# The level sets of a differentiable function is a manifold

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?

• Isn't this just the implicit function theorem? – Paul Siegel Jan 30 '17 at 12:11
• $f^{-1}(x)$ 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 $x\in M$. – Jochen Wengenroth Jan 30 '17 at 12:55
• Of course. I'm overloading $x$, should have $c \in \mathbb R$ as you say. – user8948 Jan 30 '17 at 13:15
• 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. – user8948 Jan 30 '17 at 13:33

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.