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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.

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    $\begingroup$ This is unfortunately false; consider the smooth map $f:\mathbb{R}^3\to\mathbb{R}^2$ given by $f(x,y,z)=(x,0)$. Then $f^{-1}(0,0)=\{0\}\times\mathbb{R}^2$ is an embedded submanifold of $\mathbb{R}^3$, but $(0,0)$ is not a regular value of $f$. $\endgroup$
    – Ken
    Commented Mar 3, 2020 at 4:33
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    $\begingroup$ In general, the mere condition that a map has a constant rank guarantees that the inverse image is an embedded submanifold. (See Theorem 5.12 of J.Lee's Introduction to Smooth Manifolds, for example.) $\endgroup$
    – Ken
    Commented Mar 3, 2020 at 4:38
  • $\begingroup$ @Ken what if $f$ is surjective? I will add onto, thanks for your answering though $\endgroup$
    – user318946
    Commented Mar 3, 2020 at 5:21
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    $\begingroup$ Here's how bad things can be working over $\Bbb R$ (but less so over $\Bbb C$). Suppose $f\colon M\to\Bbb R$ has $0$ as a regular value, so that $f^{-1}(0)$ is a lovely submanifold of $M$. Now take the function $f^2$ instead. $(f^2)^{-1}(0)=M$, but $0$ is a critical value. $\endgroup$
    – Ted Shifrin
    Commented Mar 3, 2020 at 6:54

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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.

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  • $\begingroup$ Is that any chance to add some conditions to make that (modified) claim true? $\endgroup$
    – user318946
    Commented Mar 3, 2020 at 6:15
  • $\begingroup$ Well, if you add strong enough conditions it will eventually become true, but nothing at a high level of generality comes to mind. $\endgroup$
    – Eric Wofsey
    Commented Mar 3, 2020 at 6:18

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