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 '17 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 '17 at 12:55
  • $\begingroup$ Of course. I'm overloading $x$, should have $c \in \mathbb R$ as you say. $\endgroup$ – user8948 Jan 30 '17 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 '17 at 13:33

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 my lecture notes: http://euclid.ucc.ie/Mckay/analysis/analysis.pdf

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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