Regular value theorem

Remarks

The set of points where $df_x$ is surjective is open, so $f$ is a submersion on a neighbourhood of $f^{-1}(y)$. In charts the claim becomes the implicit function theorem. Near each $x$ there are coordinates in which $f$ is the projection onto the first $n$ coordinates, and the fibre is then a coordinate slice. Example: $f \colon \R^{n+1} \to \R$, $f(x) = \abs{x}^2$, has $df_x(v) = 2 \langle x, v \rangle$, which is onto for $x \neq 0$. So $1$ is a regular value and the unit sphere $S^n = f^{-1}(1)$ is a submanifold of dimension $n$. The hypothesis is needed: $0$ is not a regular value of $g(x, y) = xy$, and $g^{-1}(0)$ is the union of the two axes, which is not a submanifold.

Going deeper

For a student

The theorem is the main tool for producing examples of manifolds. For instance, the orthogonal group $O(n)$ is the fibre over the identity of the map $A \mapsto A^T A$ into symmetric matrices. The identity is a regular value, so $O(n)$ is a manifold of dimension $n(n-1)/2$. By Sard's theorem almost every value is regular, so the theorem applies to almost every fibre. It is the special case of transversality to a point, and the same proof gives that the preimage of a submanifold under a transverse map is a submanifold.