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I have seen stronger propositions that imply this, but both their statement and their proofs require more advanced tools than I'd like to use in my text, which is aimed at a general scientific audience.

I want to prove: With $f\colon \mathbb R^k \to \mathbb R$ smooth (infinitely differentiable) and $x$ a non-critical point (i.e. $\nabla f(x) \neq 0$), then the level set $f^{-1}(x)$ is a manifold. Preferrably providing a procedure to construct charts.

It seems that there should be a proof using undergraduate calculus only. Any ideas?

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    $\begingroup$ Isn't this just the implicit function theorem? $\endgroup$
    – Paul Siegel
    Jan 30, 2017 at 12:11
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    $\begingroup$ $f^{-1}(x)$ for $x\in \mathbb R^n$ makes no sense for $f:\mathbb R^n\to\mathbb R$. You probably mean $M=f^{-1}(c)$ and $\nabla f(x)\neq 0$ for all $x\in M$. $\endgroup$
    – Jochen Wengenroth
    Jan 30, 2017 at 12:55
  • $\begingroup$ Of course. I'm overloading $x$, should have $c \in \mathbb R$ as you say. $\endgroup$
    – user8948
    Jan 30, 2017 at 13:15
  • $\begingroup$ Somehow I never came across the implicit function theorem as such neither in the econ-level calculi (where we did plenty of examples, of course), neither in undergraduate or graduate real analysis. That's exactly what I'm trying to prove. $\endgroup$
    – user8948
    Jan 30, 2017 at 13:33

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Proof of the implicit function theorem in several variables calculus requires the contraction mapping theorem, so is probably not suitable for your audience. You need to use an iterative method and take a limit. You can look for a complete proof in Spivak, Calculus on Manifolds.

If you just replace one of the coordinate functions by $f$, you get a chart, but the proof that it is a chart requires the implicit function theorem.

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  • $\begingroup$ The link doesn't work anymore. $\endgroup$
    – Yola
    Jan 27, 2021 at 19:39

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