Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 15 / 4 / b / Solution

Put $m=\dim M$ and $n=\dim N$. The hypothesis says that $q$ is a regular value of $F$. Since its preimage is nonempty and the differential of a smooth map is surjective there, $m\ge n$.

We use the following finite-dimensional inverse function theorem: if a smooth map $H:O\to {\mathbb{R}}^m$, with $O\subset {\mathbb{R}}^m$ open, has invertible derivative at $a$, there are open neighborhoods of $a$ and $H(a)$ on which $H$ is a diffeomorphism, with smooth inverse.

Fix $p\in F^{-1}(q)$. Choose coordinate charts centered at $p$ and $q$, and shrink the source neighborhood so that its image under $F$ lies in the target chart. Its coordinate expression is $f=(f^1,\dots ,f^n)$ with $f(0)=0$ and derivative of rank $n$. Choose $n$ domain coordinates giving an invertible $n\times n$ minor, and relabel them as $x^1,\dots ,x^n$. DefineIts derivative at zero has block formso it is invertible. The inverse function theorem makes $u=H(x)$ a new coordinate chart near $p$. In this chart $F=q$ is exactly $u^1=\cdots =u^n=0$, because the target chart sends $q$ to zero. These are slice charts for an embedded submanifold; reordering coordinates if desired puts the free coordinates first. Giving the level set the subspace topology, these restricted charts are smoothly compatible, as in part (a). This proves the regular level set theorem rather than merely invoking it.

The level set is an embedded submanifold of dimension $\boxed{m-n}$. The same coordinates show $T_{p}Z=\ker d{F}_p$: tangent vectors to the slice have zero first $n$ components, which are exactly the components of $d{F}_p$. The empty-dimensional case $m=n$ gives a discrete zero-dimensional embedded submanifold.