It's a well-known theorem (Corollary 8.10 in Lee Smooth) that given a smooth map of manifolds $\phi:M\rightarrow N$ and a regular value $p\in N$ of $\phi$, the level set $\phi^{-1}(p)\subset M$ is a closed embedded submanifold. Is the converse true? That is, given an embedded submanifold $S\subset M$, is there necessarily a manifold $N$, smooth map $\phi:M\rightarrow N$, and regular value $p\in N$ of $\phi$ such that $S=\phi^{-1}(p)$?
Prop. 8.12 in Lee Smooth shows that this is true locally; specifically,
Let $S$ be a subset of a smooth $n$-manifold $M$. Then $S$ is an embedded $k$-submanifold of $M$ if and only if every point $p\in S$ has a neighborhood $U\subset M$ such that $U\cap S$ is a level set of a submersion $\phi:U\rightarrow\mathbb{R}^{n-k}$.
(and any level set of a submersion is of course the level set of a regular value). I feel like this is the kind of question where, if there is a counterexample, it probably is very simple, but I wasn't able to come up with one.
Best Answer
A closed submanifold $S\subset M$ of codimension $k$ is the inverse image of a regular value of a smooth map $f:M\rightarrow S^k$ if and only if it has trivial normal bundle.
One implication is explained in evgeniamerkulova's answer (and still holds with $S^k$ replaced by any other manifold of dimension $k$).
For the other one, pick a tubular neighbourhood $U$ of $S$ in $M$. Then $U$ is diffeomorphic to $S\times\mathbb{R}^k$, because $S$ has trivial normal bundle. You can now define a map $U\rightarrow \mathbb{R}^k$ by $(x,v)\mapsto v$ and extend it to $M$ by mapping the complement of $U$ to infinity. You can then approximate the resulting map by a smooth map that has $0$ as a regular value and whose zero set is $S$. (This is exercise 4.6.5 in Hirsch's Differential Topology).