direction, but I could be wrong.
For a rock bottom counterexample take $$g(x,y) = \begin{cases} x & \quad\text{if $x \le 0$} \\ 2x &\quad\text{if $x \ge 0$} \end{cases} $$ so $g^{-1}(t)$ is a vertical line for all $t$, which is as "smooth" as you can get I suppose.
But $g(x,y)$ is not differentiable at any value where $x=0$.
I'm sure you can pump this up to examples which are considerably worse behaved.
The point is that you are not imposing any control on $g$ in any direction that is "transverse" to the foliation, so its unreasonable to expect smoothness of $g$.