Converse of the regular value theorem

differential-topologysmooth-manifolds

Given a smooth surjective map between two manifolds $M,N$, $f:M\to N$ with $\dim M=:m>\dim N=:n$. Suppose $f^{-1}(y)$ is a smooth manifold, how do I show $y$ is a regular value?

I am trying to show the Jacobian $(\frac{\partial f_i}{\partial x_j})_{m\times n}$ at each $x\in f^{-1}(y)$ has maximal rank $n$. And I am thinking maybe I can apply the inverse function in the form of: a set of smooth functions $g_1,…,g_k$ of $K$ ($\dim K=k$) with local coordinate charts $(t_1,…,t_k)$ of a point $h\in K$ forms a coordinate system about $h$ iff $\det (\frac{\partial g_i}{\partial t_j})$ is non-zero. So far I am stuck on this.

Best Answer

This is false. For instance, consider $f:\mathbb{R}^2\to\mathbb{R}$ given by $f(x,y)=x^3-x$. Then every fiber of $f$ is a finite union of vertical lines and in particular is a smooth manifold, but the derivative of $f$ vanishes when $x=\pm1/\sqrt{3}$ so not every value is a regular value.