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I am trying to figure out what the implicit function theorem is. Can anyone explain it to me? I was reading the wiki: http://en.wikipedia.org/wiki/Implicit_function_theorem but still don't understand it too well.

Why does the function $f: R^{n+m} \rightarrow R^m$ have to be continuously differentiable?

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It might be easiest to look at an example from $R^1$ to $R^1$ that's differentiable but not continuously differentiable. My favorite is $$ f(x) = \begin{cases} x^2 \sin (\frac{1}{x}) & x \ne 0 \\ 0 & x = 0 \end{cases}. $$

This function has a discontinuous derivative, but the derivative is zero at many many points arbitrarily near the origin, so the preimage of the origin ends up not being an embedded submanifold (of dimension 0) in $R^1$.

I hope this example helps you resolve at least some confusion.

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  • $\begingroup$ But $f'(0)=0$ violates the maximal rank condition. $\endgroup$ – Ted Shifrin Jan 2 '14 at 1:02
  • $\begingroup$ Right. Good point, Ted. Add $x$ to the function, i.e., $x + x^2 \sin(\frac{1}{x})$ has rank 1 at the origin, is differentiable, but not continuously differentiable. $\endgroup$ – John Hughes Jan 2 '14 at 1:18
  • $\begingroup$ But now $f^{-1}(0)$ is just $0$, certainly locally. $\endgroup$ – Ted Shifrin Jan 2 '14 at 1:23

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