Is a level set of a manifold a set of zeroes

Question

Suppose $X$ and $Y$ are manifolds of dimensions $k$ and $l$ (with $k>l$). Given $F : X \to Y$ a smooth map and $y$ a regular value in $Y$, does there exist a map $G : X \to \mathbb{R}^l$ such that $F^{-1}(y) = G^{-1}(0)$? (Note I want $G$ to be defined on all of $X$, not on a subset.)

Answer 1

YES if you replace $\mathbb R^l$ with $\mathbb S^l$ (this is basically the Pontryagin construction),

NO in general, if you require that $0$ is a regular value of $G$.

YES again if you don't require that $0$ is a regular value.

CONSTRUCTION WITH $\mathbb S^l$.

For a regular value $y$, the set $M=F^{-1}(y)$ is a submanifold of $X$. Let $N$ be a tubular neighborhoos of $Y$, so that $N$ is diffeomorphic to $M\times \mathbb R^{l}$.

The projection $g:N\to \mathbb R^{l}$ gives you the desired map so that $M=g^{-1}(0)$.

Now we fix the fact that we want $G$ defined on $X$ with values in $\mathbb S^l=\mathbb R^l\cup \infty$.

Let $B$ be the unit ball of $\mathbb R^l$ and let $N_1$ be the subset of $N$ corresponding to $M\times B$. Let $\varphi:[0,1]\to [0,\infty]$ be a smooth, monotone not decreasing map such that $\varphi=1$ on $[0,1/2]$ and $\varphi=\infty$ on $[3/4,1]$. Let $\Phi:B\to \mathbb R$ be defined as $$\Phi(x)=\varphi(||x||)$$

For $(x,v)\in N_1=M\times B$ define $G_1(x,v)=g(x,v)\Phi(v)$ and extend $G_1$ to a map $G:X\to \mathbb S^l$ by defining $G=G_1$ on $N_1$ and $G=\infty$ on $X\setminus N$.

---

COUNTEREXAMPLE WITH $\mathbb R^l$

Let $X=\mathbb S^1\times \mathbb R$ and $Y=\mathbb S^1$. Define $$F:\mathbb S^1\times\mathbb R\to \mathbb S^1\qquad F(y,t)=y$$ so $F^{-1}(y)=\{y\}\times \mathbb R$.

Now, let $G:\mathbb S^1\times \mathbb R\to \mathbb R$ be any map, and suppose that $G^{-1}(0)=\{y\}\times\mathbb R$, with $0$ a regular value. Since $G(y,t)=0$ for any $t\in\mathbb R$, then $0$ must be a regular value for the restriction $g_t$ of $G$ to $\mathbb S^1\times\{t\}$.

Since $S^1$ is compact and $\mathbb R$ is not, then $g_t$ has zero degree, hence there are at leat two points where $g_t=0$, but $G^{-1}(0)\cap \mathbb S^1\times\{t\}=(y,t)$ is a single point. Contradiction.

Remark: if you are not familiar in using the degree of a map, you can show that $g_t$ has at least two zeros by using the intermediate value theorem http://en.wikipedia.org/wiki/Intermediate_value_theorem and the fact that the derivative of $g_t$ at $y$ does not vanish. (because $0$ is a regular value of $G$)

---

WITHOUT REQUIRING $0$ IS REGULAR

As in the pontryagin construction, defined $G(x,v)=||v||^2$ on $N_1$ and $G=1$ on $X\setminus N_1$. This gives you a map $G:X\to [0,1]$ such that $G^{-1}(0)=F^{-1}(y)$. Now embeds $[0,1]$ in $\mathbb R^l$ so that $0$ goes to $0$.

Remark that $0$ is not a regular value of $G$.